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) = y0 sin2 ω t, where 'y' is measured from the lower end of unstretched spring. Then ω is: (a) √(g/y₀) (b) 1/2√(g/y₀) (c) √(2g/y₀) (d) √(g/2y₀)

Answer: (d). y = y₀sin ²ω t = y₀/2(1-cos 2ω t): SHM about y = y₀/2 with angular frequency 2ω. Mean position is the equilibrium point: ky₀/2 = mg ⇒ k/m = 2g/y₀. 2ω = √(k/m) = √(2g/y₀) ⇒ ω = √(g/2y₀) So the answer is (d).

When a particle of mass m is attached to a vertical spring of spring… | JEE Main 2020 (6 Sep, Shift 2)

Q81JEE Main 2020 · 6 SEP · SHIFT 2Difficult

Spring Systems

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) = y0 sin2 $\omega$ t, where 'y' is measured from the lower end of unstretched spring. Then $\omega$ is:

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Answer: (d)

$y = y_0\sin^2\omega t = \frac{y_0}{2}(1-\cos 2\omega t)$ : SHM about $y = \frac{y_0}{2}$ with angular frequency $2\omega$ .
Mean position is the equilibrium point: $k\frac{y_0}{2} = mg \Rightarrow \frac{k}{m} = \frac{2g}{y_0}$ .
$2\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{2g}{y_0}} \Rightarrow \omega = \sqrt{\frac{g}{2y_0}}$
So the answer is (d).

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Spring Systems question from JEE Main 2020 (6 Sep, Shift 2)

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