For a particle performing S.H.M.,which of the | vedclass

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

Question

(D) In $S.H.M.$,the displacement is given by $x = A \sin(\omega t + \phi)$.Velocity is given by $v = \frac{dx}{dt} = A\omega \cos(\omega t + \phi)$. A

Vedclass · Accepted Answer

The velocity of the particle is minimum at the mean position.

Vedclass · Answer

(D) In $S.H.M.$,the displacement is given by $x = A \sin(\omega t + \phi)$.Velocity is given by $v = \frac{dx}{dt} = A\omega \cos(\omega t + \phi)$. At the mean position $(x = 0)$,the velocity is maximum $(v_{max} = A\omega)$.Acceleration is given by $a = -\omega^2 x$. At the mean position $(x = 0)$,acceleration is minimum $(a = 0)$.The restoring force $F = -kx$ is always directed towards the mean position.The total energy $E = K.E. + P.E. = \frac{1}{2}kA^2$ remains constant at all times.Therefore,statement $D$ is incorrect because the velocity is maximum,not minimum,at the mean position.

For a particle performing $S.H.M.$,which of the following statements is $INCORRECT$?

Similar Questions

$A$ particle is performing $S.H.M.$ with a maximum velocity $V$. If the amplitude is doubled and the periodic time is reduced to $\left(\frac{1}{3}\right)^{\text{rd}}$ of its original value,then the new maximum velocity is:

$A$ particle executes $S.H.M.$ with a period of $6 \, s$ and amplitude of $3 \, cm$. Its maximum speed in $cm/s$ is

Obtain the velocity of a particle executing simple harmonic motion $(SHM)$ by considering the projection of a particle undergoing uniform circular motion.

Show that for a particle executing $SHM$,velocity and displacement have a phase difference of $\frac{\pi}{2}$.

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:

Vedclass Test Series

Exam Paper Generator

Online Exam Module