**Question:**

**When a particle of mass $\mathrm{m}$ is attached to a vertical spring of spring constant $k$ and released, its motion is described by $y(t)=y_{0} \sin ^{2} \omega \mathrm{t}$, where ‘ $y$ ‘ is measured from the lower end of unstretched spring. Then $\omega$ is :**

$\frac{1}{2} \sqrt{\frac{g}{y_{0}}}$

$\sqrt{\frac{g}{y_{0}}}$

$\sqrt{\frac{g}{2 y_{0}}}$

$\sqrt{\frac{2 g}{y_{0}}}$

Question of from chapter.

JEE Main Previous Year Sep. 06, 2020 (II)

Correct Option: 3

**Solution:**

### Related Questions

The position co-ordinates of a particle moving in a 3-D coordinate system is given by

$x=\mathrm{a} \cos \omega \mathrm{t}$

$y=a \sin \omega t$

and $z=a \omega t$

The speed of the particle is:

TTwo simple harmonic motions, as shown, are at right angles. They are combined to form Lissajous figures.

$$

\begin{aligned}

&x(t)=A \sin (a t+\delta) \\

&y(t)=B \sin (b t)

\end{aligned}

$$

Identify the correct match below

The ratio of maximum acceleration to maximum velocity in a simple harmonic motion is $10 \mathrm{~s}^{-1}$. At, $\mathrm{t}=0$ the displacement is $5 \mathrm{~m}$. What is the maximum acceleration ? The initial phase is $\frac{\pi}{4}$

A particle performs simple harmonic mition with amplitude A. Its speed is trebled at the instant that it is at a distance $\frac{2 \mathrm{~A}}{3}$ from equilibrium position. The new amplitude of the motion is :

Two particles are performing simple harmonic motion in a straight line about the same equilibrium point. The amplitude and time period for both particles are same and equal to $\mathrm{A}$ and $\mathrm{T}$, respectively. At time $\mathrm{t}=0$ one particle has displacement $A$ while the other one has displacement

$\frac{-\mathrm{A}}{2}$ and they are moving towards each other. If they cross each other at time $\mathrm{t}$, then $\mathrm{t}$ is:

A simple harmonic oscillator of angular frequency $2 \mathrm{rad}$ $\mathrm{s}^{-1}$ is acted upon by an external force $\mathrm{F}=\sin t \mathrm{~N}$. If the oscillator is at rest in its equilibrium position at $t=0$, its position at later times is proportional to:

$x$ and $y$ displacements of a particle are given as $x(t)=a \sin$ $\omega t$ and $y(t)=a \sin 2 \omega t$. Its trajectory will look like :

A body is in simple harmonic motion with time period half second $(T=0.5 \mathrm{~s})$ and amplitude one $\mathrm{cm}(A=1 \mathrm{~cm})$. Find the average velocity in the interval in which it moves form equilibrium position to half of its amplitude.

Which of the following expressions corresponds to simple harmonic motion along a straight line, where $x$ is the displacement and $\mathrm{a}, \mathrm{b}, \mathrm{c}$ are positive constants?

A particle which is simultaneously subjected to two perpendicular simple harmonic motions represented by; $x=a_{1} \cos \omega t$ and $y=a_{2} \cos 2 \omega t$ traces a curve given by: