The period of oscillation of the pendulum of a steel clock at 20^… - Vidyadip

Question

The period of oscillation of the pendulum of a steel clock at $20^{\circ} C$ is 2 seconds. If the temperature of the clock increases by $30^{\circ} C$, what will be the gain or loss in measuring time every day? The coefficient of linear expansion of steel is $1.2 \times$ $10^{-5}{ }^{\circ} C ^{-1}$.

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Answer

Given :
$\begin{aligned}\alpha & =1.2 \times 10^{-5}{ }^{\circ} C^{-1} \\\Delta T & =30-20=10^{\circ} C\end{aligned}$
Period of oscillation $=2 sec$
Using the relation $\Delta l=l a \Delta T$
$\begin{array}{l}\frac{\Delta l}{l}=\alpha \Delta T \\\frac{\Delta l}{l}=1.2 \times 10^{-5} \times 10=1.2 \times 10^{-4}\ldots\ldots (1)\end{array}$
Time period $\quad T =2 \pi \sqrt{\frac{l}{g}}\text {let}\ldots\ldots (2)$
When $\Delta l$ increased by $l$ then the period of oscillation is $T ^1$.
$\begin{aligned}T^1 & =2 \pi \sqrt{\frac{l+\Delta l}{g}} \\T^1 & =2 \pi \sqrt{\frac{l}{g}\left(1+\frac{\Delta l}{l}\right)}\ldots\ldots (3)\end{aligned}$
On dividing equation (3) by (2)
$\frac{T^1}{T}=\sqrt{1+\frac{\Delta l}{l}}=\sqrt{1+1.2 \times 10^{-4}}$
$\therefore$ Time loss in one oscillation $= T ^1- T$
Therefore, the daily loss in time
$\begin{array}{l}=\frac{T^1-T}{T} \times 24 \times 3600 \text { second } \\=\left(\frac{T^1}{T}-1\right) \times 24 \times 3600 \\=\left(\sqrt{1+1.2 \times 10^{-4}}-1\right) \times 24 \times 3600 \text { second } \\=\left(\left(1+1.2 \times 10^{-4}\right)^{\frac{1}{2}}-1\right) \times 24 \times 36000\end{array}$
On using binomial theorem, neglecting high order terms
$\begin{array}{l}\left(1+\frac{1}{2} \times 1.2 \times 10^{-4}-1\right) \times 24 \times 36000 \\=\frac{1.2 \times 10^{-4} \times 24 \times 3600}{2} \text { second } \\=5.18 \text { second }\end{array}$

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