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]:

Physics · Simple Harmonic Motion

JEE Main 2024 — 5 April, Shift 1 — Question 44

A simple pendulum doing small oscillations at a place $R$ height above earth surface has time period of $T_{1}=4 \mathrm{~s} . \mathrm{T}_{2}$ 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 $[R=$ radius of Earth]:

Option A:
$T_{1}=T_{2}$
Option B:
$2 \mathrm{~T}_{1}=3 \mathrm{~T}_{2}$
Option C:
$3 \mathrm{~T}_{1}=2 \mathrm{~T}_{2}$
Correct
Option D:
$2 \mathrm{~T}_{1}=\mathrm{T}_{2}$

Answer: C

Step-by-step solution

$\mathrm{T}_{1}=2 \pi \sqrt{\frac{\ell}{\mathrm{GM}}(2 \mathrm{R})^{2}}$

$\mathrm{T}_{2}=2 \pi \sqrt{\frac{\ell}{\mathrm{GM}}(3 \mathrm{R})^{2}}$

$\therefore \frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}=\frac{2}{3}$

Answer key and solution verified before publishing.

Practise Simple Harmonic Motion

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free

Exam: JEE Main 2024
Paper: 5 April, Shift 1
Subject: Physics
Chapter: Simple Harmonic Motion
Topic: Simple Pendulum and Angular SHM