JEE PYQ: Motion In Two Dimensions Question 11
Question 11 - 2020 (06 Sep Shift 2)
When a particle of mass m is attached to a vertical spring of spring constant $k$ and released, its motion is described by $y(t) = y_0 \sin^2 \omega t$, where ‘$y$’ is measured from the lower end of unstretched spring. Then $\omega$ is:
(a) $\frac{1}{2}\sqrt{\frac{g}{y_0}}$ (b) $\sqrt{\frac{g}{y_0}}$ (c) $\sqrt{\frac{g}{2y_0}}$ (d) $\sqrt{\frac{2g}{y_0}}$
Show Answer
Answer: (c)
Solution
$y = \frac{y_0}{2}(1 - \cos 2\omega t)$. Angular velocity $= 2\omega$, amplitude $= y_0/2$. At equilibrium: $ky_0/2 = mg$, so $\omega = \sqrt{g/(2y_0)}$.
BETA
