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(B) When the wire is cooled,it tends to contract,but since it is fixed between rigid supports,a tension $F$ is developed in the wire.The thermal strai

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$\sqrt{\frac{Y\alpha}{\rho}}$

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(B) When the wire is cooled,it tends to contract,but since it is fixed between rigid supports,a tension $F$ is developed in the wire.The thermal strain $\alpha \Delta T$ is equal to the elastic strain $\frac{F}{AY}$.$\frac{F}{AY} = \alpha \Delta T \implies F = Y A \alpha \Delta T$The frequency of the fundamental mode of transverse vibration is given by $f = \frac{1}{2L} \sqrt{\frac{F}{\mu}}$,where $\mu$ is the mass per unit length.Since $\mu = \rho A$,we substitute $F$ and $\mu$ into the frequency formula:$f = \frac{1}{2L} \sqrt{\frac{Y A \alpha \Delta T}{\rho A}} = \frac{1}{2L} \sqrt{\frac{Y \alpha \Delta T}{\rho}}$Since $L, \Delta T$ are constants,$f \propto \sqrt{\frac{Y \alpha}{\rho}}$.

$A$ metallic wire of length $L$ is fixed between two rigid supports. If the wire is cooled through a temperature difference $\Delta T$ ($Y =$ Young's modulus,$\rho =$ density,$\alpha =$ coefficient of linear expansion),then the frequency of transverse vibration is proportional to:

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