Question

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, $\alpha$ = co-efficient if linear expansion) then the frequency of transverse vibration is proportional to

• A. $\frac{\alpha }{\sqrt{\rho Y}}$
• B. $\sqrt{\frac{Y\alpha }{\rho }}$
• C. $\frac{\sigma }{\sqrt{Y\alpha }}$
• D. $\sqrt{\frac{\sigma \alpha }{Y}}$

Answer 1

Answer: (B)

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

Tension in the string is developed in the wire due to cooling.
The compression due to cooling is balanced by the expansion due to tension generated. Thus

$\frac{FL}{AY}=L\alpha \Delta T$

$⟹$ tension $F=YA\alpha \Delta T$

Frequency of vibration is given as $f=\frac{1}{2L}\sqrt{\frac{F}{\mu }}$

$⟹f\propto \sqrt{\frac{YA\alpha \Delta T}{\mu }}$

Since $\rho AL=\mu L$

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