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# CP A turntable 1.50 m in diameter rotates at 75 rpm. Two ISBN: 9780321675460 31

## Solution for problem 80P Chapter 16

University Physics | 13th Edition

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Problem 80P

CP A turntable 1.50 m in diameter rotates at 75 rpm. Two speakers, each giving off sound of wavelength 31.3 cm, are attached to the rim of the table at opposite ends of a diameter. A listener stands in front of the turntable. (a) What is the greatest beat frequency the listener will receive from this system? (b) Will the listener be able to distinguish individual beats?

Step-by-Step Solution:

Solution 80P Step 1 : In this question we need to find The greatest beat frequency the listener will receive We also need to gauge if the listener can distinguish between individual beats Consider the data given Diameter of table d = 1.50 cm Radius will be r = d/2 = 0.75 cm Speed of rotations R = 75 rpm Wavelength of sound = 31.3 cm Velocity of sound in air v = 344 m/s a Step 1 : Let us find the velocity of the round table It is given by v = r × R × 0.10472 Substituting the values v = 0.75 cm × 75 rpm v = 5.8905 m/s Hence we have velocity of round table as 5.8905 m/s Step 2 : Let us find the frequency of the sound waves in air It is given by f = v a Substitute v a 344 m/s and = 31.3 cm = 0.313 m f = 344 m/s/0.313 m f = 1099.04 Hz We have frequency in air as 1099.04 Hz Step 3 : Let us find frequency while the sound waves are travelling towards listener It is given by f1= f(v +av ) o(v a ) s v 0 velocity of observer = 0 m/s v s velocity of source = 5.8905 m/s f = 1099.04 Hz Substituting the values we get f1= 1099.04 Hz(344 m/s + 0 m/s) /(344 m/s 5.8905 m/s) f1= 1099.04 Hz × ( 1.01742 m/s) f1= 1118.18 Hz Thus we have frequency while the sound waves are travelling towards listener as 1118.18 Hz

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##### ISBN: 9780321675460

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