- Home
- Standard 11
- Physics
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 :
$\frac{\alpha }{{\sqrt {\rho Y} }}$
$\sqrt {\frac{{Y\alpha }}{\rho }} $
$\frac{\rho }{{\sqrt {Y\alpha } }}$
$\sqrt {\frac{{\rho \alpha }}{Y}} $
Solution
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
$\frac{F L}{A Y}=L \alpha \Delta T$
$\Longrightarrow$ tension $F=Y A \alpha \Delta T$
Frequency of vibration is given as $f=\frac{1}{2 L} \sqrt{\frac{F}{\mu}}$
$\Longrightarrow f \propto \sqrt{\frac{Y A \alpha \Delta T}{\mu}}$
since $\rho A L=\mu L$
$f \propto \sqrt{\frac{Y \alpha}{\rho}}$
