physics 6b
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Physics 6B. Oscillations. Prepared by Vince Zaccone 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 time - PowerPoint PPT PresentationTRANSCRIPT
Physics 6B
Oscillations
Prepared by Vince ZacconeFor 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
f1T T
1f
• Angular frequency:
Period and frequency are reciprocals
• The amplitude (A) is the maximum displacement from equilibrium.
Simple Harmonic MotionA 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 ZacconeFor 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 ZacconeFor 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 ZacconeFor 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
vkmx
kmvx
xkmv
22
2212
21
Block at rest (spring fully compressed)
Prepared by Vince ZacconeFor 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
vkmx
kmvx
xkmv
sm
mN
22
2212
21
Δx
Prepared by Vince ZacconeFor 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
km2T
Prepared by Vince ZacconeFor 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
km2T
In this case we only want ¼ of the period.
sec1.0T
sec4.0245
kg98.02T
41
mN
Prepared by Vince ZacconeFor 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 ZacconeFor 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 ZacconeFor 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 ZacconeFor 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 ZacconeFor 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 ZacconeFor 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
221springtotal2
1
Block at rest (spring fully compressed)
Mass-Spring Example
Stop the Block
Prepared by Vince ZacconeFor 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 ZacconeFor 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 ZacconeFor 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 ZacconeFor 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
T2mk
km
2T
km2T
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 ZacconeFor Campus Learning Assistance Services at UCSB
?
mg
kx22
T2mk
km
2T
km2T
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 ZacconeFor Campus Learning Assistance Services at UCSB
?
mg
kx22
T2mk
km
2T
km2T
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:
kmgxmgkx
Prepared by Vince ZacconeFor Campus Learning Assistance Services at UCSB
?
mg
kx22
T2mk
km
2T
km2T
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
kmgxmgkx
mN
sm2
Prepared by Vince ZacconeFor 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.
Prepared by Vince ZacconeFor 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:
gL2T
Prepared by Vince ZacconeFor 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:
gL2T
We can solve this for the length:22
2TgL
2T
gL
gL2T
Prepared by Vince ZacconeFor 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:
gL2T
We can solve this for the length:22
2TgL
2T
gL
gL2T
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 ZacconeFor 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:
gL2T
We can solve this for the length:22
2TgL
2T
gL
gL2T
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 ZacconeFor 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:
gL2T
We can solve this for the length:22
2TgL
2T
gL
gL2T
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
s52.48.9L2
sm2
Prepared by Vince ZacconeFor 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