Oscillations Question 6
Question 6 - 2024 (30 Jan Shift 2)
A simple pendulum is placed at a place where its distance from the earth’s surface is equal to the radius of the earth. If the length of the string is $4 \mathrm{~m}$, then the time period of small oscillations will be s. [ take $\left.\mathrm{g}=\pi^{2} \mathrm{~ms}^{-2}\right]$
Show Answer
Answer: (8)
Solution:
Acceleration due to gravity $\mathrm{g}^{\prime}=\frac{g}{4}$
$\mathrm{T}=2 \pi \sqrt{\frac{4 \ell}{\mathrm{g}}}$
$\mathrm{T}=2 \pi \sqrt{\frac{4 \times 4}{\mathrm{~g}}}$
$\mathrm{T}=2 \pi \frac{4}{\pi}=8 \mathrm{~s}$
