T is the time period of simple pendulum on the earth's surface. Its time period becomes $x$ T when taken to a height R (equal to earth's radius) above the earth's surface. Then, the value of $x$ will be :
Solution
At surface of earth time period<br/><br/>
$\mathrm{T}=2 \pi \sqrt{\frac{\ell}{\mathrm{g}}}$<br/><br/>
At height $\mathrm{h}=\mathrm{R}$<br/><br/>
$$
\begin{aligned}
& \mathrm{g}^{\prime}=\frac{\mathrm{g}}{\left(1+\frac{\mathrm{h}}{\mathrm{R}}\right)^2}=\frac{\mathrm{g}}{4} \\\\
& \therefore \,\mathrm{xT}=2 \pi \sqrt{\frac{\ell}{(\mathrm{g} / 4)}} \\\\
& \Rightarrow \mathrm{xT}=2 \times 2 \pi \sqrt{\frac{\ell}{\mathrm{g}}} \\\\
& \Rightarrow \mathrm{xT}=2 \mathrm{~T} \Rightarrow \mathrm{x}=2
\end{aligned}
$$
About this question
Subject: Physics · Chapter: Oscillations · Topic: Simple Harmonic Motion
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