## Question

Nine tuning forks are arranged in order of increasing frequency. Each tuning fork produces 4 beats per second when sounded with either of its neighbours. If the frequency of the 9^{th} tuning fork is twice that of the first, what is the frequency of the first tuning fork?

### Solution

32 Hz

Let the frequency of the first tuning fork be *V*. The frequency of the second will be (*V* + 4) and of the third will be (*V* + 8) and so on. Now *V* + 8 = *V* + (3 – 1) × 4. Therefore, the frequency of the 9^{th} tuning fork = *V* + (9 – 1) × 4 = *V* + 32. It is given that *V* + 32 = 2*V*. Hence *V* = 32 Hz.

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