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# chap 13 prob-YF 12th ed

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Discussion Questions horizontal spring Neither m nor k changes when the apparatus is taken to Mars so the period is unchanged The only difference is that in equilibrium the spring will stretch a shorter distance on Mars than on earth due to the weaker gravity 13 5 Answer no Just as for an object oscillating on a spring at the equilibrium position the speed of the pendulum bob is instan taneously not changing this is where the speed is maximum so its derivative at this time is zero But the direction of motion is changing because the pendulum bob follows a circular path Hence the bob must have a component of acceleration perpendicu lar to the path and toward the center of the circle see Section 3 4 To cause this acceleration at the equilibrium position when the string is vertical the upward tension force at this position must be greater than the weight of the bob This causes a net upward force on the bob and an upward acceleration toward the center of the circular path 13 6 Answer i The period of a physical pendulum is given by Eq 13 39 T 2ir 7d The distance d L from the pivot to the center of gravity is the same for both the rod and the PROBLEMS 447 simple pendulum as is the mass m This means that for any dis placement angle 0 the same restoring torque acts on both the rod and the simple pendulum However the rod has a greater moment of inertia rod m 2I niL 2 arid smpIe rnL 2 all the mass of the pendulum is a distance L from the pivot Hence the rod has a longer period 13 7 Answer ii The oscillations are underdamped with a decreasing amplitude on each cycle of oscillation like those graphed in Fig 13 26 If the oscillations were undamped they would continue indefinitely with the same amplitude If they were critically damped or overdamped the nose would not bob up and down but would return smoothly to the original equilibrium atti tude without overshooting 13 8 Answer i Figure 13 28 shows that the curve of amplitude versus driving frequency moves upward at all

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