A simple pendulum doing small oscillations at a place $R$ height above earth surface has time period of $\mathrm{T}_{1}=4 \mathrm{~s}$. $\mathrm{T}_{2}$ would be it's time period if it is brought to a point which is at a height 2$R$ from earth surface. Choose the correct relation [$\mathrm{R}=$ radius of Earth]:

(C) $3 \mathrm{~T}_{1}=2 \mathrm{~T}_{2}$

(A) $\mathrm{T}_{1}=\mathrm{T}_{2}$

(B) $2 \mathrm{~T}_{1}=3 \mathrm{~T}_{2}$

(D) $2 \mathrm{~T}_{1}=\mathrm{T}_{2}$

JEE Main · 2024 · 05 Apr 2024 · Shift 1 · Elasticity & SHM

A simple pendulum doing small oscillations at a place

R

height above earth surface has time period of

T_{1} = 4 s

T_{2}

would be it's time period if it is brought to a point which is at a height 2

R

from earth surface. Choose the correct relation [

R =

radius of Earth]:

Answer: (C)

Time period at height

h

T = 2 π L / g_{h}

with

g_{h} = \frac{g R ^{2}}{( R + h ) ^{2}}

, so

T \propto \frac{R + h}{R}

(i.e.

T \propto R + h

).

At

h = R

T_{1} \propto 2 R

. At

h = 2 R

T_{2} \propto 3 R

. Hence

\frac{T _{1}}{T _{2}} = \frac{2 R}{3 R} = \frac{2}{3} \Rightarrow 3 T_{1} = 2 T_{2} .

JEE Main · 2024 · 05 Apr 2024 · Shift 1 · Single correct option