25 tuning forks are arranged in series in the order of decreasing frequency. Any two successive forks…
25 tuning forks are arranged in series in the order of decreasing frequency. Any two successive forks produce 3 beats/s. If the frequency of the first tuning fork is the octave of the last fork, the frequency of the 21st fork is
72 Hz
288 Hz
84 Hz
87 Hz
Solution
According to the question, frequencies of first and last tuning forks are $2n$ and $n$, respectively.
A series of tuning forks numbered 1, 2, 3, 21, and 25 have frequencies $2n$, $(2n - 3)$, $(2n - 6)$, $(2n - 20 \times 3)$, and $(2n - 24 \times 3) = n$.
Hence, frequency in given arrangement are as follows
$\Rightarrow 2n - 24 \times 3 = n \Rightarrow n = 72\text{ Hz}$
So, frequency of 21st tuning fork,
$n_{21} = (2 \times 72 - 20 \times 3) = 84\text{ Hz}$