1.

A set of 65 tuning forks are arranged In the Increasing order of frequencies such that each gives 3 beats per second with the previous one. Find the frequency of the first tuning fork If the frequency of the last tuning fork Is one Octave above the first one.

Answer»

Let N be the frequency of the first tuning fork. As each tuning fork is giving 3 beats with the preceding one, and they are arranged in the ascending order of frequencies,

Frequency of the second tuning fork = N + 3

Frequency of the third T.F. = N + 6

= N + 3 × 2

Frequency of the fourth T.F = N+9

= N + 3 × 3

Similarly Frequency of the 65th T.F. = N + (65-1 ) × 3

= N + 64 × 3 = N + 192 …….(1)

But the frequency of the last tuning fork (65th) is one octave above that of the first one.

∴ Frequency of the 65th T.F.= 2N ………(2)

From (1) and (2) 2N = N + 192

N = 192

∴ Frequency of the first tuning fork = 192 Hz.


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