When a particle of mass $m$ is attached to a vertical spring of spring constant $k$ and released, its motion is described by $\mathrm{y}(\mathrm{t})=\mathrm{y}_{0} \sin ^{2} \omega \mathrm{t}$, where ' y ' is measured from the lower end of unstretched spring. Then $\omega$ is:

(A) $\frac{1}{2} \sqrt{\frac{g}{\mathrm{y}_{0}}}$

(B) $\sqrt{\frac{\mathrm{g}}{\mathrm{y}_{0}}}$

(D) $\sqrt{\frac{2 \mathrm{~g}}{\mathrm{y}_{0}}}$

JEE Main · 2020 · 06 Sep 2020 · Shift 2 · Elasticity & SHM

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} ω t

, where ' y ' is measured from the lower end of unstretched spring. Then

ω

is:

Answer: (C)

Rewrite the motion as

y = y_{0} sin^{2} ω t = \frac{y _{0}}{2} (1 - cos 2 ω t)

, which is simple harmonic about the mean position

y_{0} /2

with angular frequency

2 ω

. At that mean position the spring stretch just balances gravity,

m g = k \frac{y _{0}}{2}

, so

\frac{k}{m} = \frac{2 g}{y _{0}}

. The genuine oscillation frequency is

k / m

, and since that equals

2 ω

we have

2 ω = \frac{2 g}{y _{0}}

, giving

ω = \frac{g}{2 y _{0}}

JEE Main · 2020 · 06 Sep 2020 · Shift 2 · Single correct option