Physics 6B

Oscillations Examples

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?

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.

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.

Δx 2212

21 xkmv

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.

Δx

vk

mx

k

mvx

xkmv

22

2212

21

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?

Block at rest (spring fully compressed)

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

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)

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

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)

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

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)

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

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)

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

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.

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)

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

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:

221

total21 kxE

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)

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

Now we have to realize that since 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

J85.0

k

Ex

kxE

mN

total

221

total21

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

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?

?

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.

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?

?

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

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?

?

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

Plugging in the given values we get

mN18.8k

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?

?

mg

kx22

T

2mk

k

m

2

T

k

m2T

Plugging in the given values we get

mN18.8k

Now we can use Fspring=weight:

k

mgxmgkx

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

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?

?

mg

kx22

T

2mk

k

m

2

T

k

m2T

Plugging in the given values we get

mN18.8k

Now we can use Fspring=weight:

cm31m31.0

18.8

8.9kg26.0x

k

mgxmgkx

mN

sm2

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

11.46 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?

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

11.46 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.

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

11.46 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

11.46 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:

We can solve this for the length:22

2

TgL

2

T

g

L

g

L2T

g

L2T

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

11.46 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:

We can solve this for the length:

What value should we use for the period?

22

2

TgL

2

T

g

L

g

L2T

g

L2T

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

11.46 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:

We can solve this for the length:

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

22

2

TgL

2

T

g

L

g

L2T

g

L2T

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

11.46 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

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB