SIMPLE HARMONIC MOTION_JEE-Main_050424_(S1) - Competishun – Live JEE & NEET Courses and Exam Prep

JEE MAIN 2024

05-04-24 S1

Question

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]:

Select the correct option:

$\mathrm{T}_1=\mathrm{T}_2$

$2 \mathrm{~T}_1=3 \mathrm{~T}_2$

$3 \mathrm{~T}_1=2 \mathrm{~T}_2$

$2 \mathrm{~T}_1=\mathrm{T}_2$

✓ Correct! Well done.

✗ Incorrect. Try again or view the solution.

Solution

$\begin{aligned} & \mathrm{T}_1=2 \pi \sqrt{\frac{\ell}{\mathrm{GM}}(2 \mathrm{R})^2} \\ & \mathrm{~T}_2=2 \pi \sqrt{\frac{\ell}{\mathrm{GM}}(3 \mathrm{R})^2} \\ & \therefore \frac{\mathrm{~T}_1}{\mathrm{~T}_2}=\frac{2}{3}\end{aligned}$

Question Tags

JEE Main

Physics

Medium

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