Question

A sonometer wire is vibrating in resonance with a tuning fork. Keeping the tension applied same, the length of the wire is doubled. Under what conditions would the tuning fork still be in resonance with the wire?

Answer 1

When a wire of length L vibrates is resonant frequency in nth mode after stretching it by tension T, then frequency, \(v=\frac{n}{2L}\sqrt{\frac{T}{m}}\)

[Here, m = mass per unit length of stretched wire]

Let us consider two given cases:

\(v_1=\frac{n_1}{2L_1}\sqrt{\frac{T_1}{m_1}}\)&v_{2 = \(\frac{n_2}{2L_2}=\sqrt{\frac{T_2}{m_2}}\)}

From question,

T₁ = T₂ = T, m₁ = m₂ = m (as wires are same)

L₂ = 2L₁

\(\frac{v_1}{v_2}=(\frac {n_1}{2L_1}\sqrt{\frac{T}{m}})\div(\frac{n_2}{2(2L_1)}\sqrt{\frac{T}{m}})= \frac{2n_1}{n_2}\)

As turning fork is same, ∴ v₁ = v₂ 

\(\frac{2n_1}{n_2}=1 \,or\, n_2=2n_1\)

Hence, when length of wire double the number of harmonics double for same resonant frequency.