Chapter 14: Waves and SoundAnswers to Even-Numbered Conceptual Questions2. Waves passing through a field of grain are longitudinal waves—the motion of each stalk of grain is in thesame direction as the motion of the wave itself.4. This wave is longitudinal, because each cat moves in the same direction as the wave.6. The Doppler effect applies to radar waves as well as to sound waves. In particular, the ball sees a Doppler-shifted radar frequency for the waves coming from the radar gun. Then, the ball acts as a moving source forwaves of this frequency, giving a second Doppler shift to the echoes that are picked up by the radar gun.This provides a one-to-one correspondence between the final observed frequency and the speed of the ball.8. The sliding part of a trombone varies the length of the vibrating air column that produces the trombone’ssound. By adjusting this length, the player controls the resonant frequencies of the instrument. This, in turn,varies the frequency of sound produced by the trombone.10. The thicker string is used to produce the low-frequency notes. This follows because the frequency of thefundamental depends directly on the speed of waves on the string. Therefore, for a given tension, a stringwith a greater mass per length has a smaller wave speed and hence a lower frequency.12. You have just observed a series of beats between your wipers and the wipers of the other car.Solutions to Problems and Conceptual Exercises1. Picture the Problem: The image shows a wave with the given wave dimensions.Strategy: Set the wavelength equal to the horizontal crest-to-crest distance, or double the horizontal crest-to-trough distance. Set the amplitude equal to the vertical crest-to-midline distance, or half the vertical crest-to-trough distance.Solution: 1. (a) Double the horizontal crest-to-trough distance: λ =2 28 cm( )= 56 cm2. (b) Halve the vertical crest-to-trough distance: A =1213 cm( )= 6.5 cmInsight: Note the difference in wavelength and amplitude. The wavelength is the entire distance from crest to crest, butamplitude is only from the equilibrium point to the crest.2. Picture the Problem: A surfer measures the frequency and length of the waves that pass her. From this information we wish to calculate the wave speed.Strategy: Use equation 14-1 to write the wave speed as the product of the wavelength and frequency.Solution: Multiply wavelength by frequency: v =λ f = 34 m( )14 /min( )1 min60 sec⎛⎝⎜⎞⎠⎟= 7.9 m/sCopyright © 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.14 – 1Chapter 14: Waves and Sound James S. Walker, Physics, 4th EditionInsight: The wave speed can increase by either an increase in wavelength or an increase in frequency.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.14 – 2Chapter 14: Waves and Sound James S. Walker, Physics, 4th Edition3. Picture the Problem: The image shows water waves passing to a shallow region where the speed decreases. We need to calculate the wavelength in the shallow area. Strategy: Use equation 14-1 to calculate the frequency in the deep water. Then use the constant frequency and the speed in the shallow water to calculate the new wavelength.Solution: 1. Calculate the frequency: f =v1λ1=2.0 m/s1.5 m=1.333 Hz2. Calculate the new wavelength: λ2=v2f=1.6 m/s1.333 Hz=1.2 mInsight: Note that decreasing the speed, while holding the frequency constant, will decrease the wavelength.4. Picture the Problem: The speed and wavelength of a tsunami are given and we wish to calculate the frequency.Strategy: Solve equation 14-1 for the frequency.Solution: Calculate the frequency: f =vλ=750km/h( )310 km1 h3600 s⎛⎝⎜⎞⎠⎟= 6.7×10−4 HzInsight: Although the tsunami has a very high speed, the long wavelength gives the tsunami a low frequency.5. Picture the Problem: A wave of known amplitude, frequency, and wavelength travels along a string. Strategy: Multiply the time by the wave speed, where the wave speed is given by equation 14-1, to calculate the horizontal distance traveled by the wave. To find the distance traveled by a knot on the string, note that a point on the string travels up and down a distance four times the amplitude during each period. Determine the fraction of a period that is spanned by the elapsed time and multiply it by 4A to find the distance traveled by the knot.Solution: 1. (a) Calculate the distance traveled by the wave: dw=vt = λ f( )t = 27×10−2 m( )4.5 Hz( )0.50 s( )= 0.61 m2. (b) Multiply 4A by the fraction of a period, noting that f =1 T :: dk= 4A( )tT⎛⎝⎜⎞⎠⎟=4A f t =4 12×10−2 m( )4.5 Hz( )0.50 s( )=1.1 m3. (c) The distance traveled by a wave peak is independent of the amplitude, so the answer in part (a) is unchanged. The distance traveled by the knot varies directly with the amplitude, so the answer in part (b) is halved.Insight: A point on the string travels four times the wave amplitude in the same time that the crest travels one wavelength.6. Picture the Problem: Using the equation for the speed of deep water waves given in the problem we want to calculate the speed and frequency of the waves.Strategy: Insert the given data into the equation v = gλ 2π to solve for the speed of the waves. Then use equation 14-1 to calculate the wave frequency.Solution: 1. (a) Insert the frequency into the deep water velocity equation: v = gλ / 2π( )= 9.81m/s2( )4.5 m( )/ 2π( )= 2.65 m s2. (b) Solve equation 14-1 for the frequency: f =vλ=2.651m/s4.5 m= 0.59 HzInsight: Because the velocity is proportional to the square-root of the wavelength, the frequency is inversely proportional to the square-root of the wavelength. Increasing the wavelength by a factor of four will double the wave speed and cut the frequency in half. 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.14 – 3Chapter 14: Waves and