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 9th tuning Fork Is Twice That Of The First, What Is The Frequency Of The First Tuning Fork?

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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 9th tuning fork is twice that of the first, what is the frequency of the first tuning fork?

Solution

Correct option is

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 9th tuning fork = V + (9 – 1) × 4 = V + 32. It is given that V + 32 = 2V. Hence V = 32 Hz.

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