Solucionario Ondas y Fluidos French

Vibrations and Waves MP205, Assignment 1

* 1. Consider a vector z defined by the equation z = z1z2, where z1 = a + ib,z2 = c+ id.

(a) Show that the length of z is the product of the lengths of z1 and z2.

(b) Show that the angle between z and the x axis is the sum of the anglesmade by z1 and z2 separately.

(a)

z = z1z2 = (a+ ib)(c+ id) = (ac bd) + i(ad+ bc)|z1| =

pa2 + b2

|z2| =pc2 + d2

) |z1||z2| =pa2 + b2

pc2 + d2

Now we find the length of z = z1z2.

|z| =p(ac bd)2 + (ad+ bc)2 = pa2c2 + b2d2 + a2d2 + b2c2

=p(a2 + b2)(c2 + d2) =

pa2 + b2

pc2 + d2

= |z1||z2|

(b) From the diagrams we see that:

tan 1 =b

a, tan 2 =

d

c

We know that:

a = |z1| cos 1 c = |z1| cos 2b = |z1| sin 1 d = |z1| sin 2

Similarly, for z = z1z2 we see that

tan =ad+ bc

ac bd =cos 1 sin 2 + sin 1 cos 2cos 1 cos 2 sin 1 sin 2 =

sin(1 + 2)

cos(1 + 2)= tan(1 + 2)

2. Consider a vector z defined by the equation z = z1/z2, (z2 6= 0), wherez1 = a+ ib, z2 = c+ id.

(a) Show that the length of z is the quotient of the lengths of z1 and z2.[8]

(b) Show that the angle between z and the x axis is the dierence of theangles made by z1 and z2 separately. [7]

(a) This follows trivially from the fact that:

|z| =z1z2 = |z1||z2|

Alternatively we can do it out:

z =z1z2

=a+ ib

c+ id=

(a+ ib)(c id)c2 + d2

=(ac+ bd) + i(bc ad)

c2 + d2

|z| =s

(ac+ bd)2 + (bc ad)2(c2 + d2)2

=

pa2c2 + 2acbd+ b2d2 + b2c2 2abcd+ a2d2

c2 + d2

=

pa2 + b2(c2 + d2)

c2 + d2

=

pa2 + b2

pc2 + d2

c2 + d2

=

pa2 + b2pc2 + d2

=|z1||z2|

(b) Drawing out z1 and z2

We know that:

a = |z1| cos 1 c = |z1| cos 2b = |z1| sin 1 d = |z1| sin 2

Re

Im

ac+bd

bc-ad

z

c + d2 2

c + d2 2

From the diagram above and using the values of a, b, c, d we get:

tan' =

bc adc2 + d2

c2 + d2

ac+ bd

=

bc adac+ bd

=sin 1 cos 2 cos 1 sin 2cos 1 cos 2 + sin 1 sin 2

=sin(1 2)cos(1 2) = tan(1 2)

) ' = 1 2

3. Show that the multiplication of any complex number z by ei is describ-able, in geometric terms, as a positive rotation through the angle of thevector by which z is represented, without any alteration of its length. [5]

Consider the complex number z, that makes an angle with the x axis such thatz = a + ib = |z|ei. We multiply z by ei, that is, zei = |z|eiei = |z|ei(+). Wesee that there is no alteration of its length as before and after multiplication itslength is |z|. It has however undergone a positive rotation of as we can see fromthe diagrams below.

Re

Im

|z|

Re

Im

|z|

+

* 4. Would you be willing to pay 20 cents for any object valued by a mathe-matician at iicents?Evaluate Eulers relation ei = cos + i sin , at = /2. This gives ei/2 =cos /2 + i sin /2 = i.However since cos() = cos( + 2n) and sin() = sin( + 2n), for n = 0, 1, 2, ...,we know that ei = ei(+2n).So we can write i = ei(/2+2n)

We want to find the value of ii, hence

ii = (ei(/2+2n))i = ei2(/2+2n) = e(/2+2n)

Looking at this for the first 3 values of n:

n = 0: e(/2+2n) = e/2 21c n = 1: e(/2+2n) = e(/2+2) 0.04c n = 2: e(/2+2n) = e(/2+4) 0.00007c

So the answer here is maybe, it depends on your value of n!

5. (a) If z = Aei, deduce that dz = izd, and explain the meaning of thisrelation in a vector diagram. [8]

(b) Find the magnitudes and directions of the vectors (2 + ip3) and (2

ip3)2. [10]

(a)

z = Aei ,dz

d= iAei ) dz = iAeid = izd

Multiplication by i shifts phase by /2 as depicted below.

Re

Im

z

Re

Im

dz /2+

(b) We find the magnitudes (lengths) firstly.

|z1| =q22 +

p32 =

p7

z2 = (2 ip3)2 = 1 i4p3 ) |z2| =

q12 + (4p3)2 = 7

By drawing both complex numbers, we can easily calculate their directions.

Re

Im

1 2

z1 Re

Im

z2 1For z1, we have 1 = tan1(y/x) = tan1(

p3/2) = 40.89.

For z2, we have 2 = 360 tan1(y/x) = 360 tan1(4p3/1) = 278.21.

6. Given Eulers relation ei = cos + i sin , find

(a) An expression for ei. [5]

(b) The exponential representation of cos . [5]

(c) The exponential representation of sin . [5]

(a)

ei = cos() + i sin()= cos() i sin()

(b) We have that ei = cos + i sin and ei = cos i sin . Adding bothexpressions together:

ei + ei = 2 cos

) cos = 12(ei + ei)

(c) Similarly, using ei = cos + i sin and ei = cos i sin , then:ei ei = 2i sin ) sin = 1

2i(ei ei)

* 7. Justify the formulas cos = (ei+ ei)/2 and sin = (ei ei)/2i, using theappropriate series.The series expansion for ex is given by:

ex =1Xn=0

xn

n

By expanding ei and then ei we get:

ei = 1 +i

1!

2

2! i

3

3!+4

4!+ . . .

ei = 1 i1!

2

2!+

i3

3!+4

4!+ . . .

Therefore the series expansion for cos is

cos =ei + ei

2= 1

2

2!+4

4!+ . . .

which can be verified by Taylor expanding cos .The Taylor expansion a function f(x) is given by:

f(x) = f(0) +xf 0(0)1!

+x2f 00(0)

2!+ . . .+

xnfn(0)

n!

Taylor expanding cos we get:

cos = cos(0) + ( sin(0)) + 2( cos(0))

2+3(sin(0))

3+4( cos(0))

4+ ...

= 1 2

2+4

4+ ...

Which is what got above.Similarly the series expansion for sin is

sin =ei ei

2i=

3

3!+5

5!+ . . .

which can be verified by Taylor expanding sin .Taylor expanding cos we get:

sin = sin(0) + (cos(0)) +2( sin(0))

2+3( cos(0))

3+4(sin(0))

4+5(cos(0))

4...

= 3

3+5

5+ ...

Which is what got above.

8. To take successive derivatives of ei with respect to , one merely multi-plies by i:

d

d(Aei) = iAei

Show that this prescription works if the sinusoidal representation ei =cos + i sin is used. [5]

Let f() = cos + i sin . Want to show that:

d

df() = if().

Evaluating the LHS gives:

d

d(cos + i sin ) = sin + i cos

The RHS can be rewritten as:

i(cos + i sin ) = i cos + i2 sin = i cos sin Clearly on the LHS we have the same expression as on the RHS, hence

d

df() = if().

9. Using the exponential representations for sin and cos , verify the fol-lowing trigonometric identities:

(a) sin2 + cos2 = 1 [6]

(b) cos2 sin2 = cos 2 [6]

(c) 2 sin cos = sin 2 [6]

We know that cos = (ei + ei)/2 and sin = (ei ei)/2i, therefore cos2 =(e2i + 2 + e2i)/4 and sin2 = (e2i 2 + e2i)/4.(a)

sin2 + cos2 = e2i 2 + e2i

4+

e2i + 2 + e2i

4=

1

2+

1

2= 1

(b)

cos2 sin2 = e2i + 2 + e2i

4+

e2i 2 + e2i4

=e2i + e2i

2= cos 2

(c)

2 sin cos = 2

ei ei

2i

ei + ei

2

=

e2i e2i2i

= sin 2

* 10. Verify that the dierential equation d2y/dx2 = k2y has as its solutiony = A cos(kx) + B sin(kx),

where A and B are arbitrary constants. Show also that this solution canbe written in the form

y = C cos(kx+ ) = CRe[ei(kx+)] = Re[(Ce(i))eikx]

and express C and as functions of A and B.

11. A mass on the end of a spring oscillates with an amplitude of 5cm ata frequency of 1Hz (cycles per second). At t = 0 the mass is at itsequilibrium position (x = 0).

(a) Find the possible equations describing the position of the mass asa function of time, the form x = A cos(!t + ), giving the numericalvalues of A, !, and . [6]

(b) What are the values of x, dx/dt, and d2x/dt2 at t = 8/3 seconds? [6]

(a) At time t = 0 we have x = 0. Filling into the equation of motion gives0 = cos ) = /2. Frequency is 1Hz ) Period T = 1/F = 1 second.Angular frequency ! = 2/T = 2s1. Amplitude A is given in the questionas 5cm.This gives two possible equations describing the position of the mass as afunction of time:

x(t) = 5 cos2t+

2

and

x(t) = 5 cos2t

2

(Wed need an initial velocity to work out a specific case)

(b) Looking at x(t) = 5 cos2t+ 2

:

x(8/3) = 5 cos

2

8

3+

2

= 5 cos

11

6

= 5 cos

6

= 5

p3

2cm

dx/dt = !A sin(!t+ )dx

dt(8/3) = 2(5) sin

2

8

3+

2

= 10 sin

11

6

= 5 cms1

d2x/dt2 = !2A cos(!t+ )d2x

dt2(8/3) = (2)2(5) cos

2

8

3+

2

= 202 cos

6

= 102p3 cms2

12. A point moves in a circle at a constant speed of 50cm/sec. The periodof one complete journey around the circle is 6 seconds. At t = 0 the lineto the point from the center of the circle makes an angle of 30 with thex axis.

(a) Obtain the equation of the x coordinate of the point as a functionof time, in the form x = A cos(!t+ ), giving the numerical values ofA,!, and . [6]

(b) Find the values of x, dx/dt, and d2x/dt2 at t = 2 seconds. [6]

(a) Since at t = 0 the line to the point from the center of the circle makes an angleof 30 with the x axis, this means that the initial phase = /6. Now:

velocity =distance

time) 50 = 2r

T=

2A

6) A = 150

cm

! =2

T=

3s1

(b)

x = A cos(!t+ ) = (150/) cos(t/3 + /6)

x(2) =150

cos

5

6

= 150

cos6

= 75

p3

cm

dx/dt = !A sin(!t+ )dx

dt(2) =

3.150

sin

5

6

=

3.150

sin6

= 25 cms1

d2x/dt2 = !2A cos(!t+ )d2x

dt2(2) =

2

9.150

cos

5

6

=2

9.150

cos6

=

25p3cms2

Vibrations and Waves MP205, Assignment 2 Solutions

1. Express the following in the form Re[Aei(!t+)]

(a) z = sin!t+ cos!t.

(b) z = cos(!t /3) cos!t.(c) z = 2 sin!t+ 3 cos!t.

(d) z = sin!t 2 cos(!t /4) + cos!t.First recall that Aei(!t+) can be written in two ways:

Aei(!t+) = A{cos(!t) + i sin(!t)}{cos() + i sin()}= A{cos(!t) cos() sin(!t) sin()}+ Ai{cos(!t) sin() + sin(!t) cos()}

or

Aei(!t+) = A{cos(!t+ ) + i sin(!t+ )}

From this we can write Re[Aei(!t+)] as

Re[Aei(!t+)] = A cos(!t) cos() A sin(!t) sin()or

Re[Aei(!t+)] = A cos(!t+ )

For these solutions Ill do the first format.

(a) z = sin!t+ cos!t

sin(!t) + cos(!t) = A cos(!t) cos() A sin(!t) sin()

1 = A sin() 1 = A cos()A = 1

sin()A =

1

cos()A is positive so we know

sin() < 0 cos() > 0

1sin()

=1

cos()

sin()

cos()= 1

=7

4

A =1

cos()=

1

cos74

= p2) z = Re[p2ei(!t+ 74 )]

(b) z = cos(!t /3) cos!t = cos(!t /3) + cos(!t+ ).

z = cos(!t 3) cos(!t)cos(!t 3) = cos(!t) cos

3

+ sin(!t) sin

3

=

1

2cos(!t) +

p3

2sin(!t)

) z = 12cos(!t) +

p3

2sin(!t) cos(!t)

= 12cos(!t) +

p3

2sin(!t)

p3

2= A sin() 1

2= A cos()

2A = p3

sin()2A = 1


sin() < 0 cos() < 0

p3

sin()= 1

cos()

sin()

cos()=p3

=4

3

2A = 1cos()

=1

cos43

= 2) A = 1) z = Re[ei(!t+ 43 )]

(c) z = 2 sin!t+ 3 cos!t = 2 cos(!t /2) + 3 cos(!t).

z = 2 sin(!t) + 3 cos(!t)

2 = A sin() 3 = A cos()A = 2

sin()A =

3


sin() < 0 cos() > 0

2sin()

=3

cos()

sin()

cos()= 2

3

= 5.7rad

A =3

cos()=

3

cos2 tan1 23 =

p13

) z = Re[p13ei(!t+5.7)](d) z = sin!t 2 cos(!t /4) + cos!t

z = sin(!t) 2 cos!t

4

+ cos(!t)

cos!t

4

= cos(!t) cos

4

+ sin(!t) sin

4

=

1p2cos(!t) +

1p2sin(!t)

) z = sin(!t)p2 cos(!t)p2 sin(!t) + cos(!t)= (1p2) cos(!t) + (1p2) sin(!t)

(1p2) = A sin() (1p2) = A cos()

A = (1p2)

sin()A =

(1p2)cos()

A is positive so we know

sin() > 0 cos() < 0

(1p2)

sin()=

(1p2)cos()

sin()

cos()= 1

=3

4

A =(1p2)cos()

=(1p2)cos

34

= 2p2) z = Re[(2p2)ei(!t+ 34 )]

* 2. A particle is simultaneously subjected to three simple harmonic motions,all of the same frequency and in the x direction. If the amplitudes are0.25, 0.20, and 0.15 mm, respectively, and the phase dierence betweenthe first and second is 45, and between the second and third is 30, findthe amplitude of the resultant displacement and its phase relative to thefirst (0.25 mm amplitude) component.So again we want to get this in the form:

Re[Aei(!t+)] = A cos(!t) cos() A sin(!t) sin()First we need to set up our system:

x1 = 0.25 cos(!t)

x2 = 0.20 cos!t+

4

= 0.20

hcos(!t) cos

4

sin(!t) sin

4

i= 0.20

1p2cos(!t) 1p

2sin(!t)

0.14 cos(!t) 0.14 sin(!t)

x3 = 0.15 cos!t+

4+

6

= 0.15 cos

!t+

5

12

= 0.15

cos(!t) cos

5

12

sin(!t) sin

5

12

0.15 [0.26 cos(!t) 0.97 sin(!t)]= 0.039 cos(!t) 0.136 sin(!t)

Our particle is simultaneously to these three simple harmonic motions:

x = x1 + x2 + x3= 0.25 cos(!t) + 0.14 cos(!t) 0.14 sin(!t) + 0.039 cos(!t) 0.136 sin(!t)= 0.43 cos(!t) 0.28 sin(!t)

0.28 = A sin() 0.43 = A cos()A =

0.28

sin()A =

0.43


sin() > 0 cos() > 0

0.28

sin()=

0.43

cos()

sin()

cos()=

0.28

0.43= 0.65

= 0.58rad

A =0.43

cos()=

0.43

cos(0.58)= 51mm

) z = Re[51ei(!t+0.58)]

3. Two vibrations along the same line are described by the equations

y1 = A cos(10t)

y2 = A cos(12t)

Find the beat period, and draw a careful sketch of the resultant distur-bance over one beat period.

beat period T =2

|!1 !2| =2

|10 12| = 1 s

4. Find the frequency of the combined motion of each of the following:

* (a) sin(2tp2) + cos(2t).(b) sin(12t) + cos(13t /4).(c) sin(3t) cos(t).Here we use the fact that when two SHMs are quite close in frequency, they have afrequency equal to the average of the combining frequencies (but with an amplitudethat varies periodically with time).Also recall the following formulae:

f =1

T

T =2

!

) f = 1T

=!

2

(a) sin(2tp2) + cos(2t).x = x1 + x2 = sin(2t

p2) + cos(2t)

x1 = sin(2tp2)

= cos2tp2

2

) !1 = 2) f1 = !1

2=

2

2= 1

x2 = cos(2t)

) !2 = 2) f2 = !2

2=

2

2= 1

The average of these frequencies is given by:

f =f1 + f2

2=

1 + 1

2= 1

(b) sin(12t) + cos(13t /4).x = x1 + x2 = sin(12t) + cos(13t /4)

x1 = sin(12t)

= cos12t

2

) !1 = 12) f1 = !1

2=

12

2= 6

x2 = cos(13t /4)) !2 = 13) f2 = !2

2=

13

2= 6.5


f =f1 + f2

2=

6 + 6.5

2= 6.25

(c) sin(3t) cos(t).x = x1 + x2 = sin(3t) cos(t)x1 = sin(3t)

= cos3t

2

) !1 = 3) f1 = 3

2x2 = cos(t)

= cos(t+ )

) !2 = ) f2 = !2

2=

2=

1

2


f =f1 + f2

2=

32 +

12

2 4.9

* 5. Two vibrations at right angles to one another are described by the equa-tions

x = 10 cos(5t)

y = 10 cos(10t+ /3)

Construct the Lissajous figure of the combined motion.We have x = 10 cos(5t), y = 10 cos(10t+ /3).

6. Construct the Lissajous figures for the following motions:

(a) x = cos(2!t), y = sin(2!t).

(b) x = cos(2!t), y = cos(2!t /4).

(c) x = cos(2!t), y = cos(!t).

(a) We have x = cos(2!t), y = sin(2!t) = cos(2!t /2).

(b) We have x = cos(2!t), y = cos(2!t /4).

(c) We have x = cos(2!t), y = cos(!t).


1. An item of mass 1 g is hung from a spring and set in oscillatory mo-tion. At t = 0 the displacement is 43.785 cm and the acceleration is-1.7514cm/sec2. What is the spring constant?

Given m = 1 g = 1 103 kg and at t = 0, x = 43.785 cm = 0.43785 m anda = 1.7514cms2 = 0.017514 ms2.Undergoes oscillatory motion so it must satisfy an equation of the form a = !2x.

a = !2x= k

mx

k = amx

= (0.017514)(1 103)

0.43785= 4 105Nm1

* 2. A mass m hangs from a uniform spring of spring constant k.

(a) What is the period of oscillations in the system?

(b) What would it be if the mass m were hung so that: (see Figurebelow)

(1) It was attached to two identical springs hanging side by side?

(2) It was attached to the lower of two identical springs connectedend to end?

k k kk

(1)(2)

(a) F = kx, T = 2! and ! =q

km . Then

T = 2

rm

k

(b)(1) To move mass an equivalent displacement x as in part (a) we need twice theforce, since the restoring force is now twice as big i.e. k0 = 2k.

This means that

!0 =

rk0

m=

r2k

m

T 0 =2

!0= 2

rm

2k=

Tp2

(b)(2) k is inversely proportional to the length of the string.Let l be the length of the original spring, then:

k / 1l

or equivalently

k =a

lfor some constant a

The length of our spring here is 2l

k00 =a

2l=

k

2

This means that

!00 =

rk00

m=

sk2

m=

rk

2m

T 00 =2

!00= 2

r2m

k= T

p2

3. A platform is executing simple harmonic motion in a vertical directionwith an amplitude of 5cm and a frequency of 10 vibrations per second. Ablock is placed on the platform at the lowest point of its path

(a) At what point will the block leave the platform? [Hint: think of theforces on the block ]

(b) How far will the block rise above the highest point reached by theplatform?

(a) We have the following information:

A = 5cm

f =10

T =

10=

2

!! = 20

If we look at the forces on the block:

Relative to the block, the net force is 0 for as long as the block is on theplatform.Analysing the above diagram, kx is the force exerted by the platform on theblock (due to the simple harmonic motion),W is the weight of the block, and Nis the force exerted by the block on the platform (the normal force-as kx > W .If kx < W then the system would stop oscillating when the block is place onthe platform).This force N is the force keeping the block on the platform, it is NOT constant,and changes with respect to x.We want to find the point when N = 0 - this is the point the block leaves theplatform.

kx = W Nkx = W looking at the point whereN = 0

kx = mg

x =m

kg

=m

m!2g

=g

!2

=9.8

400= 0.0245m = 2.5cm

(b) We need to find the velocity of the block as it leaves:

x = A cos(!t) = 5 cos(20t)

2.5 = 5 cos(20t)

0.5 = cos(20t)

) 20t = 3

v =dx

dt= 5(20) sin(20t) = 100 sin(20t)

Looking at v when x leaves the platform:

v = 100 sin3

= 86.6cms1

Note: The comes from the fact we can take 20t = 3or 43 - it just indicatesthat in a SHM system well have the same speed at a point at 2 dierent times,just in 2 dierent directions depending on the direction of the motion.Using the formula that relates vi, vf , a and s, where:

vi is the initial velocity vf is the final velocity a is the acceleration s is the displacement

v2f = v2i + 2as

We know our initial velocity is vi = 86.6cms1 (as our motion is up we takethe plus direction), once the block leaves the platform its acceleration will justbe gravity a = 9.8, and the velocity when the block hits its highest point isjust vf = 0.

v2f = v2i + 2as

0 = (86.6)2 + 2(9.8)ss =

0.76

19.8= 0.038m = 3.8cm

Max height of the platform is just the amplitude A = 5cm. The height theblock reaches is its distance travelled on the platform 2.5cm and its distancetravelled once it left 3.8cm.This gives the height above the platform reached as:

h = (3.8 + 2.5) 5 = 1.3cm* 4. A uniform rod of length L is nailed to a post so that two thirds of its

length is below the nail. What is the period of small oscillations of therod?

If we look at the period in terms of the center of mass, where the center of mass Cis a distance h from the point of suspension, and k is the radius of gyration of thebody.In this case, the period T is given by:

T = 2

h2 + k2

gh

12

Looking at our system:

Our center of mass C will be at the center of the rod, so our h is given by

h =2L

3 L

2=

L

6

To get k, we recall that I = mk2, and that the moment of inertia of a rod (throughits center of mass) is given by I = 112mL

2.

I = mk2

) k2 = Im

=L2

12

This gives us the period:

T = 2

L2

36 +L2

12gL6

! 12

= 2

L16 +

12

g

! 12

= 2

s2L

3g

So we can think of the system as a simple pendulum of length 2L3 . The period of asimple pendulum is given by:

T = 2

sl

g

where l is the length of the pendulum.

l =2L

3

) T = 2s

2L

3g

5. A circular hoop of diameter d hangs on a nail. What is the period of itsoscillations at small amplitude?If we look at the period in terms of the center of mass, where the center of mass Cis a distance h from the point of suspension, and k is the radius of gyration of thebody.In this case, the period T is given by:

T = 2

h2 + k2

gh

12

Looking at our system:

Our center of mass will be at the center of the circle, so our h is just the radius ofthe circle, and as we have a circle our k is also just the radius.This gives us the period:

T = 2

r2 + r2

gr

12

= 2

2r

g

12

= 2

r2r

g

But 2r = d

) T = 2sd

g

We can think of the system as a simple pendulum of length d. The period of asimple pendulum is given by:

T = 2

sl

g

where l is the length of the pendulum.

l = d

) T = 2sd

g

* 6. (a) An object of mass 0.5 kg is hung from the end of a steel wire of length2 m and of diameter 0.5 mm. (Youngs modulus = 2 1011 N/m2).What is the extension of the wire?

(b) The object is lifted through a distance h (thus allowing the wire tobecome slack) and is then dropped so that the wire receives a suddenjerk. The ultimate strength of steel is 1.1109 N/m2. What is thelargest possible value of h if the wire is not to break?

(a)

Y =stress

strain=

PAll0

P = mg (0.5)(9.8) = 4.9N

A = r2 =

0.5 103

2

2= 1.96 107m2

l0 = 2m

Y = 2 1011Nm2

) 2 1011 =4.9

1.96107l2

2 1011 = 5 107

l

) l = 5 107

2 1011 = 2.5 104m = 0.25mm

(b) Ultimate strength = max value of stress.We know that:

Y =stress

strain

Y is constant, so if we have maximum stress, this corresponds to the maximumstrain.To get the maximum extension l we need to get the maximum strain, asl / strain.

Y =max stress

max strain

2 1011 = 1.1 109

max strain

) max strain = 1.1 109

2 1011 = 5.5 103

maxl

l= 5.5 103

maxl

2= 5.5 103

maxl = 0.011m

So the maximum change in length before the wire will break is 0.011m.To find the maximum height we can lift the object, we look at the energy of thesystem when lifted to a height h, and the energy of the system at the lowestpoint l0.

At the highest point (h) the total energy is just the potential energy: mgh. At the lowest point (l) the total energy is just kinetic energy: 12kx2 =

12kl

20

Using conservation of energy: mgh = 12kl2

k is just given by k = AYl0 =(1.96107)(21011)

2 = 19600.

h =kl2

2mg

=(19600)(0.011)2

2(0.5)(9.8)

= 0.24m

7. A solid steel ball is to be hung at the bottom end of a steel wire oflength 2m and radius 1mm. The ultimate strength (max stress) of steelis 1.1 109N/m2. What are:(a) the mass of the biggest ball the wire can bear

(b) the radius of the biggest ball the wire can bear [Hint: use the factthat the density of steel is given by s = 7900kgm

3]

(a) We know the ultimate strength of steel is 1.1 109N/m2, and r = 1mm =0.001m:

1.1 109 = PA

=mg

r2

=m(9.8)

(0.001)2

3455.75 = 9.8m

m = 352.6kg

(b) We know the density of steel is s = 7900kgm3, and m = V :

m = V

=4

3r3

r3 =3m

4

=3(352.6)

4(3.14)(7900)

= 0.0107

r = 0.22m

8. A metal rod, 0.5 m long, has a rectangular cross section of area 2 mm2.With the rod vertical and a mass of 60kg hung from the bottom, thereis an extension of 0.25 mm. What is the Youngs modulus ( N/m2) forthe material of the rod?

Given m = 60 kg, h = 0.25 mm = 0.25 103 m, `0 = 0.5 m and A = 2 mm2 =2 106 m2. We also know that

mg =AY h

`0) Y = mg`0

Ah=

(60)(9.8)(0.5)

(2 106)(0.25 103) = 5.88 1011 Nm2


1. (a) Show that the frequency of vibration under adiabatic conditions of acolumn of gas confined to a cylindrical tube, closed at one end, witha well-fitting but freely moving piston of mass m is given by:

! =

rAp

lm

.

(b) A steel ball of diameter 2cm oscillates vertically in a 12-liter glasstube containing air at atmospheric pressure (as pictured below). Ver-ify that the period of oscillation should be about 1 sec. (Assume adi-abatic pressure change with = 1.4. Density of steel = 7600kgm3.)

(a) Under adiabatic conditions we have :pV = constant

pV = const

ln(pV ) = const

ln(p) + ln(V ) = const

ln(p) + ln(V ) = const

Dierentiating this w.r.t. V gives:1

p

dp

dV+

V= 0

dp

dV+

p

V= 0

dp = dV pV

Now we know we can express :

force F as F = Ap (= Adp) volume V as V = Al the change in volume V as V = Ay (= dV )

F = Adp

= AdV pV

= A2yp

Al

= Apl

y

ma = Apl

y

a = Apml

y

We know a = !2y

) !2 = Apml

! =

rAp

ml

(b)

mass of ball;m = V =4

3r3 =

4

3

0.02

2

3(7600) 0.032kg

volume of tube;V = lA2 = lr2

0.012 = l

0.02

2

2) l 38.2m

Using this:

! =

rAp

ml

=

s0.022

2(1.4)(101325)

(38.2)(0.032)=p36.5 = 6.04sec1

) T = 2!

=2

6.04= 1.04 1sec

* 2. The motion of a linear oscillator may be represented by means of a graphin which x is shown as abscissa and dx/dt as ordinate. The history of theoscillator is then a curve.

(a) Show that for an undamped oscillator this curve is an ellipse.

(b) Show (at least qualitatively) that if a damping term is introducedone gets a curve spiraling into the origin.

(a) For an undamped oscillator we have x = A cos(!t+ ).

x = A cos(!t+ )

dx

dt= !A sin(!t+ )

We want to show that x = A cos(!t+) and y = dxdt = !A sin(!t+) satisfyan ellipse equation.Ellipse equation is given by:

xa

2+yb

2= 1

In our case x = A cos(!t+ ) and y = dxdt = !A sin(!t+ )x2 = A2 cos2(!t+ )

x2

A2= cos2(!t+ )

y2 = !2A2 sin2(!t+ )

y2

!2A2= sin2(!t+ )

Using:

sin2(x) + cos2(x) = 1

We can write:x2

A2+

y2

!2A2= 1 x

A

2+ y!A

2= 1

as required.

) the curve is an ellipse

(b) If we think of our ellipse in (a) in terms of a pendulum:

If we release the pendulum from the point 1, its velocity is 0 and it hasmaximum (positive) displacement

When the pendulum goes through 2 it has maximum (negative, as it isgoing in the negative direction) velocity and its displacement is 0

When the pendulum reaches 3 it has maximum (negative) displacement,and 0 velocity

Finally, when the pendulum returns through 4 it will have maximum (pos-itive) velocity, and again 0 displacement

We can see this corresponds to the points on the ellipse to the right

If we release the pendulum from the point 1, its velocity is 0 and it hasmaximum (positive) displacement, as before.

When the pendulum goes through 2 it has its maximum (negative, as it isgoing in the negative direction) velocity for that swing (although this is lessthan the maximum velocity before due to damping) and its displacementis 0

When the pendulum reaches 3 it has its maximum (negative) displacementfor this swing, but due to damping this displacement is not as great as theoriginal maximum displacement, and 0 velocity

When the pendulum returns through 4 it will have its maximum (positive)velocity for this swing, which is less than the velocity when it went throughit on 2 , and again 0 displacement

Continuing in this vein we can see the graph that corresponds to this is acurve spiralling into the origin.

3. Verify that x = Aet cos!t is a possible solution of the equation

d2x

dt2+

dx

dt+ !20x = 0,

and find and ! in terms of and !0.

Given x = Aet cos!t we dierentiate it using the product rule to find its first andsecond derivatives

dx

dt= !Aet sin(!t) Aet cos(!t)

d2x

dt2= !2Aet cos(!t) + !Aet sin(!t) + !Aet sin(!t) + 2Aet cos(!t)= (2 !2)Aet cos(!t) + 2!Aet sin(!t)

To show its a solution we sub these derivatives back into the original equation toobtain

(2 !2)Aet cos(!t) + 2!Aet sin(!t) !Aet sin(!t) Aet cos(!t)+ !20Ae

t cos(!t) = 0

(2 !2 + !20)Aet cos(!t) + (2! !)Aet sin(!t) = 0

For this to be true for all values of t the coecients of the cosine and sine functionsmust be 0. This means we must have that:

2 !2 + !20 = 0 (1)2! ! = 0 (2)

Looking first at (2):

2! ! = 0 =

2

And using this in (1):

2 !2 + !20 = 02

4 !2 + !20

2= 0

!2 + !20 2

4= 0

) !2 = !20 2

4

! =

r!20

2

4

So x = Aet cos!t is a solution when = 2 and ! =q!20 24 .

* 4. An object of mass 0.2 kg is hung from a spring whose spring constant is80 N/m. The object is subject to a resistive force given by bv, where vis its velocity in meters per second.

(a) Set up the dierential equation of motion for free oscillations of thesystem.

(b) If the damped frequency isp3/2 of the undamped frequency, what

is the value of the constant b?

(c) What is the Q of the system?

(a) We have a damped oscillator where the damping term is bv. So the sum ofthe forces is F = kx bv. By Newtons second law the sum of the forcesmust equal ma so we have ma = kx bv.

ma+ bv + kx = 0

md2x

dt2+ b

dx

dt+ kx = 0

d2x

dt2+

b

m

dx

dt+

k

mx = 0

d2x

dt2+

dx

dt+ !20x = 0

where = b/m and !0 =pk/m.

Using our values for m and k from the question we can write:

=b

m=

b

0.2

!0 =

rk

m=

r80

0.2=p400 = 20

(b) !0 is the undamped angular frequency and ! is the damped angular frequency.We have the relation ! =

p3/2!0.

From the previous question we know that !2 = !20 2/4. Therefore3

4!20 = !

20 2/4

!0 = =b

m) b = !0m = 20(0.2) = 4

(c) Q is given by the ratio of the constants !0/.

Q =!0

=20

20= 1

5. Many oscillatory systems, although the loss or dissipation mechanism isnot analogous to viscous damping, show an exponential decrease in theirstored average energy with time E = E0et. A Q for such oscillators maybe defined using the definition Q = !0 , where !0 is the natural angularfrequency.

(a) When the note middle C on the piano is struck, its energy ofoscillation decreases to one half its initial value in about 1 sec. Thefrequency of middle C is 256 Hz. What is the Q of the system?

(b) If the note an octave higher (512 Hz) takes about the same time forits energy to decay, what is its Q?

(c) A free, damped harmonic oscillator, consisting of a mass m = 0.1kgmoving in a viscous liquid of damped coecient b (Fviscous = bv),and attached to a spring of spring constant k = 0.9Nm1, is observedas it performs oscillatory motion. Its average energy decays to 1e ofits initial value in 4 sec. What is the Q of the oscillator? What isthe value of b?

(a) We know that E = E0et.Using this:

E = E0et

after 1 second; t = 1 we have:

E02

= E0e(1)

1

2= e

ln

1

2

=

ln21

=

ln (2) = ) = ln(2) = 0.69

To find Q we also need to find !0

!0 =2

T= 2f = 2(256) = 512

Now we can use Q = !0/

Q =!0

=512

0.69= 2331.15

(b) Our change in energy is the same so we still have = 0.69

!00 = 2f0 = 2(512) = 2(256)(2) = 2(2f) = 2(2f)

where f is the frequency from part (a), this gives us:

!00 = 2!0

Using this we can find our Q0 for this note:

Q0 =!00

= 2 !0

= 2Q = 2(1331.15) = 4662.3

(c) To find b, we know that E = E0et.Using this:

E = E0et

after 1 second; t = 1 we have:

E0e

1= E0e

(4)

) 1 = 4 =

1

4Using = b/m:

1

4=

b

m

) b = m4=

0.1

4= 0.025

To find Q:

Q =!0

=

rk

m

1

=

r0.9

0.1

11/4

=p94 = 3(4) = 12

6. A U-tube has vertical arms of radii r and 2r, connected by a horizontaltube of length ` whose radius increases linearly from r to 2r. The U-tubecontains liquid up to a height h in each arm. The liquid is set oscillating,and at a given instant the liquid in the narrower arm is a distance y abovethe equilibrium level.

*(a) Show that the potential energy of the liquid is given by U = 58gr2y2.

*(b) Show that the kinetic energy of a small slice of liquid in the horizontalarm (see the diagram) is given by

dK =1

2

r2dx

(1 + x/`)2

dy

dt

2.

(Note that, if liquid is not to pile up anywhere, the product velocity cross section must have the same value everywhere along the tube.)

(c) Using the result of part (b), show that the total kinetic energy of allthe moving liquid is given by

K =1

4r2(`+

5

2h)

dy

dt

2.

(Ignore any nastiness at the corners.)

(d) From (a) and (c) calculate the period of oscillations of ` = 5h/2.

(a) Potential energy is U = mgh.

Narrow column (I) gains P.E while wider column (II) loses P.E. Since radii aredierent to find height h that liquid is lifted we must find h on column (I) and hon column (II), then their average. Since no liquid is displaced the volumes mustequal. Let x be the height the liquid goes down in column (II). The volumes areVI = r2y and VII = 4r2x.Equating the volumes we find that x = (1/4)y and therefore the average height is:

average =1

2

1

4y + y

=

5

8y

Subbing this into the expression for the P.E gives U = mg(5/8)y.We now find the mass of this piece of liquid. mass=density cross sectional area,so m = r2y and the potential energy is given by

U =5

8gr2y2

(b) Told velocity cross section = constant everywhere. To find the cross sectionwe need the radius at each point.

r(x = 0) = r0 = r ) A(x = 0) = A0 = r2r(x > 0) = rx = r(1 + x/`) ) A(x > 0) = Ax = r2(1 + x/`)

Note that at x = 0, r = r and at x = `, r = 2r as desired. Product velocity by crosssection is constant implies that:

A0dy

dt= Ax

dx

dt) dx

dt=

A0dy

Axdt

Kinetic energy is (1/2)mv2 and the mass of dx = density volume = Axdx, hencethe kinetic energy of dx is:

dK =1

2Axdx

dx

dt

2=

1

2Axdx

A0Ax

2dydt

2=

1

2dx

A20Ax

dy

dt

2=

1

2dx

(r2)2

r2(1 + x/`)2

dy

dt

2=

1

2

dxr2

(1 + x/`)2

dy

dt

2(c) Use kinetic energy is (1/2)mv2 for columns (I) and (II) and integrate answer

of part (b) for x = 0 to x = ` to find kinetic energy of liquid in horizontal arm.

For column (I), x = 0 so velocity all in y direction.

KEI =1

2mv2 =

1

2r2h

dy

dt

2Similarly for column (II)

KEII =1

2mv2 =

1

24r2h

dx

dt

2,

but at this column x = ` and therefore

dx

dt=

A0dy

Axdt=

r2

r2(1 + x/`)2dy

dt=

1

4

dy

dt.

So filling this back in we get

KEII =1

2mv2 =

1

24r2h

1

4

2dydt

2.

To find KE of liquid in horizontal arm we integrate answer part (b) from x = 0 tox = `, that is

KEIII =

Z `0

dK =1

2r2

dy

dt

2 Z `0

dx

(1 + x/`)2=

1

2r2

`

2

dy

dt

2Total KE is KEI +KEII +KEIII , which is

KE =1

2r2h

dy

dt

2+

1

24r2h

1

4

2dydt

2+

1

2r2

`

2

dy

dt

2=

1

4r2

`+

5h

2

dy

dt

2

(d) Use answers to parts (a) and (c) to find the total energy, the sum of potentialand kinetic.

E = KE + PE =1

2m

dx

dt

2+

1

2kx2

We have from parts (a) and (c) that

E = KE + PE =1

4r2

`+

5h

2

dy

dt

2+

5

8gr2y2

=1

2

"1

2r2

`+

5h

2

dy

dt

2#+

1

2

5

4gr2y2

So clearly m = (1/2)r2(`+ 5h/2) and k = (5/4)gr2, so

! =

rk

m=

s5g/4

(1/2)(`+ 5h/2)=

rg

2h

when ` = 5h/2. This also means that the period T is

T =2

!= 2

s2h

g.


1. Consider how to solve the steady-state motion of a forced oscillator if thedriving force is of the form F = F0 sin(!t) instead of F0 cos(!t).

For a forced oscillator with driving force F0 sin(!t), its equation of motion is

md2x

dt2+ kx = F0 sin(!t)

To solve we assume the solution is of the form x = C sin(!t) and solve for theconstant C.This means dx/dt = !C cos(!t) and d2x/dt2 = !2C sin(!t).Filling these into the equation of motion (EOM) we find that

m!2C sin(!t) + kC sin(!t) = F0 sin(!t).Equating coecients gives

C =F0

m!2 + k =F0m

km !2

=F0m

!20 !2

since !0 =q

km .

We want to express x in terms of a sinusoidal vibration having an amplitude A, bydefining a positive quantity , and a phase at t = 0:

x = A cos(!t+ )

We have:

x = C sin(!t) = C cos!t+

2

= |C| cos

!t+

2

for C > 0) !0 > !

= |C|(1) cos!t+

2

for C < 0) ! > !0

We set A = |C| to be our (positive) amplitude, and use the fact that cos(x) =cos(x+ ):

x = A cos!t+

2

for !0 > !

= A cos!t+

2+ = A cos

!t+

3

2

for ! > !0

This gives us:

x = A cos (!t+ )

Where:

A = |C|

C =F0m

!20 !2 =

2for !0 > !

=3

2for ! > !0

2. An object of mass 0.2 kg is hung from a spring whose spring constant is80 Nm1. The body is subject to a resistive force given by bv, where vis its velocity in ms1 and b = 4 Nm1sec.

(a) Set up the dierential equation of motion for free oscillations of thesystem, and find the period of such oscillations.

(b) The object is subjected to a sinusoidal driving force given by F (t) =F0 sin(!t), where F0 = 2 N and ! = 30 sec1. In the steady state,what is the amplitude of the forced oscillation?

(c) What is the mean power input?

(d) Show that the energy dissipated against the resistive force in onecycle is 0.063J

(a) We know the mass is subject to a resistive force bv, as well as a force due tothe springkx: F = kx bv

ma+ bv + kx = 0

0.2a+ 4v + 80x = 0

a+ 20v + 400x = 0

d2x

dt2+ 20

dx

dt+ 400x = 0

Comparing this to the general form:d2x

dt2+

dx

dt+ !20x = 0

We can read o values for and !0:

= 20

!20 = 400 !0 = 20

To obtain the period of oscillation we require !:

!2 = !20 2

4= 400 400

4= 400 100 = 3

! =p300 = 10

p3

T =2

!=

2

10p3=

5p3 0.36s

(b) For a system with the equation of motion:

d2x

dt2+

dx

dt+ !20x =

F0m

cos!t

The amplitude is given by:

A(!) =F0mp

(!20 !2)2 + (!)2

From (a) we know that: !0 = 20, m = 0.2 and = 20. Were told here that! = 30 and F0 = 2, using this in our equation for the amplitude gives:

A(!) =20.2p

((20)2 (30)2)2 + ((20)(30))2

=10p

((400 900)2 + (600)2

=10p

((500)2 + 360, 000=

10p250, 000 + 360, 000

=10p

610, 000

= 0.0128m

(c) The mean power is given by:

P =F 20!02kQ

1!0! !!0

2+ 1Q2

Using the same values we used in (b), Q = !0 =2020 = 1 and k = !

20m =

(20)2(0.2) = 80 we get:

P =(2)2(20)

2(80)(1)

12030 3020

2+ 11

=4(20)

160

1562 + 1=

80

160

12536 + 1

=1

2

16136

=1

2

36

61

=18

61 0.3W

(d) The energy lost per cycle is given by:

E = P T = P2

!= (0.3)

2

30= 0.063J

* 3. A block of mass m is connected to a spring, the other end of which is fixed.There is also a viscous damping mechanism. The following observationshave been made on the system:

(1) If the block is pushed horizontally with a force equal to mg, the staticcompression of the spring is equal to h.

(2) The viscous resistive force is equal to mg if the block moves with acertain known speed u.

(a) For this complete system (including both spring and damper) writethe dierential equation governing horizontal oscillations of the massin terms of m,g, h and u.

(a) (1) tells us:

kxx=h

= mg

kh = mg

k =mg

h

(2) tells us:

bvv=u

= mg

bu = mg

b =mg

u

Using this in:

ma+ bv + kx = 0

md2x

dt2+ b

dx

dt+ kx = 0

gives:

md2x

dt2+mg

u

dx

dt+mg

hx = 0

d2x

dt2+g

u

dx

dt+

g

hx = 0

Comparing this withd2x

dt2+

dx

dt+ !20x = 0

gives us values for and !0:

=g

u

!20 =g

h) !0 =

rg

h

Answer the following for the case that u = 3pgh

(b) What is the angular frequency of the damped oscillations?

(c) After what time, expressed as a multiple ofq

hg , is the energy down

by a factor 1e?

(d) What is the Q of this oscillator?

Using u = 3pgh gives us:

=g

u=

g

3pgh

=1

3

rg

h

!20 =g

h

(b) To find the angular frequency we use: !2 = !20 2

4 .

!2 = !20 2

4

=g

h 1

9

g

4h

=g

h g

36h

=35g

36h

! =

r35g

36h

(c) The energy decreases according to E(t) = E0et

E(t) = E0et

we need to find a st:

E() = E0e1 = E0e

this tells us

1 =

=1

= 3

sh

g

So the time taken for the energy to decrease a factor of 1e is t = 3q

hg s.

(d)

Q =!0

=

pgh

13

pgh

=113

= 3

4. A mass m is subject to a resistive force bv but no springlike restoringforce.

* (a) Show that its displacement as a function of time is of the form:x = C v0 et where = bm

(b) At t 0 the mass is at rest at x = 0. At this instant a drivingforce F = F0 cos!t is switched on. Find the values of A and in thesteady-state solution x = A cos(!t )

(c) Write down the general solution [The sum of parts (a) and (b)] andfind the values of C and v0 from the conditions that x 0 and dxdt = 0at t = 0

(a)

F = bvma = bva = b

mv = v

Now, we know that a = dvdt :

dv

dt= v

1

vdv = dt

Integrating both sides givesZ1

vdv =

Zdt =

Zdt

ln(v) = t+D where D is a constantv = et+D

v(t) = eDet

Let v0 be the inital velocity at the time 0:

v(0) = eDe0 = v0

) eD = v0This gives us a final expression for v

v(t) = v0et

To get an expression for x we use the fact that v = dxdt

dx

dt= v

dx

dt= v0e

t

dx = v0etdt

Integrating both sides givesZdx =

Zv0e

tdt = v0

Zetdt

x = v0et + C where C is a constant

) x = C v0et

(b)

ma+ bv = F0 cos!t

md2x

dt2+ b

dx

dt= F0 cos!t

d2x

dt2+

b

m

dx

dt=

F0m

cos!t

Looking at the steady state solution: x = A cos(!t ), we want to obtainexpressions for A and .

Going to the complex-exponential method; our basic equation becomes:

d2z

dt2+

b

m

dz

dt=

F0mei!t

We assume the solution: z = Aei(!t), with x = Re(x).

z = Aei(!tdelta)

dz

dt= i!Aei(!t)

d2z

dt2= !2Aei(!t)

Using these in our EOM:

d2z

dt2+

b

m

dz

dt=

F0mei!t

!2Aei(!t) + bmi!Aei(!t) =

F0mei!t

!2Aei + bmi!Aei =

F0m

!2A+ bmi!A =

F0mei

!2A+ bmi!A =

F0m

(cos() + i sin())

!2A+ bmi!A =

F0m

cos() + iF0m

sin()

Equating the real and imaginary parts:

!2A = F0m

cos()

) cos() = mF0

!2A

b

m!A =

F0m

sin()

) sin() = bF0

!A

tan() =sin()

cos()=

bF0!A

mF0!2A=

b

m! = b

m

1

!=

!

We can easily work out values for cos() and sin():

cos() = !p2 + !2

sin() =p

2 + !2

Note: weve chosen the signs here to ensure we have a positive value for A.Using this to get an expression for A:

!2A = F0m

cos()

A = F0m!2

cos()

=F0m!2

!p2 + !2

=F0

m!p2 + !2

(c) The general solution is give by x = C v0 et + A cos(!t )We have the initial conditions: x(0) = dxdt (0) = 0

x(t) = C v0et + A cos(!t )

x(0) = C v0e0 + A cos(0 ) = 0

C v0+ A cos() = 0

dx

dt(t) = v0e

t !A sin(!t )dx

dt(0) = v0e

0 !A sin(0 ) = 0v0 + !A sin() = 0

v0 = !A sin()

= !A

p2 + !2

!= A !p

2 + !2

= F0m!p2 + !2

!Ap2 + !2

= F0m(2 + !2)

And to find C:

C v0+ A cos() = 0

C =v0 A cos()

= F0

m(2+!2)

+

F0

m!p2 + !2

!p2 + !2

= F0m(2 + !2)

+F0

m(2 + !2)

= 0

* 5. The graph shows the power resonance curve of a certain mechanical sys-tem when driven by a force F0 sin(!t), where F0 = constant and ! is vari-able.

(a) Find the numerical values of !0 and Q for this system.

(b) The driving force is turned o. After how many cycles of free oscil-lation is the energy of the system down to 1/e5 of its initial value?(e = 2.718) (To a good approximation, the period of free oscillationcan be set equal to 2/!0.)

(a) Here we use the fact that with width of the power-resonance curve at half-height

!0Q

= 2

Q =!02

=40

2= 20

(b) The energy decreases according to E(t) = E0et

E(t) = E0et

we need to find a st:

E() = E0e5 = E0e

this tells us

5 =

=5

=

5

2= 2.5

The time taken to complete one cycle is 2omega0 , so the time taken to complete

n cycles is n 2omega0 :

= 2.5 = n2

omega0

) n 2omega0

= 2.5

n2

40= 2.5

n(0.16) = 2.5

n =2.5

0.16= 15.6 16

6. The figure shows the mean power input P as a function of driving fre-quency for a mass on a spring with damping. (Driving force = F0 sin(!t),where F0 is held constant and ! is varied.) The Q is high enough so thatthe mean power input, which is maximum at !0, falls to half-maximumat the frequencies 0.98!0 and 1.02!0.

(a) What is the numerical value of Q?

(b) If the driving force is removed, the energy decreases according tothe equation E = E0et. What is the value of ?

(c) If the driving force is removed, what fraction of the energy is lostper cycle?

(a)

!0Q

= width of the power-resonance curve at half-height!0Q

= 1.02!0 0.98!0= 0.04!0

Q =!0

0.04!0

=1

0.04= 25

(b) From (a) we can just write down the value of :

= 0.04!0

(c) The energy decreases according to the equation E = E0et, so the fractionof energy lost is EE0 :

E

E0=

E0et

E0= et

= e0.04!0t

The time taken for one cycle is the perios T = 2!0 , so the fraction of energylost per cycle is

E

E0= e

0.04!02!0

= e0.08s

A new system is made in which the spring constant is doubled, but themass and the viscous medium are unchanged, and the same driving forceF0 sin(!t) is applied. In terms of the corresponding quantities for theoriginal system, find the values of the following:

(d) The new resonant frequency !00.(e) The new quality factor Q0.(f) The maximum mean power input P 0m.(g) The total energy of the system at resonance, E 00.

(d)

For the original system:

!0 =

rk

mFor the new system:

!00 =

rk0

m=

r2k

m=p2

rk

m=p2!0

(e)


Q =!0

For the new system:

Q0 =!00

=p2!0

=p2Q

(f)


Pmax =QF 202m!0

For the new system:

P 0max =Q0F 202m!00

=

p2QF 20

2mp2!0

=QF 202m!0

= Pmax

(g) Originally we had E = E0et, therefore E0 = Eet. As there is no k depen-dence here we see that E 00 = Ee

t = E0.


* 1. Two identical pendulums are connected by a light coupling spring. Eachpendulum has a length of 0.4 m, and they are at a place where g =9.8 ms2. With the coupling spring connected, one pendulum is clampedand the period of the other is found to be 1.25 sec exactly.

(a) With neither pendulum clamped, what are the periods of the twonormal modes?

(b) What is the time interval between successive maximum possible am-plitudes of one pendulum after one pendulum is drawn aside andreleased?

(Ill go through this first question in quite some detail)

(a) Our system is given by:

We can write a general equation for A, by looking at the case where we moveA a distance xA, and B a distance xB and seeing what forces aect A as aresult.

There are two forces aecting A here:

restoring force due to the penduluma = !20x

ma = m!20xF = m!20x

restoring force due to the springF = kx

where x is the change in spring length: xA xB

So the total force on A is:

FA = m!20xA k(xA xB)md2xAdt2

= m!20xA k(xA xB)d2xAdt2

= !20xA k

m(xA xB)

setting !c =q

km we get:

d2xAdt2

+ !20xA + !2c (xA xB) = 0

Similarly, the equation of motion for B is:d2xBdt2

+ !20xB + !2c (xB xA) = 0

Were told that when B is clamped (xB = 0), that the period of A is TA = 1.25s.

TA =2

!A

!A =2

TA=

2

1.25= 5.03

!2A = (5.03)2 = 25.27

Looking at this system:

d2xAdt2

+ !20xA + !2c (xA xB) = 0

d2xAdt2

+ !20xA + !2c (xA 0) = 0

d2xAdt2

+ !20xA + !2cxA = 0

d2xAdt2

+ (!20 + !2c )xA = 0

) !A =q!20 + !

2c

For a simple pendulum of length l, the angular frequency !0 is given by:

!0 =

rg

l

!20 =g

lSo in our case:

!20 =9.8

0.4= 24.5

Using this in our expression for !A gives:

!A =q!20 + !

2c

!A =p24.5 + !2c

!2A = 24.5 + !2c

25.27 = 24.5 + !2c!2c = 25.27 24.5 = 0.77!c = 0.88

To find the normal modes we let xA = C cos(!t) and xB = D cos(!t) andobtain an expression for ! that satisfies both equations of motion.:

xA = C cos(!t)d2xAdt2

= !2C cos(!t)

xB = D cos(!t)d2xBdt2

= !2D cos(!t)

Using this in our equations of motion:

First for A:d2xAdt2

+ !20xA + !2c (xA xB) = 0

!2C cos(!t) + !20C cos(!t) + !2c (C cos(!t)D cos(!t)) = 0!2C + !20C + !2c (C D) = 0C(!2 + !20 + !2c ) !2cD = 0

C

D=

!2c!2 + !20 + !2c

And for B:d2xBdt2

+ !20xB + !2c (xB xA) = 0

!2D cos(!t) + !20D cos(!t) + !2c (D cos(!t) C cos(!t)) = 0!2D + !20D + !2c (D C) = 0D(!2 + !20 + !2c ) !2cC = 0

C

D=!2 + !20 + !2c

!2c

Combining these results:

) CD

=C

D!2c

!2 + !20 + !2c=!2 + !20 + !2c

!2c(!2c )

2 = (!2 + !20 + !2c )2!2c = !2 + !20 + !2c!2 = !20 + !

2c !2c

) !20 = !20 + !2c + !2c= !20 + 2!

2c

!0 =q!20 + 2!

2c

=p24.5 + 2(0.77) =

p24.5 + 2(0.77) =

p26.04 = 5.1

T 0 =2

!0=

2

5.1= 1.23s

) !200 = !20 + !2c !2c= !20

!00 = !0 =p24.5 = 4.95

T 00 =2

!00=

2

4.95= 1.27s

Note: normal modes for this system are given by the following two cases:

(i) The case where A and B are pulled apart the same distance: xA = xB.

(ii) The case where A and B are pulled in the same direction the same distance:xA = xB.

Looking at (i), and the equation of motion for A:

d2xAdt2

+ !20xA + !2c (xA xB) = 0

d2xAdt2

+ !20xA + !2c (xA + xA) = 0

d2xAdt2

+ !20xA + 2!2cxA = 0

d2xAdt2

+ (!20 + 2!2c )xA = 0

) !0 =q!20 + 2!

2c

Looking at (ii), and the equation of motion for A:

d2xAdt2

+ !20xA + !2c (xA xB) = 0

d2xAdt2

+ !20xA + !2c (xA xA) = 0

d2xAdt2

+ !20xA = 0

) !00 = !0

(b) Starting with our general equations of motion:

d2xAdt2

+ !20xA + !2c (xA xB) = 0

d2xBdt2

+ !20xB + !2c (xB xA) = 0

Adding these equations:

d2(xA + xB)

dt2+ !20(xA + xB) = 0

and setting q1 = xA + xB we get:d2q1dt2

+ !20q1 = 0

so we can write q1 as:

q1 = C cos(!0t) = C cos(4.95t)

Subtracting these equations:d2(xA xB)

dt2+ !20(xA xB) + 2!2c (xA xB) = 0

d2(xA xB)dt2

+ (!20 + 2!2c )(xA xB) = 0

using !0 =p!20 + 2!

2c and q2 = xA xB we get:

d2q2dt2

+ !02q2 = 0

so we can write q2 as:

q2 = D cos(!0t) = D cos(5.1t)

We can write xA and xB in terms of q1 and q2.

q1 = xA + xB q1 = xA + xBq2 = xA xB q2 = xA + xB

q1 + q2 = 2xA q1 q2 = 2xBxA =

1

2(q1 + q2) xB =

1

2(q1 q2)

This gives us:

xA(t) =1

2(q1 + q2)

=1

2(C cos(4.95t) +D cos(5.1t))

xB(t) =1

2(q1 q2)

=1

2(C cos(4.95t)D cos(5.1t))

These are the general solutions for xA and xB. To go any further we needinitial conditions.In this case, were told one pendulum is drawn aside and released. This cor-responds to one pendulum having displacement 0 at t = 0, and one pendulumhaving some displacement (say A0) at t = 0.Lets assume A is drawn aside, so at t = 0:

xA(0) = A01

2(C cos(0) +D cos(0)) = A0

C +D = 2A0xB(0) = 0

1

2(C cos(0)D cos(0)) = 0

C D = 0C = D

Filling this into xA(0) we get:

C + C = 2A0C = A0

D = C = A0.This gives us our equations of motion:

xA(t) =1

2(A0 cos(4.95t) A0 cos(5.1t))

=A02

(cos(4.95t) cos(5.1t))We can use the trigonometric identity cosA cosB = 2 sin A+B2 sin AB2

= A0 sin(4.95 + 5.1)

2t

sin

5.1 4.95

2t

= A0 sin (5t) sin (0.08t)

xB(t) =1

2(A0 cos(4.95t) + A0 cos(5.1t))

=A02

(cos(4.95t) + cos(5.1t))

We can use the trigonometric identity cosA+ cosB = 2 cosA+B2

cosAB2

= A0 cos

(4.95 + 5.1)

2t

cos

5.1 4.95

2t

= A0 cos (5t) cos (0.08t)

) xA(t) = A0 sin (5t) sin (0.08t)) xB(t) = A0 cos (5t) cos (0.08t)

Were asked to find the time interval between maximum possible amplitudesof one pendulum in this system. Looking at xB we can note that this is simplya beat equation.If we plot xB against time we can see its motion:

Where the distance between peaks is the time interval between successive max-imum possible amplitudes.If we compare this to the motion of cos

!0+!0

2 t= cos(5t)

and the motion of cos!0!0

2 t= cos(0.08t)

We can see that the time interval between successive peaks is half the periodof the slow oscillating motion, which is given by:

T =2

|!0!0|2

=4

|!0 !0|

So the time we require is the beat period:

T 0 =1

2T =

2

|!0 !0| =2

|2(0.08)| = 39.3s

2. Two harmonic oscillators A and B, of mass m and spring constants kAand kB, respectively, are coupled together by a spring of spring constantkC. Show that the normal frequencies satisfy:

m!2 =kA + kB

2+ kC

skA kB

2

2+ k2C

Our equations of motion are:

d2xAdt2

+kAmxA +

kCm

(xA xB) = 0d2xBdt2

+kBm

xB +kCm

(xB xA) = 0

We want to find ! such that xA = C cos(!t) and xB = D cos(!t) are solutions tothese equations.

!2C cos(!t) + kAmC cos(!t) +

kCm

(C cos(!t)D cos(!t)) = 0m!2C + kAC + kC(C D) = 0

D

C=m!2 + kA + kC

kC

!2D cos(!t) + kBm

D cos(!t) +kCm

(D cos(!t) C cos(!t)) = 0m!2D + kBD + kC(D C) = 0

D

C=

kCm!2 + kB + kC

Combining these:

m!2 + kA + kCkC

=kC

m!2 + kB + kC(m!2 + kA + kC)(m!2 + kB + kC) = k2C

(m!2)2 + kAkB + kAkC + kBkC + k2C m!2kA m!2kB 2m!2kC = k2C

(m!2)2 m!2(kA + kB + 2kC) + (kAkB + kAkC + kBkC) = 0

This is just a quadratic equation for m!2, so we can solve it using the quadraticformula:

m!2 =((kA + kB + 2kC))

p(kA + kB + 2kC)2 4(1)(kAkB + kAkC + kBkC)

2(1)

=kA + kB + 2kC

pk2A + k

2B + 4k

2C + 2kAkB + 4kAkC + 4kBkC 4kAkB 4kAkC 4kBkC

2

=kA + kB

2+ kC

pk2A + k

2B 2kAkB + 4k2C

2

=kA + kB

2+ kC

r(kA kB)2 + 4k2C

4

=kA + kB

2+ kC

skA kB4

2

2+ k2C

As required.

3. Two equal masses on an eectively frictionless horizontal air track areheld between rigid supports by three identical springs, as shown. Thedisplacements from equilibrium along the line of the springs are describedby coordinates xA and xB, as shown. If either of the masses is clamped,the period T for one complete vibration of the other is 3 sec.

*(a) If both masses are free, what are the periods of the two normalmodes of the system?

(a) Our equations of motion are:

d2xAdt2

+k

mxA +

k

m(xA xB) = 0

d2xAdt2

+k

m(2xA xB) = 0

d2xBdt2

+k

mxB +

k

m(xB xA) = 0

d2xBdt2

+k

m(2xB xA) = 0

Setting !0 =q

km :

d2xAdt2

+ !20(2xA xB) = 0d2xBdt2

+ !20(2xB xA) = 0

Were told that when one mass is clamped, the period of the other is T = 3s.Lets clamp B, so xB = 0. Using this in the equation of motion for A we obtain:

d2xAdt2

+ !20(2xA xB) = 0d2xAdt2

+ 2!20xA = 0

) ! = p2!0We know the period of this system is T = 3s.

T =2

!= 3

! =2

3p2!0 =

2

3

!0 =

p2

3

To obtain the normal modes we set xA = C cos(!t) and xB = D cos(!t)andobtain an expression for ! that satisfies both equations of motion.

d2xAdt2

+ !20(2xA xB) = 0!2C cos(!t) + !20(2C cos(!t)D cos(!t)) = 0

!2C + !20(2C D) = 0C

D=

!202!20 !2

d2xBdt2

+ !20(2xB xA) = 0!2D cos(!t) + !20(2D cos(!t) C cos(!t)) = 0

!2D + !20(2D C) = 0C

D=

2!20 !2!20

Combining these:

!202!20 !2

=2!20 !2

!20(!20)

2 = (2!20 !2)2!202!20 !2!2 = 2!20 !20!0 = !0 =

p2

3

T 0 =2

!0=

2p23

= 3p2s

!00 =p3!0 =

p3

p2

3

!=

p2p3

T 00 =2

!00=

2p2p3

=p6s

Note: It is easy to check that these normal modes coincide with the two fol-lowing cases:

!0 = !0Both masses pulled a distance x in the same direction.

!0 = p3!0Both masses pulled a distance x in opposite directions.

At t = 0, mass A is at its normal resting position and mass B is pulledaside a distance of 5 cm. The masses are released from rest at this instant.

(b) Write an equation for the subsequent displacement of each mass asa function of time.

(c) What length of time (in seconds) characterises the periodic transferof the motion from B to A and back again?

(b) Starting with our equations of motion:

d2xAdt2

+ !20(2xA xB) = 0d2xBdt2

+ !20(2xB xA) = 0Adding our equations of motion gives:

d2q1dt2

+ !20q1 = 0

where q1 = xA + xB.

Subtracting our equations of motion gives:d2q2dt2

+ 3!20q2 = 0

where q2 = xA xB.

SO we can write q1 = C cos(!0t) and q2 = D cos(p3!0t). As in question 1., we

can thus write xA and xB:

xA(t) =1

2

C cos(!0t) +D cos(

p3!0t)

xB(t) =

1

2

C cos(!0t)D cos(

p3!0t)

Were told that t = 0, mass A is at its normal resting position and mass B ispulled aside a distance of 5 cm.) xA(0) = 0, xB(0) = 0.05

xA(0) =1

2(C +D) = 0

C = DxB(0) =

1

2(C D) = 0.05C D = 0.05

2C = 0.05

) C = 0.025) D = 0.025

Which gives us:

xA(t) =0.025

2

cos(!0t) cos(

p3!0t)

= 0.025 sin

!0 +

p3!0

2

!sin

!0

p3!0

2

!

= 0.025 sin (1 +

p3)!0

2

!sin

(1p3)!0

2

!xB(t) =

0.025

2

cos(!0t) + cos(

p3!0t)

= 0.025 cos

!0 +

p3!0

2

!cos

!0

p3!0

2

!

= 0.025 cos

(1 +

p3)!0

2

!cos

(1p3)!0

2

!where: !0 =

p23

(c) Again, as in 1., this is just the beat period:

T =2

|(1p3)!0|=

2

|(1p3)p23 |

=3p2

(p3 1) = 5.8s

4. Two objects, A and B, each of mass m, are connected by springs as shown.The coupling spring has a spring constant kc, and the other two springshave spring constant k0. If B is clamped, A vibrates at a frequency A of1.81 sec1. The frequency 1 of the lower normal mode is 1.14 sec1.

(a) Briefly explain why the equations of motion of A and B are given by:

md2xAdt2

= k0xA kc(xA xB)

md2xBdt2

= k0xB kc(xB xA)

(b) Putting !0 =pk0/m, show that the angular frequencies !1 and !2 of

the normal modes are given by

!1 = !0 , !2 = [!20 + (2kc/m)]

1/2,

and that the angular frequency of A when B is clamped (xB = 0always) is given by

!A = [!20 + (kc/m)]

1/2.

(c) Using the numerical data above, calculate the expected frequency(2) of the higher normal mode. (The observed value was 2.27 sec1).

(d) From the same data calculate the ratio kc/k0 of the two spring con-stants.

(a) Looking at our system:

Looking at A: There are two forces aecting A here:

restoring force due to the spring on its leftF = k0xA

where xA is the change in spring length, and k0 is the spring constant.

restoring force due to the spring on its rightF = kC(xA xB)

where xAxB is the change in spring length, and kC is the spring constant.So, the total force on A is (denoting the total force on A by FA):

FA = k0xA kC(xA xB)using F = ma

maA = k0xA kC(xA xB)and a = d

2xdt2

md2xAdt2

= k0xA kC(xA xB)

As this system is totally symmetric in A and B we can write the equation ofmotion for B by interchanging A and B in As equation of motion:

md2xBdt2

= k0xB kC(xB xA)

(b) Setting !0 =q

k0m (after dividing across by m):

d2xAdt2

= !20xA kCm

(xA xB)d2xBdt2

= !20xB kCm

(xB xA)To obtain the normal modes we set xA = C cos(!t) and xB = D cos(!t) andobtain an expression for ! that satisfies both equations of motion.

!2C cos(!t) = !20C cos(!t)kCm

(C cos(!t)D cos(!t))

!2C = !20C kCm

(C D)C

D=

kCm

!20 +kCm !2

!2D cos(!t) = !20D cos(!t)kCm

(D cos(!t) C cos(!t))

!2D = !20D kCm

(D C)C

D=

!20 +kCm !2kCm

Combining these:kCm

!20 +kCm !2

=!2 + kCm !20

kCm

(kCm

)2 = (!20 +kCm !2)2

kCm

= !20 +kCm !2

!2 = !20 +kCm kC

m) !1 = !0) !2 =

r!20 +

2kCm

Note: It is easy to check that these normal modes coincide with the two fol-lowing cases:

!0 = !0Both masses pulled a distance x in the same direction.

!0 =q!20 +

2kCm

Both masses pulled a distance x in opposite directions.

When B is clamped, xB = 0, using this in the equation of motion for A:

d2xAdt2

= !20xA kCm

(xA xB)d2xAdt2

= !20xA kCm

xA

d2xAdt2

= (!20 +kCm

)xA

) !A =r!20 +

kCm

(c) Were told that A = 1.81s1 and 1 = 1.14s1:

=1

T=

!

2) ! = 2!1 = 21 = 2(1.14) = 7.16

!1 = !0 = 7.16

!A = 2A = 2(1.81) = 11.37

!A =

r!20 +

kCm

= 11.37

!20 +kCm

= 129.34

(7.16)2 +kCm

= 129.34

k

m= 129.34 51.27 = 78.07

Now obtaining a numerical value for !2 is simple:

!2 =

r!20 +

2kCm

=p(7.16)2 + 2(78.07) =

p207.41 = 14.4

Using this in = !2 gives us:

2 =!22

=14.4

2= 2.29s1

(d) We want the ratio kCk0 :

kCk0

=kCmk0m

=kCm

!20=

78.07

51.27= 1.52


* 1. A string of length 1 m has a fundamental node frequency of = 5Hz. Ifthis string is plucked transversely and is then touched at a point in thecentre, what frequencies persist?For a stretched string, the permitted stationary vibrations are given by:

n =n

2L

T

12

= n1

where:

1 =1

2L

T

12

where:

L is the length of the string T is the tension in the string is the mass per unit length = mL

1 is given by:

1 =1

2L

T

12

= 5Hz (as given in the question)

So our frequencies are given by:

n = 5Hz, 10Hz, 15Hz, 20Hz, 25Hz, 30Hz, ...

After the string is plucked and touched at a point in the centre, we eectively halfthe length of the string: L0 = L2 .Our new permitted frequencies are:

0n =n

2L0

T

12

=n

2L2

T

12

=n

L

T

12

= 2n= 10Hz, 20Hz, 30Hz, 40Hz, 50Hz, 60Hz, ...

The only frequencies that persist (p) are the frequencies both systems have incommon:

p = 10Hz, 20Hz, 30Hz, 40Hz, 50Hz, 60Hz, ...

2. A uniform string of length 2.5 m and mass 0.01 kg is placed under atension 10 N.

(a) What is the frequency of its fundamental mode?

(b) If the string is plucked transversely and is then touched at a point0.5 m from one end, what frequencies persist?

(a) Again, for a stretched string, the permitted stationary vibrations are given by:

n =n

2L

T

12

= n1

where:

1 =1

2L

T

12

where:

L is the length of the string T is the tension in the string is the mass per unit length = mL

The frequency of its fundamental mode, 1, is given by:

1 =1

2L

T

12

=1

2L

TmL

! 12=

1

2L

TL

m

12

=1

2

TL

mL2

12

=1

2

T

mL

12

=1

2

10

(0.01)(2.5)

12

=1

2(400)

12

=1

2(20)

= 10Hz

(b) The frequency of the fundamental mode for a string of length L = 2.5m is10Hz So our frequencies are given by:

n = 10Hz, 20Hz, 30Hz, 40Hz, 50Hz, 60Hz, ...

After the string is plucked and touched at a point 0.5 m from one end, our newlength is L0 = 2m = 4L5 .

Our new permitted frequencies are:

0n =n

2L0

T

12

=n

24L5

T

12

=5n

8L

T

12

=5

4n

= 12.5Hz, 25Hz, 37.5Hz, 50Hz, 62.5Hz, 75Hz, 87.5Hz, 100Hz, ...

The only frequencies that persist (p) are the frequencies both systems have incommon:

p = 50Hz, 100Hz, 150Hz, 200Hz, 250Hz, 300Hz, ...

* 3. A string of length L and total mass M is stretched to a tension T . Whatare the frequencies of the three lowest normal modes of oscillation of thestring for transverse oscillations? Compare these frequencies with thethree normal mode frequencies of three masses each of mass M/3 spacedat equal intervals on a massless string of tension T and total length L.

(a) As in the previous question we use that the normal mode frequencies are givenby:

n =n

2L

sT

with =

m

Lso 1 =

1

2

rT

mL

Three lowest are A = n1 with n = 1, 2, 3, that is:

1 =1

2L

sT

2 =

1

L

sT

3 =

3

2L

sT

(b) For N particles on a string each of mass m and a distance ` apart:

!n = 2!0 sin

n

2(N + 1)

with !0 =

T

m`

Therefore n =!n2

= 20 sin

n

2N + 2

with 0 =

1

2

rT

m`

We have that m = M/3 and ` = L/4 so:

0 =1

2

r12T

ML=

p12

12

rT

ML=

p12

1 = 1.10261

This means that all of the normal mode frequencies are given by:

0n = 2(1.10261) sin

n

2N + 2

The three lowest normal mode frequencies are when n = 1, 2, 3:

01 = 2(1.10261) sin8

= 0.841

02 = 2(1.10261) sin2

8

= 1.551

03 = 2(1.10261) sin3

8

= 2.031

4. A uniform rod is clamped at the center, leaving both ends free.

(a) What are the natural frequencies of the rod in longitudinal vibration?

(b) What is the wavelength of the nth mode?

(a) We solve the wave equation for a rod fixed at the middle, that is fixed at x = 0and free at both ends x = L/2. The wave equation is given by:

@2

@x2=

1

v2@2

@t2

with solutions of the form (x, t) = f(x) cos(!t) where f(x) = A sin(!x/v).We use boundary conditions to find a solution. Fixed at x = 0 means zero displace-ment at x = 0 and so f(x = 0) = 0. Filling in gives A sin(0) = 0 ) 0 = 0, sosatisfied.Free at x = L/2 means zero stress here and so Y @/@x = 0.

for@

@x= 0 must have

df

dx= 0

df

dx=

!

vcos!xv

need

!

vcos

!xv

L/2

= 0 ) cos!L2v

= 0

cos is an even function so cos(A) = cos(A) and we have:

cos

!L

2v

= 0 ) !L

2v=

n 1

2

) !n = 2(n 1/2)v

L

!n =(2n 1)

L

sY

) n = (2n 1)

2L

sY

since the speed for rigid bodies is given in terms of the Youngs modulus Y and thedensity as v =

pY/.

(b) Recall that c = f so in this notation we have v = nn.

n =v

n=

2L

2(n 1/2) =L

(n 1/2)

5. Derive the wave equation for vibrations of an air column. Your finalresult should be

@2

@x2=

K

@2

@t2

where is the displacement from the equilibrium position, is the massdensity, and K is the elastic modulus.This derivation is almost identical to the derivation of the wave equation for longi-tudinal vibrations of a rod, but using K the elastic modulus instead of Y Youngsmodulus.

We consider the equation of motion of a thin slice of air, which in the undisturbedstate is contained between x and x+x.Then, this slice is shifted and stretched, it is pulled in opposite directions by theforces F1 and F2.The length of the slice (originally x) has increased by .So our average stress in this case is given by K xThe stress at x is therefore K @@xSimilarly, the stress at x+x is K @@x +

@2@x2x.

Taking to be the cross sectional area:

F1 = K@

@x

F2 = K@

@x+K

@2

@x2x

F2 F1 = K@2

@x2x

We now apply Newtons law to the material lying between x and x+xTaking the density to be , then the mass of this area is x.

The acceleration is given by @2

@x2

This gives us:

F = ma = F2 F1x

@2

@x2= K

@2

@x2x

@2

@x2= K

@2

@x2

@2

@x2=K

@2

@x2

as required.

* 6. A laser can be made by placing a plasma tube in an optical resonantcavity formed by two highly reflecting flat mirrors, which act like rigidwalls, see figure. The purpose of the plasma tube is to produce light byexciting normal modes of the cavity.

(a) What are the normal mode frequencies of the resonant cavity? (Ex-press your answer in terms of the distance L between the mirrorsand the speed of light c.)

(b) Suppose that the plasma tube emits light centered at frequency0 = 5 1014 Hz with a spectral width , as shown in the sketch.The value of is such that all normal modes of the cavity whosefrequency is within 1.0 109 Hz of 0 will be excited by the plasmatube.

i. How many modes will be excited if L = 1.5 m?

ii. What is the largest value of L such that only one normal modewill be excited (so that the laser will have only one output fre-quency)?

(a) We solve the wave equation for light (massless like string) fixed at both ends,that is fixed at x = 0 and at x = L. The normal mode frequencies are givenby:

n =nv

2L=

nc

2Lsince the speed of light v = c.

(b)(i) We find the position of the modes in terms of the mode frequency. WithL = 1.5 m n = nc/3 and so n = 3n/c.

when n = 0 : n =3(5 1014)3 108 = 5, 000, 000

when n = 0 + : n =3(5 1014 + 1 109)

3 108 = 5, 000, 010

when n = 0 : n = 3(5 1014 1 109)

3 108 = 4, 999, 990

Between 0 + and 0 we have n = 5, 000, 010 5, 000, 000 = 10 modes.Between 0 and 0 we have n = 5, 000, 000 4, 999, 990 = 10 modes.So when L = 1.5 m, n = 20 modes plus the mode located at n = 0 gives a totalof 21 modes.

(b)(ii) n = nc/2L so n = 2nL/c.

when n = 0 : n =2(5 1014)L

3 108 = 3, 333, 333L

when n = 0 + : n =2(5 1014 + 1 109)L

3 108 = 3, 333, 340L

We want the value of L for there to be only one excited mode, at n = 0. So wantdierence between n at 0 and n at 0 +=1.So we want: (3, 333, 340 3, 333, 333)L = 6.666L = 1 ) L = 0.15 m.


1. Find the Fourier series for the following functions (0 x L).(a) y(x) = Ax(L x)

* (b) y(x) = A sin(x/L)

(c)

y(x) =

A sin(2x/L), 0 x L/20, L/2 x L

(a) y(x) = Ax(L x)

y(x) =1Xn=0

Bn sinnx

L

Bn =

2

L

Z L0

y(x) sinnx

L

dx

=2

L

Z L0

(Ax(L x)) sinnx

L

dx

=2A

L

Z L0

hLx x2 sinnx

L

idx

Using integration by parts: Zudv = uv

Zvdu

We set:

u = Lx x2 dv = sinnx

L

dx

du = (L 2x) dx v = cosnxL

nL

v = L cosnxL

n

Using this in our integral:Z L0

hLx x2 sinnx

L

idx =

"(Lx x2)

L cos

nxL

n

!#L0

Z L0

L cos

nxL

n

!(L 2x) dx

= (L2 L2) L cos

nLL

n

! 0

+L

n

Z L0

(L 2x)cosnx

L

dx

=L

n

Z L0

(L 2x)cosnx

L

dx

Again integrating by parts:

u = L 2x dv = cosnx

L

dx

du = 2dx v = sinnxL

nL

v =L sin

nxL

n

Using this in our integral:Z L0

(L 2x)cosnx

L

dx =

"(L 2x)

L sin

nxL

n

!#L0

Z L0

L sin

nxL

n

!(2) dx

= (L 2L) L sin

nLL

n

! (L 0)

L sin (0)

n

+

2L

n

Z L0

sinnx

L

dx

= (L 2L)L sin (n)

n

0 + 2L

n

"cos

nxL

nL

#L0

= 0 2L2

(n)2

hcosnx

L

iL0

2L2

(n)2

cos

nL

L

cos (0)

= 2L

2

(n)2[cos (n) 1]

= 2L2

(n)2[(1)n 1]

Going back to our Bn:

Bn =2A

L

Z L0

hLx x2 sinnx

L

idx

=2A

L

L

n

Z L0

(L 2x)cosnx

L

dx

=2A

n

Z L0

(L 2x)cosnx

L

dx

= 4AL2

(n)3[(1)n 1]

for n even:

= 4AL2

(n)3[1 1] = 0

for n odd:

= 4AL2

(n)3[1 1] = 8AL

2

(n)3

So our Fourier series is:

y(x) =1Xn=0

Bn sinnx

L

where:

Bn =

(8AL2

(n)3 n is odd

0 n is even

(b) y(x) = A sin(x/L)

y(x) =1Xn=0

Bn sinnx

L

Bn =

2

L

Z L0

y(x) sinnx

L

dx

=2

L

Z L0

A sin

xL

sinnx

L

dx

=2A

L

Z L0

hsinxL

sinnx

L

idx

Using the relation:

2 sinA sinB = cos(A B) cos(A+B)This gives:

Bn =2A

L

Z L0

hsinxL

sinnx

L

idx

=A

L

Z L0

cos

(1 n)x

L

cos

(1 + n)x

L

dx

=A

L

24sin(1n)x

L

(1n)

L

sin(1+n)x

L

(1+n)

L

35L0

= A

24sin(1n)x

L

(1 n)

sin(1+n)x

L

(1 + n)

35L0

= A

24sin(1n)L

L

(1 n)

sin(1+n)L

L

(1 + n)

sin (0)(1 n) +

sin (0)

(1 + n)

35= A

sin ((1 n))(1 n)

sin ((1 + n))

(1 + n) 0 + 0

= A

sin ((1 n))(1 n)

sin ((1 + n))

(1 + n)

We can see that n = 1 will require us to divide by 0, so well deal with the casen = 1 seperately. Our current Bn is therefore for all n 6= 1.

=A

sin ((1 n))

1 n sin ((1 + n))

1 + n

Now, sin(n) = 0 when n 2 Z.n 2 Z) (1 n) 2 Z) sin ((1 n)) = 0n 2 Z) (1 + n) 2 Z) sin ((1 + n)) = 0

So, for n 6= 1:

Bn =A

sin ((1 n))

1 n sin ((1 + n))

1 + n

= 0

For n = 1,

B1 =A

L

Z L0

cos

(1 1)x

L

cos

(1 + 1)x

L

dx

=A

L

Z L0

cos (0) cos

2x

L

dx

=A

L

Z L0

1 cos

2x

L

dx

=A

L

"x sin

2xL

2L

#L0

=A

L

"x L sin

2xL

2

#L0

=A

L

"L 0 L sin

2LL

2

+L sin (0)

2

#

=A

L

L L sin (2)

2+ 0

=

A

L[L]

= A


y(x) =1Xn=0

Bn sinnx

L

where:

Bn =

0 n 6= 1A n = 1

(c)

y(x) =

A sin(2x/L), 0 x L/20, L/2 x L

y(x) =1Xn=0

Bn sinnx

L

Bn =

2

L

Z L0

y(x) sinnx

L

dx

=2

L

Z L2

0

A sin

2x

L

sinnx

L

dx+

2

L

Z LL2

(0) sinnx

L

dx

=2A

L

Z L2

0

sin

2x

L

sinnx

L

dx

Using the relation:

2 sinA sinB = cos(A B) cos(A+B)This gives:

Bn =2A

L

Z L2

0

sin

2x

L

sinnx

L

dx

=A

L

Z L2

0

cos

(2 n)x

L

cos

(2 + n)x

L

dx

=A

L

24sin(2n)x

L

(2n)

L

sin(2+n)x

L

(2+n)

L

35L2

0

= A

24sin(2n)x

L

(2 n)

sin(2+n)x

L

(2 + n)

35L2

0

= A

24sin(2n)L2

L

(2 n)

sin(2+n)L2

L

(2 + n)

sin (0)(2 n) +

sin (0)

(2 + n)

35= A

24sin(2n)

2

(2 n)

sin(2+n)

2

(2 + n)

0 + 035

= A

24sin(2n)

2

(2 n)

sin(2+n)

2

(2 + n)

35We can see that n = 2 will require us to divide by 0, so well deal with the casen = 2 seperately. Our current Bn is therefore for all n 6= 2.

=A

24sin(2n)

2

2 n

sin(2+n)

2

2 + n

35

Now, sin(n) = 0 when n 2 Z.

n 2 Z, and n even: ) (2 n)2

2 Z

) sin(2 n)

2

= 0

n 2 Z, and n even: ) (2 + n)2

2 Z

) sin(2 + n)

2

= 0

So, for n even:

Bn = 0

For n odd:

sin

(2 n)

2

= sin

n

2

We use sin(x) = cos

x 2

= cos

n

2

2

= cos

2 n

2

= cos

(1 n)

2

n is odd, so n 1 is even ) (1n)2 is an integer.

sin

(2 n)

2

= (1) (1n)2 for n odd

(This is a useful identity to know)This gives us:

Bn =A

24sin(2n)

2

2 n

sin(2+n)

2

2 + n

35=

A

"(1) (1n)22 n

(1) (1+n)22 + n

#

=A

242(1) (1n)2 (1) (1+n)2

+ n

(1) (1n)2 + (1) (1+n)2

4 n2

35using (1)n = (1)n:

(1) (1+n)2 = (1) (1n)2= (1) (1n)2 1

using (1)n1 = (1)n:= (1) (1n)2

Which gives us:

Bn =A

242(1) (1n)2 + (1) (1n)2

+ n

(1) (1n)2 (1) (1n)2

4 n2

35=

4A(1) (1n)2 (4 n2)

For n = 2,

B2 =A

L

Z L2

0

cos

(2 2)x

L

cos

(2 + 2)x

L

dx

=A

L

Z L2

0

cos (0) cos

4x

L

dx

=A

L

Z L2

0

1 cos

4x

L

dx

=A

L

"x sin

4xL

4L

#L2

0

=A

L

"x L sin

4xL

4

#L2

0

=A

L

2664L2 0L sin

4(L2 )

L

4

+L sin (0)

4

3775=

A

L

L

2 L sin (2)

4+ 0

=

A

L

L

2

=

A

2


y(x) =1Xn=0

Bn sinnx

L

where:

Bn =

8>:A2 n = 20 n even, n 6= 24A(1) (1n)2

(4n2) n odd

2. Satisfy yourself that the following equations can all be used to describethe same progressive wave:

(a) y = A sin2 (x vt)

(b) y = A sin (2(kx t))(c) y = A sin (2[(x/) (t/T )])

(d) y = A sin (!(t x/v))(e) y = AIm{exp[i2(kx t)]}

(a) y = A sin2 (x vt)

Well use this equation as our base equation, and attempt to rewrite the fol-lowing equations as this equation.

(b) y = A sin 2(kx t)y = A sin (2(kx t))

Using the identites:

k =1

=v

we can rewrite our equation as:

y = A sin (2(kx t))= A sin

2

1

x

v

t

= A sin

2

(x vt)

= (a)

(c) y = A sin 2[x (t/T )]

y = A sin 2[x

(t/T )]

Using the identity:

T =1


y = A sin 2[(x/) (t/T )]= A sin 2[

x

t1

]

= A sin 2[x

t]

= A sin

2

(x vt)

= (a)

(d) y = A sin!(t x/v)y = A sin!(t x/v)

Using the identity:

sin(x) = sin(x)! =

2

T= 2 = 2

v


y = A sin 2 v(t x/v)

= A sin 2v

(x/v t)

= A sin

2

(x vt)

= (a)

(e) y = AIm{exp[i2(kx t)]}y = AIm{exp[i2(kx t)]}

Using the identity:

exp[i2(kx t)] = cos 2(kx t) + i sin 2(kx t)Im{exp[i2(kx t)]} = sin 2(kx t)


y = A sin 2(kx t)= (b)

= (a)

3. The equation of a transverse wave traveling along a string is given byy = 0.3 sin (0.125x 25t), where y and x are in centimeters and t is inseconds.

(a) Find the amplitude, wavelength, wave number, frequency, period,and velocity of the wave.

(b) Find the maximum transverse speed of any particle in the string.

(a)

y = 0.3 sin ((0.125x 25t))= 0.3 sin (2(0.0625x 12.5t))= 0.3 sin (2(0.0625x 12.5t))

So we have something of the form:

y = A sin (2(kx t))or:

y = 0.3 sin (2(0.0625x 12.5t))= 0.3 sin (2(0.0625)(x 200t))= 0.3 sin

2

16(x 200t)

something of this form:

y = A sin

2

(x vt)

Now we can read o the solutions:

A = 0.3cm

= 16cm

k = 0.0625

= 12.5Hz

T =1

=

1

12.5= 0.08s

v = 200cms1

(b)

y = 0.3 sin ((0.125x 25t))dy

dt= 0.3(25) cos ((0.125x 25t))= 7.5 cos ((0.125x 25t))

This takes its maximum value when cos ((0.125x 25t)) = 1:vmax = 7.5

max speed = |vmax| = 7.5ms1= 23.56ms1

* 4. What is the equation for a transverse wave travelling in the negative x di-rection with amplitude 0.003m, frequency 5 sec1, and speed 3000m/sec?The equation for a transverse wave travelling in the negative x direction is:

y = A sin

2

(x+ vt)

So we need values for A, , and v. From the information given we have A = 0.003mand v = 3000ms1, so we just need to get a value for .

=v

) = v

=3000

5= 600m

Filling in these values gives:

y = 0.003 sin

2

600(x+ 3000t)

5. What is the equation for a longitudinal wave travelling in the positive x

direction with amplitude 0.02m, period 1.25 sec, and speed 560m/sec?The equation for a longitudinal wave travelling in the positive x direction is:

y = A sin

2

(x vt)

So we need values for A, , and v. From the information given we have A = 0.02mand v = 560ms1, so we just need to get a value for .

T =1

=v

) T = v

) = Tv= (1.25)(560)

= 700m

Filling in these values gives:

y = 0.02 sin

2

700(x 560t)

* 6. A wave of frequency 20 sec1 has a velocity of 80 m/sec.

(a) How far apart are two points whose displacements are 30 apart inphase?

(b) At a given point, what is the phase dierence between two displace-ments occurring at times separated by 0.01 sec?

(a) Two points one wavelength apart are 2 apart in phase.30 = 6 rad

=v

) = v

=80

20= 4m

So two points whose displacements are 2 apart in phase are 4m apart.

2 , 41

6(2), 1

6(4)

3, 4

6= 0.67m

So two points whose displacements are 3 apart in phase are 0.67m apart.

(b) The phase dierence between two displacements occurring at times separatedby T s are 2 apart in phase. (where T is the period.)

T =1

Documents

Solucionario Ondas y Fluidos French