Which one of the following statements is true | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) In simple harmonic motion $(S.H.M.)$,the displacement is given by $y = A \sin(\omega t + \phi)$.The velocity is $v = \frac{dy}{dt} = A\omega \cos(

Vedclass · Accepted Answer

When $v$ is maximum,$a$ is zero.

Vedclass · Answer

(D) In simple harmonic motion $(S.H.M.)$,the displacement is given by $y = A \sin(\omega t + \phi)$.The velocity is $v = \frac{dy}{dt} = A\omega \cos(\omega t + \phi)$.The acceleration is $a = \frac{dv}{dt} = -A\omega^2 \sin(\omega t + \phi) = -\omega^2 y$.At the mean position $(y = 0)$,the speed $v$ is maximum $(v_{max} = A\omega)$ and the acceleration $a$ is zero $(a = -\omega^2(0) = 0)$.At the extreme positions $(y = \pm A)$,the speed $v$ is zero and the acceleration $a$ is maximum $(a_{max} = \pm \omega^2 A)$.Therefore,when $v$ is maximum,$a$ is zero.

Which one of the following statements is true for the speed $v$ and the acceleration $a$ of a particle executing simple harmonic motion?

Similar Questions

The displacement of a particle varies with time as $x = 12 \sin \omega t - 16 \sin^3 \omega t$ (in $cm$). If its motion is $S.H.M.$,then its maximum acceleration is

Write the formula of instantaneous acceleration of $SHM$ particle along $X$-axis.

What is the force constant of a particle executing $SHM$?

Obtain the formula for the acceleration of a particle from the formula of displacement of $SHM$.

$A$ particle is executing $S.H.M.$ Then the graph of acceleration as a function of displacement is

Vedclass Test Series

Exam Paper Generator

Online Exam Module