move to the left until it reaches x=−A and then begin to move to the right. 
After the block is released from x=A, it will remain at rest. move to the left until it reaches equilibrium and stop there. move to the left until it reaches x=−A and stop there. move to the left until it reaches x=−A and then begin to move to the right. 
halved. 
If the period is doubled, the frequency is unchanged. doubled. halved. 
f = 10 Hz 
An oscillating object takes 0.10 s to complete one cycle; that is, its period is 0.10 s. What is its frequency f? Express your answer in hertz. 
T = 0.025 s 
If the frequency is 40 Hz, what is the period T ? Express your answer in seconds. 
both R and Q 
Which points on the x axis are located a distance A from the equilibrium position? R only Q only both R and Q 
K and P 
Suppose that the period is T. Which of the following points on the t axis are separated by the time interval T? K and L K and M K and P L and N M and P 
T = 0.02s 
Now assume for the remaining Parts G – J, that the x coordinate of point R is 0.12 m and the t coordinate of point K is 0.0050 s. What is the period T ? 
t = 0.01 s 
Now assume for the remaining Parts G – J, that the x coordinate of point R is 0.12 m and the t coordinate of point K is 0.0050 s. How much time t does the block take to travel from the point of maximum displacement to the opposite point of maximum displacement? Express your answer in seconds. 
d = 0.48 m 
Now assume for the remaining Parts G – J, that the x coordinate of point R is 0.12 m and the t coordinate of point K is 0.0050 s. What distance d does the object cover during one period of oscillation? 
d = 0.36 m 
Now assume for the remaining Parts G – J, that the x coordinate of point R is 0.12 m and the t coordinate of point K is 0.0050 s. What distance d does the object cover between the moments labeled K and N on the graph? Express your answer in meters. 
f = 1.25×10^14 Hz 
Calculate the oscillation frequency f of the H2 molecule. Use meff=m/2 as the "effective mass" of the system, where m in the mass of a hydrogen atom. Take the mass of a hydrogen atom as 1.008 u, where 1u=1.661×10^−27kg. Express your answer in hertz. 
Based on length: 

H 
Beginning the instant the pendulum is released, select the graph that best matches the angular position vs. time graph for the pendulum. 
E 
Beginning the instant the pendulum is released, select the graph that best matches the angular velocity vs. time graph for the pendulum 
F 
Beginning the instant the pendulum is released, select the graph that best matches the angular acceleration vs. time graph for the pendulum. 
The spring stretches more for a heavier weight. 
Place a 50 g weight on spring #1, and release it. Eventually, the weight will come to rest at an equilibrium position, with the spring somewhat stretched compared to its original (unweighted) length. At this point, the upward force of the spring balances the force of gravity on the weight. With the weight in its equilibrium position, how does the amount the spring is stretched depend on the mass of the weight? The stretch does not depend on mass. The spring stretches more for a heavier weight. The spring stretches less for a heavier weight. 
Both when the spring is most compressed and when the spring is most stretched 
When is the elastic potential energy of the spring a maximum? When the spring is most stretched When the spring isn’t stretched or compressed When the spring is most compressed Both when the spring is most compressed and when the spring is most stretched 
When the spring is at its unweighted length (when it isn’t stretched or compressed) 
When is the kinetic energy of the mass a maximum? When the spring is at its unweighted length (when it isn’t stretched or compressed) When the spring is most stretched When the spring is most compressed Both when the spring is most compressed and when the spring is most stretched 
When the mass is at the equilibrium position 
Select Earth in the menu box so that there is now a force of gravity. Now the total energy of the mass/spring system is the sum of the kinetic energy, the elastic potential energy, and the gravitational potential energy. When is the kinetic energy a maximum? (It may help to watch the simulation in slow motion – 1/16 time.) When the spring is at its unweighted length (its equilibrium position without the weight attached) When the spring is most stretched or most compressed When the mass is at the equilibrium position 
The frequency decreases as the mass increases. 
How does the frequency of oscillation depend on the mass of the weight? The frequency decreases as the mass increases. The frequency is independent of the mass. The frequency increases as the mass increases. 
The frequency is independent of the amplitude. 
How does the frequency depend on the amplitude of oscillation? The frequency increases as the amplitude increases. The frequency is independent of the amplitude. The frequency decreases as the amplitude increases. 
The frequency increases as the spring constant increases. 
How does the frequency of oscillation depend on the spring constant? The frequency decreases as the spring constant increases. The frequency is independent of the spring constant. The frequency increases as the spring constant increases. 
Potential energy 
Drag the pendulum to an angle (with respect to the vertical) of 30∘, and then release it. When the pendulum is at−30∘, what form(s) of energy does it have? Check all that apply. Kinetic energy Potential energy Thermal energy 
at 0∘ 
Drag the pendulum to an angle (with respect to the vertical) of 30∘, and then release it. Where is the pendulum swinging the fastest? at−30∘ at 15∘ at 0∘ at 30∘ 
The acceleration is never equal to zero as it swings back and forth. 
Drag the pendulum to an angle (with respect to the vertical) of 30∘, and then release it. Select to show the acceleration vector. With the pendulum swinging back and forth, at which locations is the acceleration equal to zero? The acceleration is zero when the angle is 0∘. The acceleration is zero when the angle is either+30∘ or−30∘. The acceleration is never equal to zero as it swings back and forth. 
The tension is greater than the force of gravity. 
With the pendulum swinging back and forth, how does the tension of the rope compare to the force of gravity when the angle is 0∘? The tension is less than the force of gravity. The tension is greater than the force of gravity only if it is swinging really fast. The tension is greater than the force of gravity. The tension is equal to the force of gravity. 
The tension is zero at the angles+90∘ and−90∘. 
Drag the pendulum to an angle (with respect to the vertical) of 90∘, and then release it. With the pendulum swinging back and forth, where is the tension equal to zero? The tension is zero at the angles+90∘ and−90∘. The tension is zero when the angle is +45∘ and −45∘. The tension is zero when the angle is 0∘. The tension is never zero. 
2.0 s 
Set the length of the pendulum to 1.0 m and the mass to 1.0 kg. Click Reset, and then drag the pendulum to an angle (with respect to the vertical) of 30∘ and release it. What is the period of oscillation? 1.5 s 4.0 s 20.0 s 1.0 s 0.5 s 2.0 s 
The period is longer when the initial angle is greater. 
How does the period of oscillation depend on the initial angle of the pendulum when released? (Be sure to measure the period for initial angles much greater than 30∘.) The period is longer when the initial angle is greater. The period is shorter when the initial angle is greater. The period is independent of the initial angle. 
The period is independent of the pendulum’s mass. 
How does the period of the pendulum depend on mass? A heavier pendulum has a shorter period. The period is independent of the pendulum’s mass. A heavier pendulum has a longer period. 
A longer pendulum has a longer period. 
Now, keep the mass fixed to any value you choose and measure the period for several different pendulum lengths. How does the period of the pendulum depend on the length? A longer pendulum has a longer period. A longer pendulum has a shorter period. The period is independent of the length of the pendulu 
The period of oscillation is shorter on planets with a higher value of g. 
How does the period of oscillation depend on the value of g? The period of oscillation is longer on planets with a higher value of g. The period of oscillation is independent of the value of g. The period of oscillation is shorter on planets with a higher value of g. 
(kA^2)/8 
What is the system’s potential energy when its kinetic energy is equal to 3/4E? kA^2 ( kA^2)/2 ( kA^2)/4 (kA^2)/8 
±√k/m*(A/√3) 
What is the object’s velocity when its potential energy is 2/3E? ±√(k/m)A ±√2/3√(k/m)A ±√k/m*(A/√3) ±√k/m*(A/√6) 
A 
Which moment corresponds to the maximum potential energy of the system? A B C D 
A 
Which moment corresponds to the minimum kinetic energy of the system? A B C D 
moving toward equilibrium. 
Consider the block in the process of oscillating. If the kinetic energy of the block is increasing, the block must be at the equilibrium position. at the amplitude displacement. moving to the right. moving to the left. moving away from equilibrium. moving toward equilibrium. 
C 
Which moment corresponds to the maximum kinetic energy of the system? A B C D 
C 
Which moment corresponds to the minimum potential energy of the system? A B C D 
D 
At which moment is K=U? A B C D 
K = 3/8kA^2 
Find the kinetic energy K of the block at the moment labeled B. Express your answer in terms of k and A. 
Based on LoL L0=20 cm L=5 > L0=15 L=5 = L0=20 L=10 > L0=10 L=5 = L0=15 L=10 = L0=10 L=5 
Different mass crates are placed on top of springs of uncompressed length L0 and stiffness k. (Figure 1) The crates are released and the springs compress to a length L before bringing the crates back up to their original positions. 
It decreases the amplitude. 
At the instant you pass through the equilibrium position of the spring, you drop the sandbag out of the cart onto the ground.What effect does dropping the sandbag out of the cart at the equilibrium position have on the amplitude of your oscillation? It increases the amplitude. It decreases the amplitude. It has no effect on the amplitude. 
It has no effect on the maximum speed 
At the instant you pass through the equilibrium position of the spring, you drop the sandbag out of the cart onto the ground. What effect does dropping the sandbag out of the cart at the equilibrium position have on the maximum speed of the cart? It increases the maximum speed. It decreases the maximum speed. It has no effect on the maximum speed. 
It has no effect on the amplitude. 
Instead of dropping the sandbag as you pass through equilibrium, you decide to drop the sandbag when the cart is at its maximum distance from equilibrium. What effect does dropping the sandbag at the cart’s maximum distance from equilibrium have on the amplitude of your oscillation? It increases the amplitude. It decreases the amplitude. It has no effect on the amplitude. 
It increases the maximum speed. 
Instead of dropping the sandbag as you pass through equilibrium, you decide to drop the sandbag when the cart is at its maximum distance from equilibrium. What effect does dropping the sandbag at the cart’s maximum distance from equilibrium have on the maximum speed of the cart? It increases the maximum speed. It decreases the maximum speed. It has no effect on the maximum speed. 
Mastering Physics 11
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