Question

When a particle of mass $m$ is attached to a vertical spring of spring constant $k$ and released, its motion is described by $y\left(t\right)=y_0{\sin }^2\omega t,$ where ${}^{\prime }{y}^{\prime }$ 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. $\sqrt{\frac{2g}{y_0}}$

Answer 1

The correct option is C $\sqrt{\frac{g}{2y_0}}$

Equation of motion,

$y=y_0{\sin }^2\omega t$

$\Rightarrow y=\frac{y_0}{2}(1-\cos 2\omega t)(\because {\sin }^2\omega t=\frac{1-\cos 2\omega t}{2})$

$\Rightarrow y-\frac{y_0}{2}=\frac{-y_0}{2}\cos 2\omega t........\left(1\right)$

And general equation of motion is given by,

$y=A\cos 2\omega t..........\left(2\right)$

Comparing eq.$\left(2\right)$ with eq.$\left(1\right)$, we get

$\therefore$ Amplitude, $A=\frac{y_0}{2}$

Angular velocity $=2\omega$

For equilibrium of mass,

$\frac{k{y}_0}{2}=mg\Rightarrow \frac{k}{m}=\frac{2g}{y_0}$

Also, spring constant $k=m(2\omega {)}^2$

$\Rightarrow 2\omega =\sqrt{\frac{k}{m}}=\sqrt{\frac{2g}{y_0}}$

$\therefore \omega =\frac{1}{2}\sqrt{\frac{2g}{y_0}}=\sqrt{\frac{g}{2y_0}}$

Hence, option $\left(C\right)$ is correct.