Question

A sonometer wire of length L₁ is in unison with a tuning fork of frequency n. When the vibrating length of the wire is reduced to L₂ , it produces x beats per second with the fork. Show that n = x₂ . \(\frac{L_2}{L_1-L_2}\)

Answer 1

The fundamental frequency of vibration of a wire of length L₁ , mass per unit length m and under tension T is

n₁ = \(\frac1{2L_1}\)\(\sqrt{ \frac Tm}\) = n… (1)

since it is in unison with a tuning fork of frequency n. When the vibrating length of the wire is L₂ , its fundamental frequency is

n₂ = \(\frac1{2L_2}\)\(\sqrt{ \frac Tm}\) = n… (2)

T and m remaining constant.

∴ \(\frac{n_{2}}{n_{1}} = \frac{L_{1}} {L_{2}}\)… (3)

Since L₂ < L₁ , n₂ > n₁ so that n₂ – n₁ = x 

∴ n₂ = n₁ + x …. (4) 

Substituting for n₂ in Eq. (3),

\(\frac{n_{1}+x}{n_{1}}\) = \(\frac{L_1}{L_2}\) or \(\frac{x}{n_{1}}\) = \(\frac{L_{1}-L_2}{L_{2}}\)

∴ n₁ = n = x \(\frac{L_2}{L_{1}-L_2}\)

which is the required expression.