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
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)...
Related questions
Question
In Example 14.2, at 0.5 s, the force on the glider is _______ and its acceleration is ________.
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to the right, to the left |
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to the left, to the right |
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zero, zero |
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to the right, to the right |
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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.
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1.0 s |
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2.0 s |
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3.0 s |
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4.0 s |
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5.0 s |
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6.0 s |
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7.0 s |
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8.0 s |
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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.
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1.0 s |
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2.0 s |
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3.0 s |
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4.0 s |
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5.0 s |
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6.0 s |
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7.0 s |
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8.0 s |
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none of the above |
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-
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Give it a try here.
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