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Q.

If the difference in frequencies between tuning forks is represented by beats, what would be the frequency of the 16th fork if 64 tuning forks are arranged in such a way that each fork produces 4 beats per second with the next one, and the frequency of the last fork is double that of the first fork?

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a

308 Hz

b

316 Hz

c

312 Hz

d

322 Hz

answer is C.

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Detailed Solution

detailed_solution_thumbnail

The frequencies are in A.P.

fn=fo+4n

Given that the 64th fork is octave of the first.

2fo=fo+634⇒fo=252Hz

So the frequency of 16th fork is given by 

f16=252+15(4)=312Hz

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