Wednesday, April 18, 2007

Sound – Two Questions on Beats

Questions similar to the following one have appeared in various entrance tests:

A set of 31 tuning forks are so arranged that each gives 5 beats per second with the previous one. If the frequency of the last tuning fork is double that of the first, the frequency of the third tuning fork is

(a) 250 Hz (b) 220 Hz (c) 180 Hz (d) 160 Hz (e) 150 Hz

Don’t be scared by the relatively large number of forks. This is a very simple question.

If n1, n3 and n31 are the frequencies of the 1st, 3rd and 31st forks, we have

n31 = 2n1 since the frequency of the last fork is double that of the first.

Further, n31 = n1 + 30×5 since there are 30 increments in frequency (each of 5 Hz) from the first fork to the 31st fork.

Thus we have 2n1 = n1 + 150 from which n1 = 150 Hz.

The frequency of the 3rd fork, n3 = n1 + 2×5 = 150 + 10 = 160 Hz.

Now, consider the following MCQ:

When a tuning fork is excited, molecules of air vibrate in accordance with the equation x = A cos(512πt). When this tuning fork and another identical tuning fork loaded with a little wax are excited together, 4 beats are heard. The frequency of the second fork loaded with wax is

(a) 516 Hz (b) 508 Hz (c) 384 Hz (d) 260 Hz (e) 252 Hz

When the first tuning fork is excited, the vibrations of the air molecules are simple harmonic with angular frequency ω = 512π as is evident from the form of the equation, x = A cos(512πt).

The linear frequency of vibration of the first fork is n = ω/2π = 512π/2π = 256 Hz.

The frequency of the 2nd tuning fork before loading with wax was therefore 256 Hz. After loading with wax, its frequency is lowered. Since the beat frequency is 4 Hz, its frequency (after loading) is 256 – 4 = 252 Hz.

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