Time period of simple pendulum on earth's surface is ' $\mathrm{T}$ '. Its time period becomes '…

Time period of simple pendulum on earth's surface is ' $\mathrm{T}$ '. Its time period becomes ' $\mathrm{xT}$ ' when taken to a height $\mathrm{R}$ (equal to earth's radius) above the earth's surface. Then the value of ' $x$ ' will be
  1. 4
  2. 2
  3. $\frac{1}{2}$
  4. $\frac{1}{4}$

Solution

$\mathrm{T}=2 \pi \sqrt{\frac{l}{\mathrm{~g}}}$ At a height 'h' from earth's surface, $\begin{array}{ll} & \mathrm{xT}=2 \pi \sqrt{\frac{l}{g_{\mathrm{h}}}} \\ \therefore \quad & \mathrm{x}=\sqrt{\frac{\mathrm{g}}{\mathrm{g}_{\mathrm{h}}}} \quad \ldots . .(\mathrm{i}) \\ & \text { Now, } \mathrm{g}_{\mathrm{h}}=\frac{\mathrm{GM}}{(\mathrm{R}+\mathrm{h})^2} \\ \therefore \quad & \mathrm{g}_{\mathrm{h}}=\frac{\mathrm{GM}}{4 \mathrm{R}^2} \\ \therefore \quad & \mathrm{g}_{\mathrm{h}}=\frac{\mathrm{g}}{4} \\ \therefore \quad & \text { From equations (i) and (ii), } \\ & \mathrm{x}=\sqrt{\frac{\mathrm{g}}{\mathrm{g} / 4}}=\sqrt{4}=2 \end{array}$ .

Asked in: MHT CET 2023 (12 May Shift 1)

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