Question

The time period of a pendulum is given by $T=2\pi \sqrt{\dfrac{l}{g}}$. If the temperature rises by $\Delta \theta$ and coefficient of linear expansion of the wire is $\alpha$, find out the change in time period
$A)\text{ }\alpha \Delta \theta T$
$B)\text{ }\dfrac{1}{2}\alpha \Delta \theta T$
$C)\text{ 2}\alpha \Delta \theta T$
$D)\text{ }\dfrac{1}{2}\alpha \Delta \theta$

Answer 1

This problem can be solved by using the direct formula for the change in length due to a rise in temperature of a body in terms of its initial length, coefficient of thermal expansion and the rise in temperature. We can use this in the formula for the time period to find out the change in the time period.

Formula used:
$\Delta L=L\alpha \Delta \theta$

Complete answer:
We will use the formula for the change in length of a body due to change in its temperature, in terms of its coefficient of thermal expansion and initial length.
The change in length $\Delta L$ of a body of initial length $L$ due to a change in temperature $\Delta \theta$ is given by
$\Delta L=L\alpha \Delta \theta$ --(1)
Where $\alpha$ is the coefficient of thermal expansion of the body.
Now, let us analyze the question.
It is given that the initial length of the pendulum is $l$.
The acceleration due to gravity is $g$.
The initial time period of the pendulum is $T=2\pi \sqrt{\dfrac{l}{g}}$ --(2)
The rise in temperature of the pendulum wire is $\Delta \theta$.
The coefficient of thermal expansion of the wire is $\alpha$.
Let the new length of the wire, that is, the new length of the pendulum be $l'$ and the change in the length of the wire due to the rise in temperature be $\Delta l=l'-l$.
Let the new time period of the pendulum be $T'$ and the change in the time period of the pendulum due to the rise in temperature be $\Delta T=T'-T$.
Now, using (1), we get
$\Delta l=l\alpha \Delta \theta$ --(3)
Also,
$\Delta l=l'-l$ --(4)
Putting (4) in (3), we get
$l'-l=l\alpha \Delta \theta$
$\therefore l'=l+l\alpha \Delta \theta =l\left( 1+\alpha \Delta \theta \right)$ --(5)
Now, using (2), we get
$T'=2\pi \sqrt{\dfrac{l'}{g}}$
Putting (5) in the above equation, we get
$T'=2\pi \sqrt{\dfrac{l\left( 1+\alpha \Delta \theta \right)}{g}}$ --(6)
Now, $\Delta T=T'-T$
Putting (2) and (6) in the above equation, we get
$\Delta T=2\pi \sqrt{\dfrac{l\left( 1+\alpha \Delta \theta \right)}{g}}-2\pi \sqrt{\dfrac{l}{g}}=2\pi \sqrt{\dfrac{l}{g}}\left( \sqrt{1+\alpha \Delta \theta }-1 \right)$ --(7)
Now,
$\because \alpha \ll 1$, $\sqrt{1+\alpha \Delta \theta }={{\left( 1+\alpha \Delta \theta \right)}^{\dfrac{1}{2}}}=1+\dfrac{1}{2}\alpha \Delta \theta$ [By binomial expansion, if$\left( \text{ }x\ll 1,{{\left( 1+x \right)}^{n}}=1+nx, \right)$]
Putting this in (7), we get
$\Delta T=2\pi \sqrt{\dfrac{l}{g}}\left( 1+\dfrac{1}{2}\alpha \Delta \theta -1 \right)=2\pi \sqrt{\dfrac{l}{g}}\left( \dfrac{1}{2}\alpha \Delta \theta \right)$
Using (2) in the above equation, we get
$\Delta T=T\dfrac{1}{2}\alpha \Delta \theta =\dfrac{1}{2}\alpha \Delta \theta T$
Therefore, we have got the required expression for the change in the time period of the pendulum.

Therefore, the correct answer is $B)\text{ }\dfrac{1}{2}\alpha \Delta \theta T$.

Note:
This is a complex problem that involves two seemingly different topics, that is the time period of a pendulum and thermal expansion. However, one must realize upon seeing the question that the temperature change changes one of the variables (length) in the equation of the time period and must proceed to thereby, find a relation for the change in the variable. Students must also take note that we have made a good approximation during the calculation by using binomial expansion. This is accurate enough as the thermal expansion coefficient is usually much smaller in comparison to unity.
This problem could also have been solved in a different way by finding out the relative change in the time period as a function of the relative change in the length by using the formula that $\dfrac{\Delta T}{T}=\dfrac{\Delta \left( 2\pi \right)}{2\pi }+\dfrac{1}{2}\dfrac{\Delta l}{l}-\dfrac{1}{2}\dfrac{\Delta g}{g}=0+\dfrac{1}{2}\dfrac{\Delta l}{l}-0=\dfrac{1}{2}\dfrac{\Delta l}{l}$. This in fact, would have yielded a simpler solution.