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 n_{1}, n_{3} and n_{31} are the frequencies of the 1^{st}, 3^{rd} and 31^{st} forks, we have

n_{31} = 2n_{1} since the frequency of the last fork is double that of the first.

Further, n_{31} = n_{1} + 30×5 since there are 30 increments in frequency (each of 5 Hz) from the first fork to the 31^{st} fork.

Thus we have 2n_{1} = n_{1} + 150 from which n_{1} = 150 Hz.

The frequency of the 3^{rd} fork, n_{3} = n_{1} + 2×5 = 150 + 10 = 160 Hz.

**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 2^{nd} 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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