Question

If temperature of a pendulum clock changes from $\theta_{1}$ to $\theta_{2}$, the fractional change in the period of a pendulum clock is :

• A. $\frac{1}{2} \alpha\left(\theta_{2}-\theta_{1}\right)^{2}$
• B. $2 \alpha\left(\theta_{2}-\theta_{1}\right)$
• C. $\frac{1}{2} \alpha\left(\theta_{2}-\theta_{1}\right)$
• D. $2 \alpha\left(\theta_{2}-\theta_{1}\right)^{2}$

Answer 1

Answer: (C)

We use
$T=2 \pi \sqrt{\frac{l}{g}}$
$\frac{T^{\prime}}{T}=\sqrt{\frac{l^{\prime}}{l}}=\sqrt{\frac{l+\Delta l}{l}}$
$=\sqrt{\frac{l+\alpha l \Delta \theta}{l}}$
$\Rightarrow \frac{T^{\prime}}{T} =(1+\alpha \Delta \theta)^{1 / 2}$
$\Rightarrow T' =T(1+r \wedge \theta)^{1 / 2}$
As $\alpha$ is small,
$\Rightarrow T^{\prime}=T\left[1+\frac{1}{2} \alpha \Delta \theta\right]$
$\Rightarrow \Delta T=T^{\prime}-T=\frac{1}{2} \alpha T \Delta \theta$
Fraction change in time period,
$\frac{\Delta T}{T} =\frac{1}{2} \alpha \Delta \theta$
$=\frac{1}{2} \alpha\left(\theta_{2}-\theta_{1}\right)$