Physics 6B

Oscillations

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For Campus Learning Assistance Services at UCSB

Definitions of quantities describing periodic motion

• Period (T): time required for a motion to go through a complete cycle

• Frequency (f): number of oscillations per unit timeStandard unit for frequency is Hertz (Hz)

1 Hz = 1 cycle/second

f1

T T1

f

• Angular frequency:

Period and frequency are reciprocals

• The amplitude (A) is the maximum displacement from equilibrium.

Simple Harmonic Motion

A spring exerts a restoring force that is proportional to the displacement from equilibrium:

• Period of a mass on a spring:

• Total energy in simple harmonic motion:

• Simple harmonic motion occurs when the restoring force is proportional to the displacement from equilibrium.

Simple Harmonic Motion

Equations for Simple Harmonic Motion

• Position as a function of time:

• Velocity as a function of time:

• Acceleration as a function of time:

• Note that v(t) is just the derivative of x(t), and a(t) is the derivative of v(t)

Energy in Simple Harmonic Motion

• Potential energy as a function of time:

• Kinetic energy as a function of time:

½ kA2 _

Example 1A 0.98 kg block slides on a frictionless horizontal surface with speed 1.32m/s. The block encounters an unstretched spring with force constant 245N/m, as shown.

a) how far is the spring compressed before the block comes to rest?b) how long is the block in contact with the spring before it comes to rest?c) how far is the spring compressed when the kinetic energy of the block is equal to the potential energy stored in the spring?

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

Example 1A 0.98 kg block slides on a frictionless horizontal surface with speed 1.32m/s. The block encounters an unstretched spring with force constant 245N/m, as shown.

a) how far is the spring compressed before the block comes to rest?b) how long is the block in contact with the spring before it comes to rest?c) how far is the spring compressed when the kinetic energy of the block is equal to the potential energy stored in the spring?

We can use energy conservation for the first part, setting the initial kinetic energy of the block equal to the final potential energy stored in the spring.

Block at rest (spring fully compressed)

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

Example 1A 0.98 kg block slides on a frictionless horizontal surface with speed 1.32m/s. The block encounters an unstretched spring with force constant 245N/m, as shown.

a) how far is the spring compressed before the block comes to rest?b) how long is the block in contact with the spring before it comes to rest?c) how far is the spring compressed when the kinetic energy of the block is equal to the potential energy stored in the spring?

We can use energy conservation for the first part, setting the initial kinetic energy of the block equal to the final potential energy stored in the spring.

Δx

2212

21 xkmv

Block at rest (spring fully compressed)

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

Example 1A 0.98 kg block slides on a frictionless horizontal surface with speed 1.32m/s. The block encounters an unstretched spring with force constant 245N/m, as shown.

a) how far is the spring compressed before the block comes to rest?b) how long is the block in contact with the spring before it comes to rest?c) how far is the spring compressed when the kinetic energy of the block is equal to the potential energy stored in the spring?

We can use energy conservation for the first part, setting the initial kinetic energy of the block equal to the final potential energy stored in the spring.

Δx

vk

mx

k

mvx

xkmv

22

2212

21

Block at rest (spring fully compressed)

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

Example 1A 0.98 kg block slides on a frictionless horizontal surface with speed 1.32m/s. The block encounters an unstretched spring with force constant 245N/m, as shown.

a) how far is the spring compressed before the block comes to rest?b) how long is the block in contact with the spring before it comes to rest?c) how far is the spring compressed when the kinetic energy of the block is equal to the potential energy stored in the spring?

Block at rest (spring fully compressed)

We can use energy conservation for the first part, setting the initial kinetic energy of the block equal to the final potential energy stored in the spring.

cm3.8m083.032.1245

kg98.0x

vk

mx

k

mvx

xkmv

sm

mN

22

2212

21

Δx

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

Example 1A 0.98 kg block slides on a frictionless horizontal surface with speed 1.32m/s. The block encounters an unstretched spring with force constant 245N/m, as shown.

a) how far is the spring compressed before the block comes to rest?b) how long is the block in contact with the spring before it comes to rest?c) how far is the spring compressed when the kinetic energy of the block is equal to the potential energy stored in the spring?

For part b) we can use the formula for the period of oscillation of a mass-on-a-spring:

Δx

v=1.32

v=1.32

k

m2T

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

Block at rest (spring fully compressed)

Example 1A 0.98 kg block slides on a frictionless horizontal surface with speed 1.32m/s. The block encounters an unstretched spring with force constant 245N/m, as shown.

a) how far is the spring compressed before the block comes to rest?b) how long is the block in contact with the spring before it comes to rest?c) how far is the spring compressed when the kinetic energy of the block is equal to the potential energy stored in the spring?

For part b) we can use the formula for the period of oscillation of a mass-on-a-spring:

Δx

v=1.32

v=1.32

k

m2T

In this case we only want ¼ of the period.

sec1.0T

sec4.0245

kg98.02T

41

mN

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

Block at rest (spring fully compressed)

Example 1A 0.98 kg block slides on a frictionless horizontal surface with speed 1.32m/s. The block encounters an unstretched spring with force constant 245N/m, as shown.

a) how far is the spring compressed before the block comes to rest?b) how long is the block in contact with the spring before it comes to rest?c) how far is the spring compressed when the kinetic energy of the block is equal to the potential energy stored in the spring?

Part c) is easiest to understand using energy.

We know that Etotal = Kinetic + Potential.

Δx

v=1.32

v=1.32

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

Block at rest (spring fully compressed)

Example 1A 0.98 kg block slides on a frictionless horizontal surface with speed 1.32m/s. The block encounters an unstretched spring with force constant 245N/m, as shown.

a) how far is the spring compressed before the block comes to rest?b) how long is the block in contact with the spring before it comes to rest?c) how far is the spring compressed when the kinetic energy of the block is equal to the potential energy stored in the spring?

Part c) is easiest to understand using energy.

We know that Etotal = Kinetic + Potential.

We can also calculate the total energy from the given initial speed:

Δx

v=1.32

v=1.32

J85.032.1kg98.0mvE2

sm

212

021

total

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

Block at rest (spring fully compressed)

Example 1A 0.98 kg block slides on a frictionless horizontal surface with speed 1.32m/s. The block encounters an unstretched spring with force constant 245N/m, as shown.

a) how far is the spring compressed before the block comes to rest?b) how long is the block in contact with the spring before it comes to rest?c) how far is the spring compressed when the kinetic energy of the block is equal to the potential energy stored in the spring?

Part c) is easiest to understand using energy.

We know that Etotal = Kinetic + Potential.

We can also calculate the total energy from the given initial speed:

Δx

v=1.32

v=1.32

J85.032.1kg98.0mvE2

sm

212

021

total

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

Now we have to realize that when the kinetic and potential energies are equal, they are also each equal to half of the total energy.

Block at rest (spring fully compressed)

Example 1A 0.98 kg block slides on a frictionless horizontal surface with speed 1.32m/s. The block encounters an unstretched spring with force constant 245N/m, as shown.

a) how far is the spring compressed before the block comes to rest?b) how long is the block in contact with the spring before it comes to rest?c) how far is the spring compressed when the kinetic energy of the block is equal to the potential energy stored in the spring?

Part c) is easiest to understand using energy.

We know that Etotal = Kinetic + Potential.

We can also calculate the total energy from the given initial speed:

Δx

v=1.32

v=1.32

J85.032.1kg98.0mvE2

sm

212

021

total

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

Now we have to realize that when the kinetic and potential energies are equal, they are also each equal to half of the total energy.

Since we want to find the compression distance, we should use the formula involving potential energy:

Block at rest (spring fully compressed)

Example 1A 0.98 kg block slides on a frictionless horizontal surface with speed 1.32m/s. The block encounters an unstretched spring with force constant 245N/m, as shown.

a) how far is the spring compressed before the block comes to rest?b) how long is the block in contact with the spring before it comes to rest?c) how far is the spring compressed when the kinetic energy of the block is equal to the potential energy stored in the spring?

Part c) is easiest to understand using energy.

We know that Etotal = Kinetic + Potential.

We can also calculate the total energy from the given initial speed:

Δx

v=1.32

v=1.32

J85.032.1kg98.0mvE2

sm

212

021

total

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

Now we can realize that when the kinetic and potential energies are equal, they are also each equal to half of the total energy.

Since we want to find the compression distance, we should use the formula involving potential energy:

cm6m06.0245

J85.0k

Ex

kxUE

mN

total

221

springtotal21

Block at rest (spring fully compressed)

Mass-Spring Example

Stop the Block

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Example 2A 0.260kg mass is attached to a vertical spring. When the mass is put into motion, its period is 1.12sec. How much does the mass stretch the spring when it is at rest in its equilibrium position?

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

?

Example 2A 0.260kg mass is attached to a vertical spring. When the mass is put into motion, its period is 1.12sec. How much does the mass stretch the spring when it is at rest in its equilibrium position?

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

?

mg

kx

Example 2A 0.260kg mass is attached to a vertical spring. When the mass is put into motion, its period is 1.12sec. How much does the mass stretch the spring when it is at rest in its equilibrium position?

If we consider all the forces acting on the mass when it is hanging at rest we see that the weight must cancel the spring force.

So if we can find the spring constant k, we can solve for x.

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

?

mg

kx

If we consider all the forces acting on the mass when it is hanging at rest we see that the weight must cancel the spring force.

So if we can find the spring constant k, we can solve for x.

Use the formula for the period of a mass-spring system:

22

T

2mk

k

m

2

T

k

m2T

Example 2A 0.260kg mass is attached to a vertical spring. When the mass is put into motion, its period is 1.12sec. How much does the mass stretch the spring when it is at rest in its equilibrium position?

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

?

mg

kx22

T

2mk

k

m

2

T

k

m2T

Plugging in the given values we get mN18.8k

Example 2A 0.260kg mass is attached to a vertical spring. When the mass is put into motion, its period is 1.12sec. How much does the mass stretch the spring when it is at rest in its equilibrium position?

If we consider all the forces acting on the mass when it is hanging at rest we see that the weight must cancel the spring force.

So if we can find the spring constant k, we can solve for x.

Use the formula for the period of a mass-spring system:

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

?

mg

kx22

T

2mk

k

m

2

T

k

m2T

Plugging in the given values we get mN18.8k

Example 2A 0.260kg mass is attached to a vertical spring. When the mass is put into motion, its period is 1.12sec. How much does the mass stretch the spring when it is at rest in its equilibrium position?

If we consider all the forces acting on the mass when it is hanging at rest we see that the weight must cancel the spring force.

So if we can find the spring constant k, we can solve for x.

Use the formula for the period of a mass-spring system:

Now we can use Fspring=weight:

k

mgxmgkx

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

?

mg

kx22

T

2mk

k

m

2

T

k

m2T

Plugging in the given values we get mN18.8k

Example 2A 0.260kg mass is attached to a vertical spring. When the mass is put into motion, its period is 1.12sec. How much does the mass stretch the spring when it is at rest in its equilibrium position?

If we consider all the forces acting on the mass when it is hanging at rest we see that the weight must cancel the spring force.

So if we can find the spring constant k, we can solve for x.

Use the formula for the period of a mass-spring system:

Now we can use Fspring=weight:

cm31m31.0

18.8

8.9kg26.0x

k

mgxmgkx

mN

sm2

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

The Simple Pendulum

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

Period of a Pendulum

• Period of a simple pendulum:

• Period of a physical pendulum:

• A simple pendulum with small amplitude exhibits simple harmonic motion

You want to construct a simple pendulum with a 3.25kg bob that will take 1.13sec to swing from a maximum angle of 9.5° to its lowest point after being released from rest. How long should this pendulum be?

θHere is a diagram of the pendulum.

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For Campus Learning Assistance Services at UCSB

You want to construct a simple pendulum with a 3.25kg bob that will take 1.13sec to swing from a maximum angle of 9.5° to its lowest point after being released from rest. How long should this pendulum be?

θHere is a diagram of the pendulum. We have a formula for the period of this pendulum:

g

L2T

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

You want to construct a simple pendulum with a 3.25kg bob that will take 1.13sec to swing from a maximum angle of 9.5° to its lowest point after being released from rest. How long should this pendulum be?

θHere is a diagram of the pendulum. We have a formula for the period of this pendulum:

g

L2T

We can solve this for the length:22

2

TgL

2

T

g

L

g

L2T

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

You want to construct a simple pendulum with a 3.25kg bob that will take 1.13sec to swing from a maximum angle of 9.5° to its lowest point after being released from rest. How long should this pendulum be?

θHere is a diagram of the pendulum. We have a formula for the period of this pendulum:

g

L2T

We can solve this for the length:22

2

TgL

2

T

g

L

g

L2T

What value should we use for the period?

We are given a time of 1.13s to go from max angle to the lowest point.

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

You want to construct a simple pendulum with a 3.25kg bob that will take 1.13sec to swing from a maximum angle of 9.5° to its lowest point after being released from rest. How long should this pendulum be?

θHere is a diagram of the pendulum. We have a formula for the period of this pendulum:

g

L2T

We can solve this for the length:22

2

TgL

2

T

g

L

g

L2T

What value should we use for the period?

We are given a time of 1.13s to go from max angle to the lowest point.

This is only ¼ of a full cycle.

So we multiply by 4: T = 4.52s

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

You want to construct a simple pendulum with a 3.25kg bob that will take 1.13sec to swing from a maximum angle of 9.5° to its lowest point after being released from rest. How long should this pendulum be?

θHere is a diagram of the pendulum. We have a formula for the period of this pendulum:

g

L2T

We can solve this for the length:22

2

TgL

2

T

g

L

g

L2T

What value should we use for the period?

We are given a time of 1.13s to go from max angle to the lowest point.

This is only ¼ of a full cycle.

So we multiply by 4: T = 4.52s

Now we can plug in to get our answer:

m07.52

s52.48.9L

2

sm2

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For Campus Learning Assistance Services at UCSB

Damped Oscillations

• Oscillations where there is a nonconservative (i.e. friction) force are called damped.

• Underdamped: the amplitude decreases exponentially with time:

• Critically damped: no oscillations; system relaxes back to equilibrium in minimum time

• Overdamped: also no oscillations, but slower than critical damping

• The frequency of oscillation is also affected.

 

Damped Oscillations

Driven Oscillations

• An oscillating system may be driven by an external force

• This force may replace energy lost to friction, or may cause the amplitude to increase greatly at resonance

• Resonance occurs when the driving frequency is equal to the natural frequency of the system

Driven Oscillations

• The Amplitude of the oscillation depends on the driving frequency.

𝐴=𝐹 0

√𝑚2 (𝜔02−𝜔2 )2+𝑏2𝜔2