JEE PYQ: Oscillations Question 16
Question 16 - 2021 (26 Feb 2021 Shift 1)
Assume that a tunnel is dug along a chord of the earth, at a perpendicular distance $(R/2)$ from the earth’s centre, where $R$ is the radius of the Earth. The wall of the tunnel is frictionless. If a particle is released in this tunnel, it will execute a simple harmonic motion with a time period:
(1) $2\pi\sqrt{\frac{R}{g}}$
(2) $\frac{1}{2\pi}\sqrt{\frac{g}{R}}$
(3) $\frac{2\pi R}{g}$
(4) $\frac{g}{2\pi R}$
Show Answer
Answer: (1)
Solution
The restoring acceleration is $a = \frac{g}{R}x$. $T = 2\pi\sqrt{\frac{R}{g}}$ (independent of chord position).
BETA
