Question

A metallic wire of length l is held between two rigid supports. If the wire is cooled through a temperature t. (Y$=$Young's modulus of elasticity of wire, $\rho =$ density, $\alpha =$ thermal coefficient linear expansion). Then the frequency of oscillation is proportional to.

• A. $\frac{1}{l}$
• B. $\sqrt{Y}$
• C. $\sqrt{\frac{\alpha }{\rho }}$
• D. $\sqrt{t}$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $\frac{1}{l}$
B $\sqrt{\frac{\alpha }{\rho }}$
C $\sqrt{Y}$
D $\sqrt{t}$

We know that $f=\frac{1}{2l}\sqrt{\frac{T}{\mu }}$

where $T$ is tension and $\mu$ mass per unit length

Now, $f=\frac{1}{2l}\sqrt{\frac{T}{\rho A}}$

Now, $T/A$ is stress and starin,$\Delta l/l=\alpha t$

stress = $Y$ strain

$T/A=Y\alpha t$

$\therefore f=\frac{1}{2l}\sqrt{\frac{Y\alpha t}{\rho }}$

$ALLANWERSARECORRECT$