A sonometer wire resonates with a given tunin | vedclass

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Question

(A) The frequency of a vibrating string in a sonometer is given by $f = \frac{n}{2L} \sqrt{\frac{T}{\mu}}$,where $n$ is the number of antinodes,$L$ is

Vedclass · Accepted Answer

$25$

Vedclass · Answer

(A) The frequency of a vibrating string in a sonometer is given by $f = \frac{n}{2L} \sqrt{\frac{T}{\mu}}$,where $n$ is the number of antinodes,$L$ is the length of the wire,$T$ is the tension $(T = Mg)$,and $\mu$ is the linear mass density.In the first case,$n_1 = 5$ and $T_1 = 9g$. Thus,$f = \frac{5}{2L} \sqrt{\frac{9g}{\mu}}$.In the second case,$n_2 = 3$ and $T_2 = Mg$. Thus,$f = \frac{3}{2L} \sqrt{\frac{Mg}{\mu}}$.Since the tuning fork is the same,the frequency $f$ remains constant. Equating the two expressions:$\frac{5}{2L} \sqrt{\frac{9g}{\mu}} = \frac{3}{2L} \sqrt{\frac{Mg}{\mu}}$$5 \sqrt{9g} = 3 \sqrt{Mg}$$5 	imes 3 \sqrt{g} = 3 \sqrt{Mg}$$5 \sqrt{g} = \sqrt{Mg}$Squaring both sides: $25g = Mg$$M = 25 \, kg$.

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