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# ch10 - CHAPTER 10 SIMPLE HARMONIC MOTION AND ELASTICITY...

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CHAPTER 10 SIMPLE HARMONIC MOTION AND ELASTICITY CONCEPTUAL QUESTIONS ____________________________________________________________________________________________ 1 . REASONING AND SOLUTION A horizontal spring is attached to an immovable wall. Two people pull on the spring. They then detach it from the wall and pull on opposite ends of the horizontal spring. They pull just as hard in each case. Suppose that each person pulls with a force P . When the spring is attached to the wall, both people pull on the same end, so that they pull with a total force 2 P . The wall exerts a force –2 P at the other end of the spring. The tension in the spring has magnitude 2 P . When the two people pull on opposite ends of the horizontal spring, one person exerts a force P at one end of the spring, and the other person exerts a force – P at the other end of the spring. The tension in the spring has magnitude P . The tension in the spring is greater when it is attached to the immovable wall; therefore, the spring will stretch more in this case. ____________________________________________________________________________________________ 2 . REASONING AND SOLUTION Identical springs are attached to a box in two different ways, as shown in the drawing in the text. Initially, the springs are unstrained. The box is then pulled to the right and released. The displacement of the box is the same in both cases. For the box on the left, each spring is stretched and its displacement is +∆ x , where the plus sign indicates a displacement to the right. According to Equation 10.2, the restoring force exerted by each spring on the box is ( ) x F k x = − +∆ . Thus, the net force due to both springs is . ( ) 2 x F k Σ = − +∆ x For the box on the right, one spring is stretched and the other is compressed. However, the displacement of each spring is still +∆ x . The net force exerted on the box is ( ) ( ) ( ) 2 x F k x k x k x Σ = − +∆ +∆ = − +∆ . Thus, both boxes experience the same net force. ____________________________________________________________________________________________ 3 . REASONING AND SOLUTION Figures 10.11 and 10.14 show the velocity and the acceleration, respectively, of the shadow of a ball that undergoes uniform circular motion. The shadow undergoes simple harmonic motion. a. The velocity of the shadow is given by Equation 10.7: sin x v A ω θ = − . The velocity of the shadow will be zero when θ = 0 or π rad. From Figure 10.11, we see that the velocity equals zero each time the shadow reaches the right and left endpoints of its motion (that is, when the ball crosses the x axis). b. The acceleration of the shadow is given by Equation 10.9: 2 cos x a A ω θ = − . The acceleration of the shadow will be zero when θ is π /2 or 3 π /2. From Figure 10.14, we see

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