QJEEM21Phfbeb: JEE Main 2021 Physics - Class 11 JEEM21Phfbeb Solution: A mass of 5 kg is connected to a spring. The potential energy…

From the potential energy curve,

${U_{\max }} = {1 \over 2}k{A^2}$

$10 = {1 \over 2}k{(2)^2}$

$\Rightarrow$ k = 5 N/m

The length of the simple pendulum, L = 4 m

Time period of spring,

$T = 2\pi \sqrt {{k \over m}}$

Time period of simple pendulum,

$T = 2\pi \sqrt {{l \over g}}$

The time period of simple pendulum is same as the time period of the spring oscillation.

$\Rightarrow$ $2\pi \sqrt {{l \over g}}$ = $2\pi \sqrt {{k \over m}}$

Substituting the values in the above equations, we get

$2\pi \sqrt {{4 \over g}}$ = $2\pi \sqrt {{5 \over 5}}$

g = 4m/s²

$\therefore$ The acceleration due to the gravity on the planet is 4 m/s².