Question

A wire of uniform cross-section is suspended vertically from fixed point with a load attached to the other end. What will be the ratio of change in frequency to the altered new frequency due to arise of temperature t ${}^o$C? The coefficient of linear expansion of wire is $\alpha$ and the rise of temperature t is small.

• A. $\frac{\alpha t}{4}$
• B. $\frac{\alpha t}{2}$
• C. $\frac{3\alpha t}{4}$
• D. $\alpha$t

Answer 1

Answer: (B)

$\frac{\alpha t}{2}$

Let $n_1$ and $n_2$ be frequencies at temps $T_1$ and $T_2$ respectively.
$T_2-T_1=T$ as given
$n_1=\frac{v_1}{2l_1}=\frac{1}{2l_1}\sqrt{\frac{T}{{\mu }_1}}$
$n_2=\frac{v_2}{2l_2}=\frac{1}{2l_2}\sqrt{\frac{T}{{\mu }_2}}$
$l_2=l_1(1+\alpha t)$ where $\alpha$ is the coefficient of linear expansion
$\therefore \frac{n_1}{n_2}=\frac{l_2}{l_1}\sqrt{\frac{{\mu }_2}{{\mu }_1}}=\frac{l_2}{l_1}\sqrt{\frac{m{l}_1}{m{l}_2}}=\sqrt{\frac{l_2}{l_1}}=\sqrt{1+\alpha T}=1+\frac{1}{2}\alpha T$
$\therefore \frac{n_1}{n_2}-1=1+\frac{1}{2}\alpha T-1$
$\Rightarrow \frac{n_1-n_2}{n_2}=\frac{1}{2}\alpha T$