A sonometer wire resonates with a given tunin | vedclass

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Question

(A) The frequency of a sonometer wire is given by the formula $n = \frac{p}{2L} \sqrt{\frac{T}{\mu}}$,where $p$ is the number of loops (antinodes),$L$

Vedclass · Accepted Answer

$25$

Vedclass · Answer

(A) The frequency of a sonometer wire is given by the formula $n = \frac{p}{2L} \sqrt{\frac{T}{\mu}}$,where $p$ is the number of loops (antinodes),$L$ is the length,$T$ is the tension $(T = Mg)$,and $\mu$ is the linear mass density.Since the frequency $n$,length $L$,and linear mass density $\mu$ remain constant,we have $p \propto \sqrt{T} \propto \sqrt{M}$.For the first case: $p_1 = 5$ and $M_1 = 9 \ kg$.For the second case: $p_2 = 3$ and $M_2 = m$.Using the ratio: $\frac{p_1}{p_2} = \sqrt{\frac{M_1}{M_2}}$.Substituting the values: $\frac{5}{3} = \sqrt{\frac{9}{m}}$.Squaring both sides: $\frac{25}{9} = \frac{9}{m}$.Solving for $m$: $m = \frac{9 	imes 9}{25} = \frac{81}{25} = 3.24 \ kg$. Wait,re-evaluating the relationship: $p_1 \sqrt{M_1} = p_2 \sqrt{M_2}$.$5 \sqrt{9} = 3 \sqrt{m} \implies 5 	imes 3 = 3 \sqrt{m} \implies 5 = \sqrt{m} \implies m = 25 \ kg$.

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