An object is executing simple harmonic motion with an angular frequency $\omega$. If the maximum velocity is $v_{\max}$,then the maximum acceleration of the object is:

  • A
    $\omega^2 v_{\max}$
  • B
    $\omega v_{\max}$
  • C
    $\omega \sqrt{v_{\max}}$
  • D
    $3 \omega v_{\max}$

Explore More

Similar Questions

$A$ particle executes simple harmonic motion with an amplitude of $4 \ cm$. At the mean position,the velocity of the particle is $10 \ cm/s$. The distance of the particle from the mean position when its speed becomes $5 \ cm/s$ is $\sqrt{\alpha} \ cm$,where $\alpha = $ . . . . . . .

From the given displacement-time graph of an oscillating particle,find the maximum velocity of the particle.

Difficult
View Solution

$A$ $S.H.M.$ has amplitude $a$ and time period $T$. The maximum velocity will be

$A$ particle executes simple harmonic motion with a periodic time of $0.05 \, s$ and an amplitude of $4 \, cm$. What is its maximum velocity?

The plot of velocity $(v)$ versus displacement $(x)$ of a particle executing simple harmonic motion is shown in the figure. The time period of oscillation of the particle 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