A set of 20 tuning forks is arranged in a series of increasing frequencies. If each fork gives 4 beats with respect to the preceding fork and the frequency of the last fork is twice the frequency of the first, then the frequency of last fork is _________ Hz.

Solution

Physics sir · 3 steps

The forks form an AP with common difference 4, so count the gaps between them carefully: 20 forks, 19 gaps.

1. Beats as difference. The number of beats per second equals the difference of the two frequencies. So each fork is 4 Hz higher than the one before it, and the frequencies form an arithmetic progression.
2. Count the gaps. Going from fork 1 to fork 20 takes 19 steps of 4 Hz each. If the first fork has frequency f, the last one is f + 19 × 4 = f + 76.

   f + 76 = 2f ⇒ f = 76 Hz

3. So. The first fork is at 76 Hz, so the last fork is at 2 × 76 = 152 Hz.

The common slip: 160. Counting 20 gaps instead of 19 makes the spread 80 Hz, giving a first fork of 80 Hz and a last fork of 160.

The idea to keep. In a chain of n forks with a constant beat count b between neighbours, the extremes differ by (n-1)b. Always count gaps, not forks.

Answer from the NTA final answer key. Solution by PrepWiser.