Question 288: Gravitation

Question

A simple pendulum has a periodic time ' $\mathrm{T}_1$ ' when it is on the surface of earth of radius ' $R$ '. Its periodic time is ' $\mathrm{T}_2$ ' when it is taken to a height ' $R$ ' above the earth's surface. The value of $\frac{T_2}{T_1}$ is

Accepted Answer

$$\mathrm{T}_1=2 \pi \sqrt{\frac{l}{\mathrm{~g}}}$$
At a height ' $h=2 R$ ' from earth's surface,
$$\begin{aligned}
\mathrm{T}_2 & =2 \pi \sqrt{\frac{l}{\mathrm{~g}_{\mathrm{h}}}} \
	herefore \quad \frac{\mathrm{~T}_2}{\mathrm{~T}_1} & =\sqrt{\frac{\mathrm{g}}{\mathrm{~g}_{\mathrm{h}}}}\quad	ext{.... (i)}
\end{aligned}$$
Now, $\mathrm{g}_{\mathrm{h}}=\frac{\mathrm{GM}}{(\mathrm{R}+\mathrm{h})^2}$
$	herefore \quad \mathrm{g}_{\mathrm{h}}=\frac{\mathrm{GM}}{4 \mathrm{R}^2}$ $\ldots .(\because \mathrm{R}+\mathrm{h}=2 \mathrm{R})$
$	herefore \quad g_h=\frac{g}{4}$
$$\begin{aligned}
&	herefore \quad 	ext { From equations (i) and (ii), }\
&\frac{T_2}{T_1}=\sqrt{\frac{\mathrm{g}}{\mathrm{~g} / 4}}=2
\end{aligned}$$

Answer

$\sqrt2$

Answer

1

Answer

2

Answer

$\frac{1}{2}$

Gravitation

A simple pendulum has a periodic time ' \mathrm{T}_1 ' when it is on the surface of earth of radius ' R '. Its periodic time is ' \mathrm{T}_2 ' when it is taken to a height ' R ' above the earth's surface. The value of \frac{T_2}{T_1} is

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