Question

A metal wire with density $\rho$ and Young's modulus Y is stretched between rigid support At temperature T the speed of transverse wave is found to be $v_1$ When temperature is increased to $T+\Delta T$ the speed decreases to $v_2<v_1$ If the change in cross section area of the wire is ignored then the coefficient of linear expansion of the wire is

• A. $\frac{(v_1^2-v_2^2)\rho }{4Y\Delta T}$
• B. $\frac{2(v_1^2-v_2^2)\rho }{Y\Delta T}$
• C. $\frac{(v_1^2-v_2^2)\rho }{Y\Delta T}$
• D. $\frac{(v_1^2-v_2^2)\rho }{2Y\Delta T}$

Answer 1

Answer: (C)

$\frac{(v_1^2-v_2^2)\rho }{Y\Delta T}$

Let $A$ be the cross sectional area of the wire.

Using $\Delta L=L\alpha \Delta T$ and $Y=\frac{FL}{A\Delta L}$

$\therefore$ Change in force due to change in temperature $\Delta F=YA\alpha \Delta T$

Speed of wave at temperature $T$, $v_1=\sqrt{\frac{F}{\mu }}$ where $\mu =\rho A$

$⟹$ $F=\mu v_1^2$ ...........(1)

$\therefore$ Speed of wave at temperature $T+\Delta T$, $v_2=\sqrt{\frac{F-\Delta F}{\mu }}$

$⟹$ $F-\Delta F=\mu v_2^2$

OR $\mu v_1^2-YA\alpha \Delta T=\mu v_2^2$

OR $\alpha =\frac{\rho A(v_1^2-v_2^2)}{YA\Delta T}$ $⟹$ $\alpha =\frac{\rho (v_1^2-v_2^2)}{Y\Delta T}$