Question

A man has an antique pendulum clock of 1832 which bears the signature of the purchaser. He does not want to replace it in the fond memory of his great grandparents. It ticks off one second in each side to side swing. It keeps correct time at ${{20}^{0}}C$. The pendulum shaft is made of steel and its mass can be ignored as compared to the mass of the bob. Linear expansion coefficient of steel is $1.2\times {{10}^{-5}}{{/}^{0}}C$. How many seconds the clock will gain or lose in a day at ${{10}^{0}}C$?
A. gains 5.2s
B. loses 5.2s
C. gains 10.4s
D. loses 10.4s

Answer 1

Hint: We know that there is extension in length in cables or wires made up of metals with increase in the temperature. So, here as T increases the length of the pendulum also increases. The period T of the pendulum is given by $T=2\pi \sqrt{\dfrac{l}{g}}$. So, as temperature increases, the length of the pendulum increases and hence the time period also increases. This means the clock will tick slower, now we will see accordingly in the same manner what is given in the problem.

Complete step by step answer:
We know the time period of a simple pendulum is given by $T=2\pi \sqrt{\dfrac{l}{g}}$ where l is the length of the pendulum and g is the acceleration due to gravity. Also, with the increase in temperature length increases and in that case time period is given by $T=2\pi \sqrt{\dfrac{l-\Delta l}{g}}$.
Using the Fourier theorem, we can write it as.
$T'=2\pi \sqrt{\dfrac{l}{g}}(1-\dfrac{\Delta l}{2l})$
Given a linear expansion coefficient of steel is $1.2\times {{10}^{-5}}{{/}^{0}}C$.
So, at ${{20}^{0}}C$, $T=2\pi \sqrt{\dfrac{l}{g}}$---(1)
Also, at ${{10}^{0}}C$, $T'=2\pi \sqrt{\dfrac{l}{g}}(1-\dfrac{\Delta l}{2l})$--(2)
$\Delta T=T-T' \\
\Rightarrow\Delta T=\dfrac{\Delta l\times T}{2l}\\
\Rightarrow\Delta T=\dfrac{1.2\times {{10}^{-14}}\times 24\times 3600}{2}\\
\Rightarrow\Delta T =5.184s$
$\therefore\Delta T \approx 5.2s$

Since the time period has decreased, it gains time. So, the correct option is A.

Note: While substituting the values we have to be careful that all the units are in SI. The pendulum moves under the influence of restoring force. The tension in the string is always directed towards the centre. We have used the Fourier theorem to solve the square root because the change in the length is very small and hence as such we can neglect the term involving square root of such.