Chapter 13: Oscillations About EquilibriumAnswers to Even-Numbered Conceptual Questions2. The person’s shadow undergoes periodic motion, with the same period as the period of the Ferris wheel’srotation. In fact, if we take into account the connection between uniform circular motion and simpleharmonic motion, we can say that the shadow exhibits simple harmonic motion as it moves back and forthon the ground.4. Recall that the maximum speed of a mass on a spring is vmax=ωA, where ω = k / m. It follows that themaximum kinetic energy is Kmax=12mvmax2=12m kA2/ m( )=12kA2. Note that the mass cancels in our finalexpression for the maximum kinetic energy. Therefore, the larger mass moves more slowly by just the rightamount so that the kinetic energy is unchanged.6. The constant A represents the amplitude of motion; the constant B is the angular frequency. Noting that theangular frequency is ω =2πf , we have that the frequency is f =ω / 2π =B / 2π.8. The period of a pendulum is independent of the mass of its bob. Therefore, the period should be unaffected.Solutions to Problems and Conceptual Exercises1. Picture the Problem: The sketch shows the cart traveling up and back, thus completing one cycle on the track.Strategy: One period is the time for the cart to move down the track and back. The frequency is the inverse of the period.Solution: 1. Divide twice track length by the velocity to obtain the period: T =2 5.0 m( )0.85 m/s=11.8 s =12 s2. Invert the period to determine the frequency:Insight: The full period of oscillation is complete when the cart returns to its starting position with the same speed and direction of motion with which it began.2. Picture the Problem: The rocking chair completes one full cycle or oscillation each time it returns back to its original position. Strategy: The period is the time for one cycle. The frequency is the inverse of the period, or the number of cycles per second.Solution: 1. Divide the total time by the number of cycles to determine the period: T =tn=21 s12 cycles=1.75s2. Invert the period to determine the frequency: f =1T=11.75s= 0.57 HzInsight: Since period and frequency are inverses of each other, when the period is greater than a second, the frequency will be less than a hertz.Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. Noportion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.13 – 1Chapter 13: Oscillations About Equilibrium James S. Walker, Physics, 4th Edition3. Picture the Problem: As the bobber moves up and down its motion is periodic. One period is the amount of time for the bob to drop down and rise back up to its original position.Strategy: We can find the period by taking the inverse of the period.Solution: Invert the period to obtain the frequency: T =1f=12.6 Hz= 0.38 sInsight: Because frequency is the inverse of period, when the frequency is greater than one hertz, the period will necessarily be less than a second.4. Picture the Problem: As the basketball is dribbled it moves up and down in periodic motion. One period is the time forthe ball to drop and return to the player’s hand. Strategy: The time for one dribble is the period, or the inverse of the frequency. Multiplying the period by the number of dribbles will give the total time.Solution: Multiply the inverse of the frequency by the number of dribbles to obtain the total time: t =12T =12f=121.77 Hz= 6.78 sInsight: The greater the frequency, the more dribbles possible over a given time.5. Picture the Problem: A heart beats in a regular periodic pattern. To measure your heart rate you typically count the number of pulses in a minute.Strategy: Convert the time from minutes to seconds to obtain the frequency in hertz. The period is obtained from the inverse of the frequency.Solution: 1. Multiply the heart rate by the correct conversion factor to get the frequency in Hz. f = 74 beatsmin⎛⎝⎜⎞⎠⎟1 min60 s⎛⎝⎜⎞⎠⎟=1.2 Hz2. Invert the frequency to obtain the period: T =1f=11.23 Hz= 0.81 sInsight: Typical resting heartbeats have frequencies around one hertz and periods of about one second.6. Picture the Problem: When you measure your pulse you typically count the number of pulses per minute. The numberof beats per minute, or frequency, can change depending on whether you are resting or exercising.Strategy: Convert the time rate from per second to per minute to obtain the frequency in beats/min. Solution: 1. (a) Multiply the frequency by the conversion factor to obtain the heart rate: 1.45 beatss⎛⎝⎜⎞⎠⎟60 smin⎛⎝⎜⎞⎠⎟= 87.0 beats min2. (b) The number of beats per minute will increase, because more beats per second means more beats per minute.3. (c) Multiply the new frequency by the conversion factor to obtain the heart rate: 1.55 beatss⎛⎝⎜⎞⎠⎟60 smin⎛⎝⎜⎞⎠⎟= 93.0 beats minInsight: Increasing the frequency in any set of units (such as beat/sec or beat/min) will increase the frequency in any other set of units. Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. Noportion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.13 – 2Chapter 13: Oscillations About Equilibrium James S. Walker, Physics, 4th Edition7. Picture the Problem: As gasoline burns inside the engine of a car, it causes the pistons to expand, which in turn causes the crankshaft to rotate. Increasing the gas in the engine (revving) causes the crankshaft to rotate faster. The frequency of the car’s engine is measured as the number of times the crankshaft rotates per minute. Strategy: The frequency is given in units of rev/min which can be converted to hertz. The period can then be found by inverting the frequency. Reverse the process to convert a period back into a frequency in hertz.Solution: 1. (a) Convert f to hertz: f = 2700 revmin⎛⎝⎜⎞⎠⎟1 min60 s⎛⎝⎜⎞⎠⎟= 45 Hz2. Invert the frequency to obtain the period: T =1f=145 Hz= 0.022 s3. (b) Invert the new period and convert seconds to minutes to obtain the rpm: f =1T=1 rev0.044 s×60 smin=1400 rpmInsight: Since period and frequency are inverses of each other, a longer
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