JEE MAIN - Physics (2023 - 30th January Evening Shift - No. 17)
For a simple harmonic motion in a mass spring system shown, the surface is frictionless. When the mass of the block is $1 \mathrm{~kg}$, the angular frequency is $\omega_{1}$. When the mass block is $2 \mathrm{~kg}$ the angular frequency is $\omega_{2}$. The ratio $\omega_{2} / \omega_{1}$ is
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$1 / \sqrt{2}$
$1 / 2$
2
$\sqrt{2}$
Explanation
$$\omega = \sqrt {{K \over m}} \Rightarrow \omega \propto {1 \over {\sqrt m }}$$
$${{{\omega _2}} \over {{\omega _1}}} = \sqrt {{{{m_1}} \over {{m_2}}}} = \sqrt {{1 \over 2}} $$
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