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14.2 Understanding the motion of a glider on a spring An air-track glider oscillates horizontally on a spring at a frequency of 0.50 Hz. Suppose the glider is pulled to the right of its equilibrium position by 12 cm and then released. Where will the glider be 1.0 s after its release? What is its velocity at this point? STRATEGIZE The motion of the glider will be sinusoidal. It is released from rest at a = 12 cm, so its amplitude is 12 cm. PREPARE The frequency is 0.50 Hz, so the period is T = 1/f = 2.0 s. The glider is released at maximum extension from the equilibrium position, meaning that we can take this point to be t = 0. SOLVE 1.0 s is exactly half the period. As the graph of the motion in Figure 14.10 O shows, half a cycle brings the glider to its left turning point, 12 cm to the left of the equilibrium position. The velocity at this point is zero. FIGURE 14.10 Position-versus-time graph for the glider. х (ст) The starting point of the motion 12- T ! (s) 2.0 0.5 1.0 /1.5 At 1.0 s, the glider has completed half of one cycle. -12 - We've updated our read aloud feature! Give it a try here.

14.2 Understanding the motion of a glider on a spring An air-track glider oscillates horizontally on a spring at a frequency of 0.50 Hz. Suppose the glider is pulled to the right of its equilibrium position by 12 cm and then released. Where will the glider be 1.0 s after its release? What is its velocity at this point? STRATEGIZE The motion of the glider will be sinusoidal. It is released from rest at a = 12 cm, so its amplitude is 12 cm. PREPARE The frequency is 0.50 Hz, so the period is T = 1/f = 2.0 s. The glider is released at maximum extension from the equilibrium position, meaning that we can take this point to be t = 0. SOLVE 1.0 s is exactly half the period. As the graph of the motion in Figure 14.10 O shows, half a cycle brings the glider to its left turning point, 12 cm to the left of the equilibrium position. The velocity at this point is zero. FIGURE 14.10 Position-versus-time graph for the glider. х (ст) The starting point of the motion 12- T ! (s) 2.0 0.5 1.0 /1.5 At 1.0 s, the glider has completed half of one cycle. -12 - We've updated our read aloud feature! Give it a try here.

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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In Example 14.2, at 0.5 s, the force on the glider is _______ and its acceleration is ________.

   

to the right, to the left
   

to the left, to the right
   

zero, zero
   

to the right, to the right
   

to the left, to the left

 

In Example 14.2, at which of the following times the glider is at its starting position, that is x = A? Select all apply.

   

1.0 s
   

2.0 s
   

3.0 s
   

4.0 s
   

5.0 s
   

6.0 s
   

7.0 s
   

8.0 s
   

none of the above

In Example 14.2, at which of the following times the glider's Kinetic Energy is the maximum? Select all apply.

   

1.0 s
   

2.0 s
   

3.0 s
   

4.0 s
   

5.0 s
   

6.0 s
   

7.0 s
   

8.0 s
   

none of the above
EXAMPLE 14.2 Understanding the motion of a glider on a spring An air-track glider oscillates horizontally on a spring at a frequency of 0.50 Hz. Suppose the glider is pulled to the right of its equilibrium position by 12 cm and then released. Where will the glider be 1.0 s after its release? What is its velocity at this point? STRATEGIZE The motion of the glider will be sinusoidal. It is released from rest at a = 12 cm, so its amplitude is 12 cm. PREPARE The frequency is 0.50 Hz, so the period is T = 1/f = 2.0 s. The glider is released at maximum extension from the equilibrium position, meaning that we can take this point to be t = 0. SOLVE 1.0 s is exactly half the period. As the graph of the motion in Figure 14.10O shows, half a cycle brings the glider to its left turning point, 12 cm to the left of the equilibrium position. The velocity at this point is zero. FIGURE 14.10 Position-versus-time graph for the glider. x (ст) LZThe starting point of the motion 12 T t (s) 2.0 0.5 1.0 1.5 At 1.0 s, the glider has completed half of one cycle. -12- We've updated our read aloud feature! Give it a try here.
Transcribed Image Text:EXAMPLE 14.2 Understanding the motion of a glider on a spring An air-track glider oscillates horizontally on a spring at a frequency of 0.50 Hz. Suppose the glider is pulled to the right of its equilibrium position by 12 cm and then released. Where will the glider be 1.0 s after its release? What is its velocity at this point? STRATEGIZE The motion of the glider will be sinusoidal. It is released from rest at a = 12 cm, so its amplitude is 12 cm. PREPARE The frequency is 0.50 Hz, so the period is T = 1/f = 2.0 s. The glider is released at maximum extension from the equilibrium position, meaning that we can take this point to be t = 0. SOLVE 1.0 s is exactly half the period. As the graph of the motion in Figure 14.10O shows, half a cycle brings the glider to its left turning point, 12 cm to the left of the equilibrium position. The velocity at this point is zero. FIGURE 14.10 Position-versus-time graph for the glider. x (ст) LZThe starting point of the motion 12 T t (s) 2.0 0.5 1.0 1.5 At 1.0 s, the glider has completed half of one cycle. -12- We've updated our read aloud feature! Give it a try here.
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