Question

A pendulum clock gains $5s$ a day if the temperature is $15^\circ C$ and loses $10s$ a day if the temperature is $30^\circ C.$ What will be value of $x$ if $\alpha =\frac{x}{T}$ . here $\alpha $ is the coefficient of linear expansion of the metal of the pendulum shaft and $\left(\right.T\left.\right)$ is the time period of the pendulum.

Answer 1

Fractional change in time period of pendulum is,
$\frac{\Delta T}{T}=\frac{1}{2}\alpha \Delta t$
Given, when $t_{1}=15^\circ C,\Delta T=5s$
For linear expansion,
$\frac{\Delta T}{T}=\frac{1}{2}\alpha \left(t_{0} - t_{1}\right)$
$\therefore \frac{5}{T}=\frac{1}{2}\alpha \left(t_{0} - 15\right)$
$\ldots .\left(\right.i\left.\right)$
Similarly, when $t_{2}=30^\circ C,\Delta T=10s$
$\therefore \frac{10}{T}=\frac{1}{2}\alpha \left(30 - t_{0}\right)....\left(i i\right)$
Dividing equation $\left(\right.ii\left.\right)$ by $\left(\right.i\left.\right)$ ,
$\frac{10}{5}=\frac{\left(30 - t_{0}\right)}{\left(t_{0} - 15\right)}$
$\therefore 2t_{0}-30=30-t_{0}$
$\therefore t_{0}=20^\circ C$
Substituting it in equation $\left(\right.i\left.\right)$ ,
$\frac{5}{T}=\frac{1}{2}\alpha \left(\right.20-15\left.\right)$
$\therefore \alpha =\frac{2}{T}$
$\therefore x=2$