Question

Two wires of the same material and same cross section are stretched on a sonometer. One wire is loaded with 1.5 kg and another is loaded with 6 kg. The vibrating length of first wire is 60 cm and its fundamental frequency of vibration is the same as that of the second wire. Calculate vibrating length of the other wire.

 

Answer 1

Given : m₁ = m₂ = m, L₁ = 60 cm = 0.6 m, T₁ = 1.5 kg=14.7 N, T₂ =6 kg=58.8 N

n₁ = \(\frac{1}{2L_1}\sqrt{\frac{T_1}{m}}\) and  n₂ = \(\frac{1}{2L_2}\sqrt{\frac{T_2}{m}}\)

But n₁ = n₂

∴ \(\frac{1}{2L_1}\sqrt{\frac{T_1}{m}}\)  = \(\frac{1}{2L_2}\sqrt{\frac{T_2}{m}}\)

L₂ = \(\sqrt{\frac{T_2}{T_1}}×L_1\)  = \(\sqrt{\frac{58.8}{14.7}}×0.6\)= 1.2 m

The vibrating length of the second wire is 1.2 m.