$A$ particle is performing $S.H.M.$ about its mean position with an amplitude $a$ and periodic time $T$. The speed of the particle when its displacement from the mean position is $\frac{a}{3}$ will be:

  • A
    $\frac{2 \pi a}{T}$
  • B
    $\frac{4 \sqrt{2} \pi a}{3 T}$
  • C
    $\frac{4 \pi^2 a}{3 T}$
  • D
    $\frac{\sqrt{3} \pi^2 a}{2 T}$

Explore More

Similar Questions

If a simple pendulum oscillates with an amplitude of $50\, mm$ and time period of $2\, s$,then its maximum velocity is .... $m/s$.

The equations for the displacements of two particles in simple harmonic motion are $y_1=0.1 \sin \left(100 \pi t+\frac{\pi}{3}\right)$ and $y_2=0.1 \cos \pi t$ respectively. The phase difference between the velocities of the two particles at a time $t=0$ is

$A$ particle is performing simple harmonic motion with angular frequency $\omega$ and amplitude $A$. If $a$ is acceleration and $v$ is speed at any instant,then the graph showing the correct variation between $v^2$ and $x^2$ is (where $x$ is displacement):

$A$ particle performs linear $SHM$. At a particular instant,the velocity of the particle is $u$ and the acceleration is $a_1$. At another instant,the velocity is $V$ and the acceleration is $a_2$ $(0 < a_1 < a_2)$. The distance between the two positions is:

The velocity of a particle executing a simple harmonic motion is $13 \ m/s$, when its distance from the equilibrium position $(Q)$ is $3 \ m$ and its velocity is $12 \ m/s$, when it is $5 \ m$ away from $Q$. The frequency of the simple harmonic motion is

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo