Transcript Example 1 - UCSB Campus Learning Assistance Services

Physics 6B
Oscillations
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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 time
Standard unit for frequency is Hertz (Hz)
1 Hz = 1 cycle/second
1
T
f
1
f
T
Period and frequency are reciprocals
•Angular frequency:
•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:
Simple Harmonic Motion
• Simple harmonic motion occurs when the
restoring force is proportional to the
displacement from equilibrium.
• Period of a mass on a spring:
• Total energy in 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 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)
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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
2
 1 k x 
2
Δ
x
2
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
x 
2
2
 1 k x 
2
Δ
x
2
mv 2
m

 x 
v
k
k
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
x 
2
2
x 
 1 k x 
2
Δ
x
2
mv 2
m

 x 
v
k
k


0.98kg
 1.32 m  0.083m  8.3cm
N
s
245
Block at rest (spring fully compressed)
m
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  2
m
k
Δx
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  2
m
k
Δx
In this case we only want ¼ of the period.
T  2
0.98kg
 0.4 sec
245 N
m
1 T  0.1sec
4
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.
Δx
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:

Δx

2
Etotal  1 mv02  1 0.98kg 1.32 m  0.85J
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:

Δx

2
Etotal  1 mv02  1 0.98kg 1.32 m  0.85J
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:

Δx

2
Etotal  1 mv02  1 0.98kg 1.32 m  0.85J
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:
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:

Δx

2
Etotal  1 mv02  1 0.98kg 1.32 m  0.85J
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:
1E
 Uspring  1 kx2
2 total
2
x
Etotal

k
0.85J
245 N
 0.06m  6cm
m
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Mass-Spring Example
Stop the Block
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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
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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?
?
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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
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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:
2
m
m
T
 2 
T  2
     k  m 
k
k
 2 
T
2
?
kx
mg
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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:
2
m
m
T
 2 
T  2
     k  m 
k
k
 2 
T
2
N
Plugging in the given values we get k  8.18 m
?
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:
2
m
m
T
 2 
T  2
     k  m 
k
k
 2 
T
2
N
Plugging in the given values we get k  8.18 m
?
kx
mg
Now we can use Fspring=weight:
kx  mg  x 
mg
k
Prepared by Vince Zaccone
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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:
2
m
m
T
 2 
T  2
     k  m 
k
k
 2 
T
2
N
Plugging in the given values we get k  8.18 m
?
kx
mg
Now we can use Fspring=weight:
kx  mg  x 
x
mg
k
0.26kg 9.8 m2 
s 

 0.31m  31cm
8.18 N
m
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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
• A simple pendulum with small amplitude
exhibits simple harmonic motion
• Period of a simple pendulum:
• Period of a physical pendulum:
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?
Here is a diagram of the pendulum.
θ
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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?
Here is a diagram of the pendulum. We have a
formula for the period of this pendulum:
T  2
θ
L
g
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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?
Here is a diagram of the pendulum. We have a
formula for the period of this pendulum:
T  2
θ
L
g
We can solve this for the length:
2
L
L T 
 T 
T  2
     L  g 
g
g  2 
 2 
2
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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?
Here is a diagram of the pendulum. We have a
formula for the period of this pendulum:
T  2
θ
L
g
We can solve this for the length:
2
L
L T 
 T 
T  2
     L  g 
g
g  2 
 2 
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.
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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?
Here is a diagram of the pendulum. We have a
formula for the period of this pendulum:
T  2
θ
L
g
We can solve this for the length:
2
L
L T 
 T 
T  2
     L  g 
g
g  2 
 2 
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 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?
Here is a diagram of the pendulum. We have a
formula for the period of this pendulum:
T  2
θ
L
g
We can solve this for the length:
2
L
L T 
 T 
T  2
     L  g 
g
g  2 
 2 
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 ¼ of a full cycle.
So we multiply by 4: T = 4.52s
Now we can plug in to get our answer:
2
 4.52s 
L   9.8 m2 
  5.07m
s  2 

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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
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