Chapter 10Simple Harmonic Motion

1) Hooke’s Law• spring force prop. to (-)displacement

Fx = −kx

Force exerted by spring

displacement from equilibrium

spring constant(units: N/m)

restoring force(force opposes displacement)

FxApplied = kx

Is the spring accelerating when it is extended a distance x?

Image reprinted with permission of John Wiley and Sons, Inc.

Ans: No. So there must be another force acting on it.

A car is hauling a 92 kg trailer, to which it is connected by a spring.

The spring constant is 2300 N/m.

The trailer accelerates with an acceleration of 0.30 m/s2.

By how much does the spring stretch?

Example

An object is attached to the lower end of a 100-coil spring that is hanging from the ceiling. The spring stretches by 0.160 m. The spring is then cut into two identical springs of 50 coils each. As the drawing shows, each spring is attached between the ceiling and the object. By how much does each spring stretch?

Homework

A 0.70-kg block is hung from and stretches a spring that is attached to the ceiling. A second block is attached to the first one, and the amount that the spring stretches from its unstrained length triples. What is the mass of the second block?

Example

2) Simple harmonic motion (SHM)Motion of an object that is acted on by the spring force:

t = 0: extend spring and release.

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Fx < 0, x > 0, v = 0

�

Fx > 0, x < 0, v = 0

�

Fx = 0, x = 0, v > 0

�

Fx = 0, x = 0, v < 0

Result is Simple Harmonic Motion (SHM).Plot the position x of the ball versus time:

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Amplitude: largest extension of spring.

• Dynamic definition: motion produced by a spring-like force (prop. to negative displacement)

F = −kx

• Kinematic definition: accel. prop. to negative displacement

F = ma ⇒ ma = −kx

a = −kmx

3) Description of the motion: x(t)• Consider circular motion

projection on one axis gives oscillatory motion in 1d-like mass on a spring?

x = Acosθ ax = −ac cosθ

ax = − acAx SHM

Position Acceleration

}

A projection of circular motion is an example of SHM

A projection of circular motion is an example of SHM

x = Acosθω = Δθ

Δt⇒θ =ωt

x = Acosωt

By analogy, for a masson a spring:

x = Acosωt

x

Equilibrium

Time dependence

4) Interpretation of A and ω in (a) Amplitude, A

x = Acosωt

Maximum displacement (when cos ωt = 1)

x

A

x = A when t = 0arbitrary phase selection

(b) Frequency

From circular motion, ω = θt

= 2πT

= 2π f

ω = 2π f ω = 2πT

frequency of motion (cycles per second)

period of motion(time for one cycle)

(c) Frequency, mass, and spring constant

x

Equilibrium

ax = − acAx a = −k

mx

ac =v2

A=

ωA( )2A

=ω 2A

ω = acA

ω = km

SummaryIf a force F = -kx acts on mass m, the resulting motion can be described by

x = Acosωt

Amplitudex = A at t = 0

Angular frequency ω = 2π f = 2πT

given by ω = km

5) Position, velocity, acceleration in SHM

• Position (displacement)

x = Acosωt

period T: the time required to complete one cycle

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

amplitude A: the maximum displacement

• Velocity

v = −Aω sinωt}

vmaxvx = −vT sinθ = −Aω sinωt

• Acceleration

a = −Aω 2 cosωt

ax = −ac cosθ = −Aω 2 cosωt}amax

Summary: displacement, velocity and acceleration in SHM

x = Acosωt

v = −Aω sinωt

a = −Aω 2 cosωt

Amplitude, Ax = A at t = 0

ω = 2π f = 2πT

Angular freq.

given by

ω = k m

6) PhaseIf starting position (t = 0) is defined for some value of x other than x = A

x = Acos(ωt +φ)

phase anglex

3π/2

For φ = -π/2 (or 3π/2)

x = Asin(ωt)

(still SHM)

A 0.80 kg object is attached to one end of a spring and the system is set into SHM. Shown is the displacement of the object. Determine (a) the amplitude A of the motion, (b) the angular frequency ω, (c) the spring constant k, (d) the speed of the object at t = 1.0 s and (e) the magnitude of the acceleration at t = 1.0 s.

Example

A spring at any angle:

Fnet = −kx if x measured from equilibrium

Therefore, SHM about new equilibrium

7) Energy in SHM(a) Elastic potential energy

A compressed spring can do work

Elastic potential energy is the external work to compress a spring by distance x:

Recall gravitational potential energy is work required to lift an object a distance h:

PEgravity = mgh

Wext = Fext x Fext = kxbut

⇒Wext = 12 kx

2

PEelastic = 12 kx

2

Fext = 12 kxso

ΔPEelastic = 12 kx f

2 − 12 kx0

2

(b) Total mechanical energy (incl gravity and elastic PE)

E = KE + PE

E = 12 mv

2 +mgh + 12 kx

2

}if there are no dissipative forcesE = const

(c) Distribution of energy in SHM

PE→ KE→ PE→ KE→…E = 1

2 mv2 + 1

2 kx2

E = 12 kA

2x = A;v = 0x = 0;v = vmax

E = 12 mvmax

2

⇒ vmax = A k m = Aω

(as before)

12 mv

2max = 1

2 kA2

ExampleWhat are PE and KE at x = A/2

A 0.20-kg ball is attached to a vertical spring. The spring constantis 28 N/m. When released from rest, how far does the ball fallbefore being brought to a momentary stop by the spring?

Example

8) Simple pendulum

Force (or torque) is restoring, therefore oscillation

F = −mgsinθ

For small angles, sinθ ≈θ = sL

F = − mgLs (like F = −kx)

a = − gLs (using F = ma)

⇒ SHM (almost)

θ = θmax cosωt

x

Equilibrium

a = −kmx

ω = km

a = − gLs

ω = gL

α = − gLθ

• Independent of amplitude (isochronous)• Independent of mass• (for small angles)

f = ω2π

=g L2π

Frequency

Period T = 1f= 2π L g

ω = gL

ω = 2π f

ExampleA “seconds” pendulum is typically used in a grandfather’s clock, and has a period of 2 s (one second for each swing). What is the length of a simple seconds pendulum?

The length of a simple pendulum is 0.79 m and the mass of the particle (the “bob”) at the end of the cable is 0.24 kg. The pendulum is pulled away from its equilibrium position by an angle of 8.50° and released from rest. Assume that friction can be neglected and that the resulting oscillatory motion is simple harmonic motion. (a)  What is the angular frequency of the motion? (b) Using the position of the bob at its lowest point as the reference level, determine the total mechanical energy of the pendulum as it swings back and forth. (c) What is the bob's speed as it passes through the lowest point of the swing?

Example

A spring is hung from the ceiling. A 0.450-kg block is then attached to the free end of the spring. When released from rest, the block drops 0.150 m before momentarily coming to rest, after which it moves back upward.(a) What is the spring constant of the spring? (b) Find the angular frequency of the block's vibrations.

Example

A person bounces up and down on a trampoline, while always staying in contact with it. The motion is simple harmonic, and it takes 1.90 s to complete one cycle. The height of each bounce above the equilibrium position is 45.0 cm. Determine (a) the amplitude and (b) the angular frequency of the motion. (c) What is the maximum speed attained by the person?

Example

A spring is compressed by 0.0620 m and is used to launch an object horizontally with a speed of 1.50 m/s. If the object were attached to the spring, at what angular frequency (in rad/s) would it oscillate?

Homework