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When a tuning fork of frequency 341 is sounded with another tuning fork, six beats per second are heard. When the second tuning fork is loaded with wax and sounded with the first tuning fork, the number of beats is two per second. The natural frequency of the second tuning fork is

a
334
b
339
c
343
d
347

detailed solution

Correct option is D

nA = Known frequency = 341 Hz, nB = ?x = 6 bps, which is decreasing (i.e. x↓) after loading (from 6 to 1 bps)Unknown tuning fork is loaded so nB↓Hence nA – nB↓ = x↓       ... (i)       →         Wrong            nB↓  – nA = x↓      ... (ii)      →        Correct⇒ nB = nA + x = 341 + 6 = 347 Hz.

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Similar Questions

A metal wire of diameter 1.5 mm is held on two knife edges separated by a distance 50 cm the tension in the wire is 100 N the wire vibrating with its fundamental frequency and vibrating tuning fork together produces 5 beats per second. The tension in the wire is then reduced to 81 N, when the two are excited, beats are heard at the same rate. Calculate frequency of the fork.


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