Chapter 14 Chapter 14 -- Simple Simple Harmonic MotionHarmonic Motion

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Paul E. Tippens, Professor of PhysicsPaul E. Tippens, Professor of Physics

Southern Polytechnic State UniversitySouthern Polytechnic State University

© 2007

Photo by Mark Tippens

A TRAMPOLINE exerts a restoring force on the jumper that is directly proportional to the average force required to displace the mat. Such restoring forces provide the driving forces necessary for objects that oscillate with simple harmonic motion.

Objectives: After finishing this Objectives: After finishing this unit, you should be able to:unit, you should be able to:

•• Write and apply Write and apply HookeHooke’’s Laws Law for objects moving for objects moving with simple harmonic motion.with simple harmonic motion.

•• Describe the motion of Describe the motion of pendulumspendulums and calculate the and calculate the length length required required to produce a given to produce a given frequency.frequency.

•• Write and apply formulas for Write and apply formulas for finding the finding the frequency frequency ff,, period period TT,, velocity velocity vv, or , or accelerationacceleration aa in in terms of terms of displacementdisplacement xx or or time time tt..

Periodic MotionPeriodic MotionSimple periodic motionSimple periodic motion is that motion in which a is that motion in which a body moves back and forth over a fixed path, body moves back and forth over a fixed path, returning to each position and velocity after a returning to each position and velocity after a definite interval of time.definite interval of time.

Amplitude A

Period, T, is the time for one complete oscillation. (seconds,s)

PeriodPeriod, T, is the time for one complete oscillation. (seconds,s)(seconds,s)

Frequency, f, is the number of complete oscillations per second. Hertz (s-1)

FrequencyFrequency, f, is the number of complete oscillations per second. Hertz (sHertz (s--11))

1fT

Example 1:Example 1: The suspended mass makes 30 The suspended mass makes 30 complete oscillations in 15 s. What is the complete oscillations in 15 s. What is the period and frequency of the motion?period and frequency of the motion?

x FF

15 s 0.50 s30 cylces

T

Period: T = 0.500 sPeriod: T = 0.500 s

1 10.500 s

fT

Frequency: f = 2.00 HzFrequency: f = 2.00 Hz

Simple Harmonic Motion, SHMSimple Harmonic Motion, SHM

Simple harmonic motionSimple harmonic motion is periodic motion in is periodic motion in the absence of friction and produced by a the absence of friction and produced by a restoring force that is directly proportional to restoring force that is directly proportional to the displacement and oppositely directed.the displacement and oppositely directed.

A restoring force, F, acts in the direction opposite the displacement of the oscillating body.

F = -kx

A restoring force, F, acts in the direction opposite the displacement of the oscillating body.

F = -kx

x FF

HookeHooke’’s Laws Law

When a spring is stretched, there is a restoring force that is proportional to the displacement.

F = -kx

The spring constant k is a property of the spring given by:

k = F

x

F

x

m

Work Done in Stretching a SpringWork Done in Stretching a Spring

F

x

m

Work done Work done ONON the spring is the spring is positivepositive; ; work work BYBY spring is spring is negative.negative.

From HookeFrom Hooke’’s law the force F is:s law the force F is:

F (x) = kx

x1 x2

FTo stretch spring from To stretch spring from

xx 11 to xto x 2 2 , work is:, work is:2 22 1½ ½Work kx kx

(Review module on work)(Review module on work)

Example 2:Example 2: A 4A 4--kg mass suspended from a kg mass suspended from a spring produces a displacement of 20 cm. spring produces a displacement of 20 cm. What is the spring constant?What is the spring constant?

F20 cm

m

The stretching force is the weight The stretching force is the weight (W = mg) of the 4(W = mg) of the 4--kg mass:kg mass:

F = F = (4 kg)(9.8 m/s(4 kg)(9.8 m/s22) = 39.2 N) = 39.2 N

Now, from HookeNow, from Hooke’’s law, the force s law, the force constant k of the spring is:constant k of the spring is:

k = =k = =FF

xx

0.2 m0.2 mk = 196 N/mk = 196 N/m

Example 2(cont.:Example 2(cont.: The mass The mass m m is now stretched is now stretched a distance of 8 cm and held. What is the a distance of 8 cm and held. What is the potential energypotential energy? (k = 196 N/m)? (k = 196 N/m)

F8 cm

m

U = 0.627 JU = 0.627 J

The potential energy is equal to The potential energy is equal to the work done in stretching the the work done in stretching the spring:spring:

2 22 1½ ½Work kx kx

0

2 2½ ½(196 N/m)(0.08 m)U kx

Displacement in SHMDisplacement in SHM

m

x = 0 x = +Ax = -A

x

• Displacement is positive when the position is to the right of the equilibrium position (x = 0) and negative when located to the left.

• The maximum displacement is called the amplitude A.

Velocity in SHMVelocity in SHM

m

x = 0x = 0 x = +Ax = +Ax = x = --AA

v (+)v (+)

•• Velocity is Velocity is positive positive when moving to the when moving to the rightright and negative when moving to the and negative when moving to the left.left.

•• It is It is zerozero at the end points and a at the end points and a maximummaximum at the midpoint in either direction (+ or at the midpoint in either direction (+ or --).).

v (v (--))

Acceleration in SHMAcceleration in SHM

m

x = 0 x = +Ax = -A

•• Acceleration is in the direction of the Acceleration is in the direction of the restoring forcerestoring force. (. (aa isis positivepositive when when xx is is negative, and negative, and negativenegative when x is positive.)when x is positive.)

• Acceleration is a maximum at the end points and it is zero at the center of oscillation.

+x-a

-x+a

F ma kx

Acceleration vs. DisplacementAcceleration vs. Displacement

m

x = 0 x = +Ax = -A

x va

Given the spring constant, the displacement, and Given the spring constant, the displacement, and the mass, the the mass, the accelerationacceleration can be found from:can be found from:

oror

Note: Acceleration is always Note: Acceleration is always opposite opposite to displacement.to displacement.

F ma kx kxam

Example 3:Example 3: A A 22--kgkg mass hangs at the end mass hangs at the end of a spring whose constant is of a spring whose constant is k = 400 N/mk = 400 N/m. . The mass is displaced a distance of The mass is displaced a distance of 12 cm12 cm and released. What is the acceleration at and released. What is the acceleration at the instant the displacement is the instant the displacement is x = +7 cmx = +7 cm??

m+x

(400 N/m)(+0.07 m)2 kg

a

a = -14.0 m/s2a = -14.0 m/s2 a

Note: When the displacement is Note: When the displacement is +7 cm+7 cm (downward), the acceleration is (downward), the acceleration is --14.0 m/s14.0 m/s22

(upward) independent of motion direction.(upward) independent of motion direction.

kxam

Example 4:Example 4: What is the What is the maximum maximum acceleration acceleration for the for the 22--kgkg mass in the previous problem? (mass in the previous problem? (A A = 12 cm= 12 cm, , k = 400 N/mk = 400 N/m))

m+x

The maximum acceleration occurs The maximum acceleration occurs when the restoring force is a when the restoring force is a maximum; i.e., when the stretch or maximum; i.e., when the stretch or compression of the spring is largest.compression of the spring is largest.

F = ma = -kx xmax = A

400 N( 0.12 m)2 kg

kAam

amax = ± 24.0 m/s2amax = ± 24.0 m/s2Maximum Maximum Acceleration:Acceleration:

Conservation of EnergyConservation of EnergyThe The total mechanical energytotal mechanical energy (U + K)(U + K) of a of a vibrating system is constant; i.e., it is the vibrating system is constant; i.e., it is the same at any point in the oscillating path.same at any point in the oscillating path.

m

x = 0 x = +Ax = -A

x va

For any two points A and B, we may write:For any two points A and B, we may write:

½mvA 2 + ½kxA

2 = ½mvB 2 + ½kxB

2½mvA2 + ½kxA

2 = ½mvB2 + ½kxB

2

Energy of a Vibrating System:Energy of a Vibrating System:

mx = 0 x = +Ax = -A

x va

•• At any other point: At any other point: U + K = U + K = ½½mvmv22 + + ½½kxkx22

U + K = U + K = ½½kAkA22 x = x = A and v = 0.A and v = 0.

•• At points At points AA and and BB, the velocity is zero and the , the velocity is zero and the acceleration is a maximum. The total energy is:acceleration is a maximum. The total energy is:

A B

Velocity as Function of Position.Velocity as Function of Position.

mx = 0 x = +Ax = -A

x va

kv Am

vmax when

x = 0:

2 2kv A xm

2 2 21 1 12 2 2mv kx kA

Example 5:Example 5: A A 22--kgkg mass hangs at the end of mass hangs at the end of a spring whose constant is a spring whose constant is k = 800 N/mk = 800 N/m. The . The mass is displaced a distance of mass is displaced a distance of 10 cm10 cm and and released. What is the velocity at the instant released. What is the velocity at the instant the displacement is the displacement is x = +6 cmx = +6 cm??

m+x

½½mvmv22 + + ½½kx kx 22 = = ½½kAkA22

2 2kv A xm

2 2800 N/m (0.1 m) (0.06 m)2 kg

v

v = ±1.60 m/sv = ±1.60 m/s

Example 5 (Cont.):Example 5 (Cont.): What is the maximum What is the maximum velocity for the previous problem? (velocity for the previous problem? (A = 10 A = 10 cm, k = 800 N/m, m = 2 kgcm, k = 800 N/m, m = 2 kg.).)

m+x

½mv2 + ½kx 2 = ½kA2

800 N/m (0.1 m)2 kg

kv Am

v = ± 2.00 m/sv = ± 2.00 m/s

0The velocity is maximum when x = 0:The velocity is maximum when x = 0:

The The reference circlereference circle compares compares the circular motion of an object the circular motion of an object with its horizontal projection.with its horizontal projection.

2f

The Reference CircleThe Reference Circle

cos(2 )x A ft

cosx A t

x = Horizontal displacement.x = Horizontal displacement.A = Amplitude A = Amplitude ((xxmaxmax ).).

= Reference angle.= Reference angle.

Velocity in SHMVelocity in SHMThe The velocityvelocity (v) of an (v) of an oscillating body at any oscillating body at any instant is the horizontal instant is the horizontal component of its component of its tangential velocity (tangential velocity (vvTT ).).

vT = R = A; 2f

v = -vT sin ; = t

v = - A sin t

v = -2f A sin 2f tv = -2f A sin 2f t

The The accelerationacceleration ((aa)) of an of an oscillating body at any instant is oscillating body at any instant is the horizontal component of its the horizontal component of its centripetal acceleration (centripetal acceleration (aacc).).

Acceleration Reference CircleAcceleration Reference Circle

a = -ac cos

= -ac cos(t)2 2 2

2; c cv Ra a RR R

R = A

a = -cos(t)

2 24 cos(2 )a f A ft

2 24a f x

The Period and Frequency as a The Period and Frequency as a Function of Function of aa and and xx..

For any body undergoing For any body undergoing simple harmonic motionsimple harmonic motion::

Since a = -4f2x and T = 1/f

12

afx

2 xTa

The frequency and the period can be found if the displacement and acceleration are known. Note that the signs of a and x will always be opposite.

The frequency and the period can be found if the displacement and acceleration are known. Note that the signs of a and x will always be opposite.

Period and Frequency as a Function Period and Frequency as a Function of Mass and Spring Constant.of Mass and Spring Constant.

For a vibrating body with an For a vibrating body with an elastic restoring force:elastic restoring force:

Recall that Recall that F = ma = F = ma = --kxkx:

12

kfm

2 mTk

The frequency f and the period T can be found if the spring constant k and mass m of the vibrating body are known. Use consistent SI units.

The frequency f and the period T can be found if the spring constant k and mass m of the vibrating body are known. Use consistent SI units.

Example 6:Example 6: The frictionless system shown The frictionless system shown below has a below has a 22--kgkg mass attached to a spring mass attached to a spring ((k = 400 N/mk = 400 N/m). The mass is displaced a ). The mass is displaced a distance of distance of 20 cm20 cm to the right and released.to the right and released. What is the frequency of the motion?What is the frequency of the motion?

mx = 0 x = +0.2 m

x va

x = -0.2 m

1 1 400 N/m2 2 2 kg

kfm

f = 2.25 Hzf = 2.25 Hz

Example 6 (Cont.):Example 6 (Cont.): Suppose the Suppose the 22--kgkg mass mass of the previous problem is displaced of the previous problem is displaced 20 cm20 cm and released (and released (k = 400 N/mk = 400 N/m). What is the ). What is the maximum acceleration? (maximum acceleration? (f = f = 2.25 Hz2.25 Hz))

mx = 0 x = +0.2 m

x va

x = -0.2 m

2 2 2 24 4 (2.25 Hz) ( 0.2 m)a f x

Acceleration is a maximum when Acceleration is a maximum when x = x = AA

a = 40 m/s2a = 40 m/s2

Example 6:Example 6: The The 22--kgkg mass of the previous mass of the previous example is displaced initially at example is displaced initially at x = 20 cmx = 20 cm and released. What is the velocity and released. What is the velocity 2.69 s2.69 s after release? (Recall that after release? (Recall that ff = 2.25 Hz= 2.25 Hz.).)

m

x = 0 x = +0.2 m

x va

x = -0.2 m

v = -0.916 m/sv = -0.916 m/s

v = -2f A sin 2f tv = -2f A sin 2f t

2 (2.25 Hz)(0.2 m)sin 2 (2.25 Hz)(2.69 s)v

(Note:

in rads) 2 (2.25 Hz)(0.2 m)(0.324)v

The minus sign means it The minus sign means it is moving to the left.is moving to the left.

Example 7:Example 7: At what time will the 2At what time will the 2--kg mass kg mass be located 12 cm to the left of x = 0? be located 12 cm to the left of x = 0? (A = 20 cm, f = 2.25 Hz)(A = 20 cm, f = 2.25 Hz)

m

x = 0 x = +0.2 m

x va

x = -0.2 m

t = 0.157 st = 0.157 s

cos(2 )x A ft

-0.12 m

10.12 mcos(2 ) ; (2 ) cos ( 0.60)0.20 m

xft ftA

2.214 rad2 2.214 rad; 2 (2.25 Hz)

ft t

The Simple PendulumThe Simple Pendulum

The period of a The period of a simple simple pendulumpendulum is given by:is given by:

mg

L2 LT

g

For small angles

12

gfL

Example 8.Example 8. What must be the length of a simple pendulum for a clock which has a period of two seconds (tick-tock)?

2 LTg

L

22 2

24 ; L = 4

L T gTg

2 2

2

(2 s) (9.8 m/s )4

L

L = 0.993 m

The Torsion PendulumThe Torsion Pendulum

The period The period TT of a of a torsion torsion pendulumpendulum is given by:is given by:

Where k’ is a torsion constant that depends on the material from which the rod is made; I is the rotational inertia of the vibrating system.

Where k’ is a torsion constant that depends on the material from which the rod is made; I is the rotational inertia of the vibrating system.

2'

ITk

Example 9:Example 9: A A 160 g160 g solid disk is attached to solid disk is attached to the end of a wire, then twisted at the end of a wire, then twisted at 0.8 rad0.8 rad and released. The torsion constant kand released. The torsion constant k’’ is is 0.025 N m/rad0.025 N m/rad. Find the period.. Find the period.

(Neglect the torsion in the wire)(Neglect the torsion in the wire)

For DiskFor Disk: I = I = ½½mRmR22

I = I = ½½(0.16 kg)(0.12 m)(0.16 kg)(0.12 m)22

== 0.00115 kg m0.00115 kg m22

20.00115 kg m2 2' 0.025 N m/rad

ITk

T = 1.35 sT = 1.35 s

Note: Period is independent of angular displacement.Note: Period is independent of angular displacement.

SummarySummary

Simple harmonic motion (SHM) is that motion in which a body moves back and forth over a fixed path, returning to each position and velocity after a definite interval of time.

Simple harmonic motion (SHM)Simple harmonic motion (SHM) is that motion in is that motion in which a body moves back and forth over a fixed which a body moves back and forth over a fixed path, returning to each position and velocity path, returning to each position and velocity after a definite interval of time.after a definite interval of time.

1fT

F

x

m

The frequency (rev/s) is the reciprocal of the period (time for one revolution).

The frequency (rev/s) is the reciprocal of the period (time for one revolution).

Summary (Cont.)Summary (Cont.)

F

x

m

Hooke’s Law: In a spring, there is a restoring force that is proportional to the displacement. Hooke’s Law: In a spring, there is a restoring force that is proportional to the displacement.

The spring constant k is defined by:

Fkx

F kx

Summary (SHM)Summary (SHM)

F ma kx kxam

mx = 0 x = +Ax = -A

x va

½mvA 2 + ½kxA

2 = ½mvB 2 + ½kxB

2½mvA2 + ½kxA

2 = ½mvB2 + ½kxB

2

Conservation of Energy:

Summary (SHM)Summary (SHM)

2 2kv A xm

2 2 21 1 12 2 2mv kx kA

0kv Am

cos(2 )x A ft

2 sin(2 )v fA ft

2 24a f x

Summary: Period and Summary: Period and Frequency for Vibrating Spring.Frequency for Vibrating Spring.

mx = 0 x = +Ax = -A

x va

12

afx

2 xTa

2 mTk

12

kfm

Summary: Simple Pendulum Summary: Simple Pendulum and Torsion Pendulumand Torsion Pendulum

2 LTg

12

gfL

L

2'

ITk

CONCLUSION: Chapter 14CONCLUSION: Chapter 14 Simple Harmonic MotionSimple Harmonic Motion