Question

A sonometer wire is in unison with a tuning fork. When its length increases by 4%, it gives 8 beat/s with the same fork. The frequency of the fork is

Answer 1

208 Hz

Answer 2

For a stretched string, the fundamental frequency is given by

$$f = \frac{1}{2L}\sqrt{\frac{T}{\mu}},$$

so $f \propto \frac{1}{L}$ (with tension $T$ and linear density $\mu$ constant).

Let the original frequency be $f$ (which is equal to the tuning fork’s frequency). When the length is increased to $1.04L$, the new frequency becomes

$$f' = \frac{f}{1.04}.$$

Since the wire now gives 8 beats per second with the fork, the difference in frequencies is

$$f - \frac{f}{1.04} = 8.$$

Simplify:

$$f\left(1 - \frac{1}{1.04}\right) = 8 \quad \text{or} \quad f\left(\frac{1.04 - 1}{1.04}\right) = 8.$$ $$\Rightarrow f\left(\frac{0.04}{1.04}\right) = 8.$$

Thus,

$$f = 8 \times \frac{1.04}{0.04} = 8 \times 26 = 208\,\text{Hz}.$$

Increase in length reduces frequency by a factor of 1.04. The beat frequency is

$$f - \frac{f}{1.04} = \frac{0.04f}{1.04} = 8,$$

leading to $f = 208\,\text{Hz}$.