A sonometer wire resonates with a given tuning fork forming standing waves with five antinodes between the two bridges when a mass of $$9$$ kg is suspended from the wire. When this mass is replaced by a mass M, the wire resonates with the same tuning fork forming three antinodes for the same positions of the bridges. The value of M is

Question

Infinity Learn · Accepted Answer

Let L be the length of the wire between the bridges and let m be the mass per unit length of the wire.

Five antinodes on a length L implies that $L = \frac{5}{2} λ_{1}$ or $λ_{1} = \frac{2 L}{5}$ . Thus in this case we have

$v_{1} = \frac{V_{1}}{λ_{1}} = \frac{5}{2 L} \sqrt{\frac{T_{1}}{m}}$ , where $T_{1} = M_{1} g$

Three antinodes on a length L implies that $L = \frac{3}{2} λ_{2}$ or $λ_{2} = \frac{2 L}{3}$ . In this case, we have $v_{2} = \frac{V_{2}}{λ_{2}} = \frac{3}{2 L} \sqrt{\frac{T_{2}}{m}}$ , where $T_{2} = M_{2} g$

Given $v_{1} = v_{2}, M_{1} = 9 kg and M_{2} = M$ . Therefore, we have

$\frac{5}{2 L} \sqrt{\frac{9 g}{m}} = \frac{3}{2 L} \sqrt{\frac{Mg}{m}}$

which gives M = 25 kg.

Stationary waves

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