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 ]:

Correct answer: (C) 3 ~T_1=2 ~T_2. Time period of simple pendulum, T=2 π √(l/(~g)) \ ~g=GM/((R+h)^2) ∴ T ∝(R+h) ⇒ T_1/(~T_2)=((R+h_1 ))/((R+h_2 ))=((R+R))/((R+2 R))=2/3 ⇒ 3 ~T_1=2 ~T_2

Q100 / 100 · BITSAT 2024 (Memory Based Paper 1) · Physics

A simple pendulum doing small oscillations at a place R height above earth surface has time period of $T_1=4 \mathrm{~s}$. $T_2$ 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 $[\mathrm{R}=$ radius of Earth $]:$

A
$T_1=T_2$
B
$2 \mathrm{~T}_1=3 \mathrm{~T}_2$
Correct answer
C
$3 \mathrm{~T}_1=2 \mathrm{~T}_2$
D
$2 \mathrm{~T}_1=\mathrm{T}_2$

Solution

Time period of simple pendulum, $\mathrm{T}=2 \pi \sqrt{\frac{l}{\mathrm{~g}}} \& \mathrm{~g}=\frac{\mathrm{GM}}{(\mathrm{R}+\mathrm{h})^2}$ $\therefore \mathrm{T} \propto(\mathrm{R}+\mathrm{h})$ $\Rightarrow \frac{\mathrm{T}_1}{\mathrm{~T}_2}=\frac{\left(\mathrm{R}+\mathrm{h}_1\right)}{\left(\mathrm{R}+\mathrm{h}_2\right)}=\frac{(\mathrm{R}+\mathrm{R})}{(\mathrm{R}+2 \mathrm{R})}=\frac{2}{3} \Rightarrow 3 \mathrm{~T}_1=2 \mathrm{~T}_2$

Hence, the correct answer is option: 'C'.

A simple pendulum doing small oscillations at a place R height above earth surface has time period of $T_1=4 \mathrm{~s}$. $T_2$ 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 $[\mathrm{R}=$ radius of Earth $]:$

Solution

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