In a set, 21 turning forks are arranged in a series of decreasing frequencies. Each tuning fork produces 4 beats per second with the preceding fork. If the first fork is an octave of the last fork, find the frequencies of the first and tenth fork.

#### Solution

Given: N = 21, x = 4, nF = 2 n_{L}

To find:

i. Frequency of first fork (n_{F})

ii. Frequency of tenth fork (n_{10})

Formula: n_{L} = n_{F} - (N – 1)x

Calculation:-

i. When tuning forks are arranged in the decreasing order of frequencies, the frequency of the p^{th} tuning fork is,

n_{L} = n_{F} - (N - 1)x = n_{1} - (21 - 1) 4

∴ n_{L} = n_{F} - 80 ............(1)

As frequency of first fork is an octave of last,

∴ n_{F} = 2n_{L}

∴ n_{L} = n_{F}/2

From equation (1),

n_{F}/2 = n_{F} - 80

∴ n_{F} - (n_{F}/2) = 80

∴ n_{F}/2 = 80

∴ n_{F} = 160 Hz

**The frequency of the first fork is 160 Hz.**

ii. For 10^{th} fork,

n_{10} = n_{1} - (10 - 1)x

= 160 - 9 * 4 = 160 - 36

n_{10} = 124Hz

**The frequency of the tenth fork is 124 Hz.**