WAVES

&

OSCILLATIONS

WAVES ARE EVERYWHERE!

Sound Waves result from periodic oscillations of air molecules, which collide with their neighbours and create a disturbance which moves at the speed of sound.

Electric and Magnetic fields, when oscillated, can create waves which carry energy. At the right frequency, we see electromagnetic waves as Light.

PERIODIC MOTION IS

EVERYWHERE

Examples of periodic motion

Earth around the sun

Elastic ball bouncing up an down

Quartz crystal in your watch,

computer clock, iPod clock, etc.

PERIODIC MOTION IS EVERYWHERE

Examples of periodic motion

Heart beat

In taking your pulse, you count 70.0

heartbeats in 1 min.

What is the period, in seconds, of your

heart's oscillations?

Period is the time for one

oscillation

T= 60 sec/ 70.0 = 0.86 s

What is the frequency?

f = 1 / T = 1.17 Hz

SOME OSCILLATIONS ARE NOT

SINUSOIDAL:

A SPECIAL KIND OF PERIODIC OSCILLATOR:

HARMONIC OSCILLATOR

WHAT DO ALL “HARMONIC OSCILLATORS”

HAVE IN COMMON?

1. A position of equilibrium

2. A restoring force, (which may be linear )

Hooke’s law spring F = -k x

In a pendulum the behavior only linear for

small angles:

F=-mg sinθ≈-mg θ where θ = s / L. In this

limit we have: F = -ks with k = mg/L

3. Inertia

4. The drag forces are reasonably small

SIMPLE HARMONIC MOTION (SHM)

Pull block to the right until x = A

After the block is released from x = A

it will move to the left until it reaches

equilibrium, continue moving to the

left due to inertia until it reaches x=-A

and stop there. Due to restoring

force moves to the right

k m

k m

k m

-A A 0(≡Xeq)

SIMPLE HARMONIC MOTION (SHM)

The time it takes the block to complete one cycle is called the

period.

Usually, the period is denoted by T and is measured in

seconds.

The frequency, denoted f, is the number of cycles that are

completed per unit of time: f = 1 / T.

In SI units, f is measured in inverse seconds, or hertz (Hz).

If the period is doubled, the frequency is

A. unchanged

B. doubled

C. halved

SHM DYNAMICS: NEWTON’S LAWS STILL

APPLY

At any given instant we know that F = ma must be true.

But in this case F = -k x and

So: -

k

x

m

F = -k x

a

xm

k

dt

xd

2

2

a differential equation for x(t) !

2

2

dt

xdmmakx

2

2

dt

xdmma

“Simple approach, guess a solution and see if it works!

SHM SOLUTION...

Try either cos ( t ) or sin ( t )

Below is a drawing of A cos ( t )

where A = amplitude of oscillation

[with = (k/m)½ and = 2 f = 2 /T ]

Both sin and cosine work so need to include

both

T = 2/

A

A

COMBINING SIN AND COSINE

SOLUTIONS

k

x

m

0

Use “initial conditions” to determine phase !

sin cos

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

= A cos ( t + )

= A (cos t cos – sin t sin )

=A cos cos t – A sin sin t

Notice that B = A cos C = -A sin tan = -C/B

WORK “INTEGRAL” Since work is force times distance, and the force

ramps up as we compress the spring further…

takes more work (area of rectangle) to compress a little

bit more (width of rectangle) as force increases (height of

rectangle)

if full distance compressed is kx, then force is kx, and

area under force “curve” is ½(base)(height) = ½(x)kx =

½kx2

area under curve is called an integral: work is integral of

force

F

orc

e

F

orc

e

F

orc

e

distance (x) distance (x) distance (x)

Area is a work: a

force (height) times

a distance (width) Force from spring

increases as it is

compressed

further

Total work done is

area of triangle

under force

curve

ENERGY OF THE SPRING-MASS

SYSTEM

Kinetic energy is always

K = ½ mv2 = ½ m(A)2 sin2(t+)

Potential energy of a spring is,

U = ½ k x2 = ½ k A2 cos2(t + )

And 2 = k / m or k = m 2

U = ½ m 2 A2 cos2(t + )

K + U = ½ m 2 A2 = ½ k A2 which is constant

x(t) = A cos( t + )

v(t) = dx/dt = -A sin( t + )

a(t) = dv/dt = -2A cos( t + )

This diagram shows how the energy transforms from potential to kinetic and back, while the total energy remains the same.

EXAMPLE:

USING CONSERVATION OF

ENERGY

M

arc

h 1

5, 2

01

3

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A - L

ectu

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2

A 500 g block on a spring is pulled a distance of 20 cm and released. The subsequent oscillations are measured to have a period of 0.80 s. At what position (or positions) is the speed of the block 1.0 m/s?

2 2 2 2kv Ax Ax

m

22

2 2(1.0 m/s)(0.20 m) 0.154 m15.4 cm(7.85 rad/s)

vxA

220.80 s so 7.85 rad/s

(0.80 s)T

T

QUESTION

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Four springs have been compressed from their equilibrium positions at x = 0.

Which system has the largest maximum speed in its oscillation?

The Simple Pendulum A simple pendulum consists of a mass m (of negligible size) suspended by a string or rod of length L (and negligible mass).

M

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24/32

The Simple Pendulum

Looking at the forces on the pendulum bob, we see that the restoring force is proportional to sin, whereas the restoring force for a spring is proportional to the displacement (which is in this case).

sinF mg mg

This is not a Hooke’s Law force.

The Small Angle Approximation

However, for small angles, sin θ and θ are approximately equal.

3 5

sin3! 5!

if 1

Fy = may = T – mg cos()

Fx = max = -mg sin()

If small then x L and sin() dx/dt = L d/dt

ax = d2x/dt2 = L d2/dt2

ax = -g = L d2/ dt2

L d2/ dt2 + g = 0

The solution of the above equation is

= cos(t + ) or = sin(t + )

with = (g/L)½

g

LT

2

2

Note that T does not depend on m, the mass of the pendulum bob.

A TORSION OSCILLATOR

Ph

ysic

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ectu

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2

The restoring torque is proportional to the angular displacement

The minus sign shows that the torque is directed opposite to the angular displacement

2

2

dt

dII

where I is the rotational inertia

Idt

d

dt

dI

2

2

2

2

This equation is similar to the equation for linear SHM i.e., xImk , ,

)cos( tm

where ω is angular frequency. Similarly

IT 2

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The Physical Pendulum

A physical pendulum is an extended solid mass that oscillates around its center of mass.

It cannot be modeled as a point mass suspended by a massless string because there is energy in the rotation of the object as its center of mass moves.

Examples:

P

In figure, a body of irregular shape is pivoted at P and is

displaced from the equilibrium position by an angle θ . The

centre of mass C lies vertically below P. The restoring torque

is given as

sinmgd

For small θ

mgd

The time period will be

2

2

4

as 2

mgdTI

mgdmgd

IT

So we can find moment of inertia of irregular object by this formula.

Simple pendulum is a special case of physical pendulum.

EXAMPLE: A COMFORTABLE

PACE

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The pace of a comfortable walk can be estimated if we model each leg as a swinging physical pendulum.

1 12

rod3 2, , and 0.9 mI ML L L

12

rod 3

2 2

/2 /2 22 2 2

3/2 /2

MLIL L LT

g g gML ML

2

2 2(0.9 m)(10 steps)51010 7.7 s

3 3(9.81 m/s)

Lt T

g

Test: 10 steps takes 6.7 s7.7 s

Note that pace period is proportional to L½.

EXAMPLE: A BLOCK ON A

SPRING A 2.00 kg block is attached to a spring as shown. The force constant of the spring is k = 196 N/m. The block is held a distance of 5.00 cm from equilibrium and released at t = 0.

(a) Find the angular frequency , the frequency f, and the period T. (b) Write an equation for x vs. time.

(196 N/m)9.90 rad/s

(2.00 kg)

k

m

(9.90 rad/s)1.58 Hz

2 2f

1/ 0.635 sT f 5.00 cm and 0A

(5.00 cm)cos(9.90 rad/s)x t

EXAMPLE: A SYSTEM IN SHM

M

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1

A mass attached to a spring, pulled 20 cm to the right, and released at t=0. It makes 15 complete oscillations in 10 s.

a. What is the period of oscillation?

b. What is the object’s maximum speed?

c. What is its position and velocity at t=0.80 s?

15 oscillations

10 s

1.5 oscillations/s1.5 Hz

f

1/ 0.667 sT f

max

2 2(0.20 m)1.88 m/s

(0.667 s)

Av

T

2 2(0.80 s)cos(0.20 m)cos 0.062 m6.2 cm

(0.667 s)

txAT

max

2 2(0.80 s)sin(1.88 m/s)sin 1.79 m/s

(0.667 s)

tvv

T

EXAMPLE: FINDING THE TIME

A mass, oscillating in simple harmonic motion, starts at x = A and has period T.

At what time, as a fraction of T, does the mass first pass through x = ½A?

111

62cos

2 23

T Tt T

1

2

2cos

tx AA

T

If the springs in your 1000 kg car compress by 10

cm (e.g., when lowered off of jacks):

then the springs must be exerting mg = 10,000 Newtons

of force to support the car

F = kx = 10,000 N, x = 0.1 m

so k = 100,000 N/m (stiff spring)

this is the collective spring constant: they all add to this

Now if you pile 400 kg into your car, how much will

it sink?

4,000 = (100,000)x, so x = 4/100 = 0.04 m = 4 cm

Could have taken short-cut:

springs are linear, so 400 additional kg will depress car

an additional 40% (400/1000) of its initial depression

EXAMPLE