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.

Correct answer: C. 3 T_1=2 T_2 Time period of simple pendulum, T=2 π l g \ and 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

BITSAT 2024Paper 1

Exam: BITSAT
Year: 2024
Paper: Paper 1
Subject: Physics
Chapter: Oscillations

BITSAT 2024 · Paper 1 · Physics · Oscillations

A simple pendulum doing small oscillations at a place R height...

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

]:

Options

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

Correct answer

C. $3 \mathrm{ T}_1=2 \mathrm{ T}_2$

Step-by-step solution

Time period of simple pendulum,

\mathrm{T}=2 \pi \sqrt{\frac{l}{\mathrm{ g}}} \ \text{ and } \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

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