A sonometer wire resonates with 4 antinodes b | vedclass

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

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(C) The frequency of a vibrating string is given by $f = \frac{n}{2L} \sqrt{\frac{T}{\mu}}$, where $n$ is the number of loops (antinodes), $L$ is the length, $T$ is the tension, and $\mu$ is the linear mass density.Since the tuning fork is the same, the frequency $f$ remains constant.Also, $L$ and $\mu$ are constant, so $n \propto \sqrt{T}$.Given $T = Mg$, we have $n \propto \sqrt{M}$.For the first case: $n_1 = 4$ and $M_1 = 1 	ext{ kg}$.For the second case: $n_2 = 2$ and $M_2 = M$.Thus, $\frac{n_1}{n_2} = \sqrt{\frac{M_1}{M_2}}$.Substituting the values: $\frac{4}{2} = \sqrt{\frac{1}{M}}$.$2 = \sqrt{\frac{1}{M}}$.Squaring both sides: $4 = \frac{1}{M}$.Therefore, $M = \frac{1}{4} 	ext{ kg} = 0.25 	ext{ kg}$.Wait, re-evaluating the relationship: $n \propto \sqrt{T} \implies n^2 \propto T \implies n^2 \propto M$.So, $n_1^2 M_2 = n_2^2 M_1$.$4^2 	imes M = 2^2 	imes 1$.$16M = 4$.$M = \frac{4}{16} = 0.25 	ext{ kg}$.Given the options provided, there might be a misunderstanding in the question's premise or options. If the number of antinodes decreases, the tension must increase. Let's re-check: $f = \frac{n}{2L} \sqrt{\frac{T}{\mu}}$. If $n$ decreases from $4$ to $2$, then $\sqrt{T}$ must increase by a factor of $2$, meaning $T$ must increase by a factor of $4$. Thus $M = 4 	imes 1 	ext{ kg} = 4 	ext{ kg}$.

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