$A$ body is vibrating in simple harmonic motion with an amplitude of $0.06\, m$ and frequency of $15\, Hz$. The maximum velocity and maximum acceleration of the body are:

  • A
    $5.65\, m/s$ and $5.32 \times 10^2\, m/s^2$
  • B
    $6.82\, m/s$ and $7.62 \times 10^2\, m/s^2$
  • C
    $8.91\, m/s$ and $8.21 \times 10^2\, m/s^2$
  • D
    $9.82\, m/s$ and $9.03 \times 10^2\, m/s^2$

Explore More

Similar Questions

The maximum velocity of a particle,executing $S$.$H$.$M$. with an amplitude $7 \ mm$ is $4.4 \ ms^{-1}$. The period of oscillation is $\left[\pi=\frac{22}{7}\right]$. (in $s$)

$A$ particle performing $S.H.M.$ starts from the equilibrium position and its time period is $12 \ s$. After $2 \ s$,its velocity is $\pi \ m/s$. The amplitude of the oscillation is: $[\sin 30^{\circ}=\cos 60^{\circ}=0.5, \sin 60^{\circ}=\cos 30^{\circ}=\sqrt{3}/2]$

$A$ point performs simple harmonic oscillation of period $T$ and the equation of motion is given by $x = A \sin(\omega t + \frac{\pi}{6})$. After the elapse of what fraction of the time period will the velocity of the point be equal to half of its maximum velocity?

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

The maximum velocity of a simple harmonic motion represented by $y = 3\sin \left( 100t + \frac{\pi}{6} \right)$ is given by

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