Question

A clock pendulum made of invar has a period of $0.53$ s, at ${20}^{\circ }C.$ If the clock is used in a climate where the temperature averages to ${30}^{\circ }C$, how much time does the clock lose in each oscillation?
(For invar, $\alpha =9{10}^{-7}{/}^{\circ }C,g=$constant)

• A. $2.25\times {10}^{-6}s$
• B. $2.5\times {10}^{-7}s$
• C. $5\times {10}^{-7}s$
• D. $1.125\times {10}^{-4}s$

Answer 1

Answer: (A)

$2.25\times {10}^{-6}s$

We know that time period of a pendulum is given by

$t=2\pi \sqrt{\frac{l}{g}}$

Thus, for small variations in $l$ and $g$, the time period varies as

$\frac{\Delta t}{t}=\frac{1}{2}(\frac{\Delta l}{l}-\frac{\Delta g}{g})$

It is given that $g$ is a constant and $t=0.5\text{s}$

We know that $\Delta l$ occurs due to change in temperature and $\Delta l=l\times \alpha \times \Delta T$

We have $\Delta T={10}^{\circ }C$

Thus, $\frac{\Delta l}{l}=\alpha \Delta T=9\times {10}^{-7}\times 10=9\times {10}^{-6}$

Thus, $\frac{\Delta t}{t}=\frac{1}{2}\frac{\Delta l}{l}=\frac{9}{2}\times {10}^{-6}$

But, $t=0.5$. Thus, $\Delta t=0.5\times 4.5\times {10}^{-6}=2.25\times {10}^{-6}\text{s}$