Q 11-13-069JEE MainJEE Main 2023 (30 Jan, Shift 2)Easy
For a simple harmonic motion in a mass spring system shown, the surface is frictionless. When the mass of the block is $1\ \text{kg}$, the angular frequency is $\omega_1$. When the mass of the block is $2\ \text{kg}$ the angular frequency is $\omega_2$. The ratio $\dfrac{\omega_2}{\omega_1}$ is
Answer: (B) $\dfrac1{\sqrt2}$
$\omega=\sqrt{\dfrac km}\propto\dfrac1{\sqrt m}$, so $\dfrac{\omega_2}{\omega_1}=\sqrt{\dfrac12}=\dfrac1{\sqrt2}$.
Solution by Sreeraj P, M.Sc Physics