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(C) The frequency $f$ of a standing wave in a wire under tension $T$ is given by $f = \frac{n}{2L} \sqrt{\frac{T}{\mu}}$,where $n$ is the number of an

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$9$

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(C) The frequency $f$ of a standing wave in a wire under tension $T$ 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,and $\mu$ is the linear mass density.The tension $T$ in the wire is provided by the spring,so $T = kx$,where $k$ is the spring constant and $x$ is the stretch.Thus,$f = \frac{n}{2L} \sqrt{\frac{kx}{\mu}}$.For the first case,$n_1 = 3$ and $x_1 = 4.0\, cm$. So,$f = \frac{3}{2L} \sqrt{\frac{k(4.0)}{\mu}}$.For the second case,$n_2 = 2$ and we need to find $x_2$. Since the frequency $f$ is the same,$f = \frac{2}{2L} \sqrt{\frac{k(x_2)}{\mu}}$.Equating the two expressions for $f$: $\frac{3}{2L} \sqrt{\frac{k(4.0)}{\mu}} = \frac{2}{2L} \sqrt{\frac{k(x_2)}{\mu}}$.Simplifying,we get $3 \sqrt{4.0} = 2 \sqrt{x_2}$.$3 	imes 2 = 2 \sqrt{x_2} \implies 6 = 2 \sqrt{x_2} \implies \sqrt{x_2} = 3$.Squaring both sides,$x_2 = 9\, cm$.

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