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```Physics 6B
Oscillations Examples
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Example 1
A 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 1
A 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 1
A 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.
1 mv 2 пЂЅ 1 k пЂЁ пЃ„ x пЂ©2
2
2
О”x
Block at rest (spring fully compressed)
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Example 1
A 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.
О”x
1 mv 2 пЂЅ 1 k пЂЁ пЃ„ x пЂ©2
2
2
2
пЂЅ
mv
k
2
пѓћ пЃ„x пЂЅ
m
k
пѓ—v
Block at rest (spring fully compressed)
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Example 1
A 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.
О”x
2
2
1
mv пЂЅ 1 k пЂЁ пЃ„ x пЂ©
2
2
2
пЃ„x пЂЅ
пЂЅ
mv
2
k
0 . 98 kg
245 N
m
m
пѓћ пЃ„x пЂЅ
пЂЁ
k
пѓ—v
пѓ— 1 . 32 m пЂЅ 0 . 083 m пЂЅ 8 . 3 cm
s
Block at rest (spring fully compressed)
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Example 1
A 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?
v=1.32
For part b) we can use the formula for the period of
oscillation of a mass-on-a-spring:
T пЂЅ 2пЃ°
m
О”x
k
Block at rest (spring fully compressed)
v=1.32
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Example 1
A 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?
v=1.32
For part b) we can use the formula for the period of
oscillation of a mass-on-a-spring:
T пЂЅ 2пЃ°
m
О”x
k
In this case we only want Вј of the period.
T пЂЅ 2пЃ°
0 . 98 kg
245 N
m
пЂЅ 0 . 4 sec
Block at rest (spring fully compressed)
v=1.32
1 T пЂЅ 0 . 1 sec
4
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Example 1
A 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?
v=1.32
Part c) is easiest to understand using energy.
We know that Etotal = Kinetic + Potential.
О”x
Block at rest (spring fully compressed)
v=1.32
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Example 1
A 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?
v=1.32
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:
пЂЁ
О”x
2
2
E total пЂЅ 1 mv 0 пЂЅ 1 пЂЁ0 . 98 kg пЂ© 1 . 32 m пЂЅ 0 . 85 J
2
2
s
Block at rest (spring fully compressed)
v=1.32
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Example 1
A 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?
v=1.32
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:
пЂЁ
О”x
2
2
E total пЂЅ 1 mv 0 пЂЅ 1 пЂЁ0 . 98 kg пЂ© 1 . 32 m пЂЅ 0 . 85 J
2
2
s
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)
v=1.32
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Example 1
A 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?
v=1.32
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:
пЂЁ
О”x
2
2
E total пЂЅ 1 mv 0 пЂЅ 1 пЂЁ0 . 98 kg пЂ© 1 . 32 m пЂЅ 0 . 85 J
2
2
s
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)
v=1.32
Since we want to find the compression distance, we
should use the formula involving potential energy:
2
1E
пЂЅ 1 kx
total
2
2
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Example 1
A 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?
v=1.32
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:
пЂЁ
О”x
2
2
E total пЂЅ 1 mv 0 пЂЅ 1 пЂЁ0 . 98 kg пЂ© 1 . 32 m пЂЅ 0 . 85 J
2
2
s
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.
Block at rest (spring fully compressed)
v=1.32
Since we want to find the compression distance, we
should use the formula involving potential energy:
2
1
E total пЂЅ 1 kx
2
2
x пЂЅ
E total
k
пЂЅ
0 . 85 J
245 N
пЂЅ 0 . 06 m пЂЅ 6 cm
m
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Example 2
A 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 2
A 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 2
A 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.
?
kx
mg
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Example 2
A 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:
T пЂЅ 2пЃ°
пѓ¦ T пѓ¶
пѓћ пѓ§
пѓ·
k
пѓЁ 2пЃ° пѓё
m
2
пЂЅ
пѓ¦ 2пЃ° пѓ¶
пѓћ k пЂЅ mпѓ§
пѓ·
k
пѓЁ T пѓё
m
?
kx
2
mg
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Example 2
A 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:
T пЂЅ 2пЃ°
пѓ¦ T пѓ¶
пѓћ пѓ§
пѓ·
k
пѓЁ 2пЃ° пѓё
m
2
пЂЅ
пѓ¦ 2пЃ° пѓ¶
пѓћ k пЂЅ mпѓ§
пѓ·
k
пѓЁ T пѓё
m
?
kx
2
Plugging in the given values we get k пЂЅ 8 . 18
N
m
mg
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Example 2
A 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:
T пЂЅ 2пЃ°
пѓ¦ T пѓ¶
пѓћ пѓ§
пѓ·
k
пѓЁ 2пЃ° пѓё
m
2
пЂЅ
пѓ¦ 2пЃ° пѓ¶
пѓћ k пЂЅ mпѓ§
пѓ·
k
пѓЁ T пѓё
m
?
kx
2
Plugging in the given values we get k пЂЅ 8 . 18
N
m
mg
Now we can use Fspring=weight:
kx пЂЅ mg пѓћ x пЂЅ
mg
k
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Example 2
A 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:
T пЂЅ 2пЃ°
пѓ¦ T пѓ¶
пѓћ пѓ§
пѓ·
k
пѓЁ 2пЃ° пѓё
m
2
пЂЅ
пѓ¦ 2пЃ° пѓ¶
пѓћ k пЂЅ mпѓ§
пѓ·
k
пѓЁ T пѓё
m
?
kx
2
Plugging in the given values we get k пЂЅ 8 . 18
N
m
mg
Now we can use Fspring=weight:
kx пЂЅ mg пѓћ x пЂЅ
mg
k
пЂЁ0 . 26 kg пЂ©пѓ¦пѓ§ 9 . 8
x пЂЅ
пѓЁ
8 . 18 N
m пѓ¶
s
2
пѓ·
пѓё
пЂЅ 0 . 31m пЂЅ 31cm
m
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 9.5В° 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 9.5В° to its lowest point after being released from rest. How long should this pendulum be?
Here is a diagram of the pendulum.
Оё
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 9.5В° to its lowest point after being released from rest. How long should this pendulum be?
Here is a diagram of the pendulum. We have a formula
for the period of this pendulum:
T пЂЅ 2пЃ°
Оё
L
g
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 9.5В° to its lowest point after being released from rest. How long should this pendulum be?
Here is a diagram of the pendulum. We have a formula
for the period of this pendulum:
T пЂЅ 2пЃ°
Оё
L
g
We can solve this for the length:
пѓ¦ T пѓ¶
T пЂЅ 2пЃ°
пѓћ
пЂЅпѓ§
пѓ·
g
g пѓЁ 2пЃ° пѓё
L
L
2
пѓ¦ T пѓ¶
пѓћ L пЂЅ gпѓ§
пѓ·
пѓЁ 2пЃ° пѓё
2
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 9.5В° to its lowest point after being released from rest. How long should this pendulum be?
Here is a diagram of the pendulum. We have a formula
for the period of this pendulum:
T пЂЅ 2пЃ°
Оё
L
g
We can solve this for the length:
пѓ¦ T пѓ¶
T пЂЅ 2пЃ°
пѓћ
пЂЅпѓ§
пѓ·
g
g пѓЁ 2пЃ° пѓё
L
L
2
пѓ¦ T пѓ¶
пѓћ L пЂЅ gпѓ§
пѓ·
пѓЁ 2пЃ° пѓё
2
What value should we use for the period?
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 9.5В° to its lowest point after being released from rest. How long should this pendulum be?
Here is a diagram of the pendulum. We have a formula
for the period of this pendulum:
T пЂЅ 2пЃ°
Оё
L
g
We can solve this for the length:
пѓ¦ T пѓ¶
T пЂЅ 2пЃ°
пѓћ
пЂЅпѓ§
пѓ·
g
g пѓЁ 2пЃ° пѓё
L
L
2
пѓ¦ T пѓ¶
пѓћ L пЂЅ gпѓ§
пѓ·
пѓЁ 2пЃ° пѓё
2
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 Вј of a full cycle.
So we multiply by 4: T = 4.52s
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 9.5В° to its lowest point after being released from rest. How long should this pendulum be?
Here is a diagram of the pendulum. We have a formula
for the period of this pendulum:
T пЂЅ 2пЃ°
Оё
L
g
We can solve this for the length:
пѓ¦ T пѓ¶
T пЂЅ 2пЃ°
пѓћ
пЂЅпѓ§
пѓ·
g
g пѓЁ 2пЃ° пѓё
L
L
2
пѓ¦ T пѓ¶
пѓћ L пЂЅ gпѓ§
пѓ·
пѓЁ 2пЃ° пѓё
2
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 Вј of a full cycle.
So we multiply by 4: T = 4.52s
Now we can plug in to get our answer:
пѓ¦
пѓ¶ пѓ¦ 4 . 52 s пѓ¶
L пЂЅ пѓ§ 9 . 8 m2 пѓ· пѓ§
пѓ·
s пѓёпѓЁ 2 пЃ° пѓё
пѓЁ
2
пЂЅ 5 . 07 m
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
```
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