A simple pendulum has a time period $T_{1}$ when on the earth's surface and $T_{2}$ when taken to a height $R$ above the earth's surface, where $R$ is the radius of the earth. The value of $\frac{T _{2}}{T _{1}}$ is:

Question

Verified by Toppr

Answer 1

Time period of a simple pendulum $= 2 π \frac{l}{g}$
$\Rightarrow g$ on earth's surface $= \frac{G M}{R ^{2}}$
$g$ on a height $R$ above earth's surface $= \frac{G M}{( 2 R ) ^{2}}$
$\Rightarrow T_{1}$ (time period on earth's surface ) $= 2 π \frac{l ( R ^{2} )}{G M}$
$T_{2}$ (time period at a height $^{'} h^{'}$ above earth's surface ) $= 2 π \frac{l ( 2 R ) ^{2}}{G M}$
$\Rightarrow \frac{T _{2}}{T _{1}} = 4 = 2 .$
Hence, the answer is $2 .$

Answer 2

Time period of a simple pendulum

= 2 π \frac{l}{g}

\Rightarrow g

on earth's surface

= \frac{G M}{R ^{2}}

g

on a height

R

above earth's surface

= \frac{G M}{( 2 R ) ^{2}}

\Rightarrow T_{1}

(time period on earth's surface )

= 2 π \frac{l ( R ^{2} )}{G M}

T_{2}

(time period at a height

^{'} h^{'}

above earth's surface )

= 2 π \frac{l ( 2 R ) ^{2}}{G M}

\Rightarrow \frac{T _{2}}{T _{1}} = 4 = 2 .

Hence, the answer is

2 .