Chapter 11

Vibrations and Waves

The Ideal Spring and Simple Harmonic Motion

xkF Appliedx =

spring constant Units: N/m

If you apply a force FxApplied to an ideal spring, it will compress or

expand by an amount x according to the relation,

The Ideal Spring and Simple Harmonic Motion

HOOKE’S LAW: RESTORING FORCE OF AN IDEAL SPRING The restoring force on an ideal spring is xkFx −=

Simple harmonic motion is a type of periodic motion where the restoring force on a mass is directly proportional to the displacement. The displacement of the mass has a sinusoidal dependence on the elapsed time, i.e. depends on time through a sine or cosine function.

A mass-spring system undergoing Simple Harmonic Motion

A mass-spring system undergoing Simple Harmonic Motion and tracing out a sinusoidal function of its displacement as time elapses

The Ideal Spring and Simple Harmonic Motion

Energy and Simple Harmonic Motion

DEFINITION OF ELASTIC POTENTIAL ENERGY The elastic potential energy is the energy that a spring has by virtue of being stretched or compressed. For an ideal spring, the elastic potential energy is

221

elasticPE kx=

SI Unit of Elastic Potential Energy: joule (J)

Since the restoring force due to an ideal spring is a conservative force, the total mechanical energy of a spring-mass system is conserved, and is

Energy and Simple Harmonic Motion

E = KE +PE = 12 mv

2 + 12 kx

2

Since v = 0 at ± xmax ⇒ E = 12 kxmax

2

∴ 12 kxmax

2 = 12 mv

2 + 12 kx

2

⇒ v = ± km

xmax2 − x2

v = ±vmax = ±xmaxkm

at x = 0

Simple Harmonic Motion and the Reference Circle

How do x, v and a depend on time quantitatively in SHM? Since F=-kx does not give constant acceleration à can’t used our constant acceleration equations. As an analogy with the sinusoidal dependence of displacement on time in SHM, consider the projection on the x-axis of the position of a ball undergoing uniform circular motion on a turntable. Record how x depends on time with a moving roll of film.

Simple Harmonic Motion and the Reference Circle

Since θ =ωt,⇒ x = Acosθ = Acosωt

Assume that the turntable is rotating at the constant angular speed ω andthe ball is at radius A.

DISPLACEMENT

Simple Harmonic Motion and the Reference Circle

period T: the time required to complete one cycle

frequency f: the number of cycles per second (measured in Hz)

Tf 1= ω =

2πT

= 2π f

amplitude A: the maximum displacement

Simple Harmonic Motion and the Reference Circle

VELOCITY

vx = −vT sinθ = −Aωvmax!sinωt

Simple Harmonic Motion and the Reference Circle

Example: The Maximum Speed of a Loudspeaker Diaphragm The motion of a loudspeaker is sinusoidal. The frequency of motion is 1.0 KHz and the amplitude is 0.20 mm. (a) What is the maximum speed of the diaphragm? (b) Where in the motion does this maximum speed occur?

Simple Harmonic Motion and the Reference Circle

vx = −vT sinθ = −Aωvmax!sinωt(a)

( ) ( )( )( )sm3.1

Hz100.12m1020.02 33max

=

××=== − ππω fAAv

(b) The maximum speed occurs midway between the ends of its motion.

x = Acosωt = 0

when ωt = π2

, 3π2

,…

vx = vmax occurs when ωt = π2

, 3π2

,…

Simple Harmonic Motion and the Reference Circle

ACCELERATION

ax = −ac cosθ = −Aω2

amax!cosωt

x = Acosωt

vx = −Aωvmax!sinωt

ax = −Aω2

amax!cosωt

Displacement

Displacement, Velocity and Acceleration for SHM vs. time

Simple Harmonic Motion and the Reference Circle

A

A

A

A

Simple Harmonic Motion and the Reference Circle

Find a relationship between vx and x for the reference circle:

x = Acosωt vx = −Aω sinωt

cos2ωt + sin2ωt =1 ⇒x2

A2+

vx2

A2ω 2 =1

∴vx = ±ω A2 − x2

Compare this equation with the equation for v as a function of x that we got earlier for the spring-mass system from conservation of energy, i.e.

v = ± km

xmax2 − x2

This suggests that the sinusoidal equations of the reference circle for x, v and a are also valid for the spring-mass system if we set

ω =km

and A = xmax

Simple Harmonic Motion and the Reference Circle

Another way to extract the frequency of vibration of the spring-mass system: Use the reference circle equations and Newton’s 2nd law.

mk

=ω

tAax ωω cos2−=tAx ωcos=

xmakxF =−=∑2ωmAkA −=−

Simple Harmonic Motion and the Reference Circle

Example: A Body Mass Measurement Device The device consists of a spring-mounted chair in which the astronaut sits. The spring has a spring constant of 606 N/m and the mass of the chair is 12.0 kg. The measured period is 2.41 s. Find the mass of the astronaut.

Simple Harmonic Motion and the Reference Circle

totalmk

=ω 2total ωkm =

Tf π

πω22 ==

( ) astrochair2total 2mm

Tkm +==π

( )( )( ) kg 77.2kg 0.12

4s 41.2mN606

2

2

2

chair2astro

=−=

−=

π

πm

Tkm

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Example: Simple Harmonic Motion of a vertical spring-mass system. A 0.20 kg ball is attached to a vertical spring. The spring constant is 28 N/m. Find the equilibrium position of the mass with respect to the unstrained position of the spring and find the amplitude of the SHM about this equilibrium position after the ball is dropped from the unstrained position.

Equilibrium position:

F =ma⇒∑ kd0 −mg = 0

∴d0 =mgk=

0.20( ) 9.8( )28

= 0.070 m

Amplitude of oscillations:Ef = E0

12 kdmax

2 =mgdmax

dmax = 2mgk= 2

0.20( ) 9.8( )28

= 0.140 m

∴A = dmax − d0 = 0.140− 0.070 = 0.070 m

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dmax

} }

A A

The Pendulum

A simple pendulum consists of a mass attached to a frictionless pivot by a cable of negligible mass.

FT = −mgsinθ

For small angles: sinθ ≈θ

Since θ = sL

⇒ FT ≈ −mgLs↔ F = −kx

⇒ k↔ mgL⇒ω =

km=

mgmL

=gL

∴ω =gL

Pendulum (small angles only)

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The Pendulum

Example: Keeping Time Determine the length of a simple pendulum that will swing back and forth in simple harmonic motion with a period of 1.00 s.

22Lg

Tf ===

ππω

( ) ( ) m 248.04

sm80.9s 00.14 2

22

2

2

===ππ

gTL

2

2

4πgTL =

Damped Harmonic Motion

In simple harmonic motion, an object oscillates with a constant amplitude. In reality, friction or some other energy dissipating mechanism is always present and the amplitude decreases as time passes. This is referred to as damped harmonic motion.

Damped Harmonic Motion

1)  simple harmonic motion – amplitude stays constant

2&3) underdamped – amplitude decreases but still oscillations 4)  critically damped – amplitude decreases to 0 without oscillations in shortest possible time

5)  Overdamped – amplitude decreases to 0 without oscillations slower than in critically damped case

Different types of damped harmonic motion: