A body is oscillating in simple harmonic moti | vedclass

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(A) The displacement equation for simple harmonic motion is given as $x = 6 \cos \left(2 \pi t + \frac{\pi}{3}\right)$.The velocity $v$ is the first d

Vedclass · Accepted Answer

$12 \pi^2$

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(A) The displacement equation for simple harmonic motion is given as $x = 6 \cos \left(2 \pi t + \frac{\pi}{3}ight)$.The velocity $v$ is the first derivative of displacement with respect to time:$v = \frac{dx}{dt} = \frac{d}{dt} \left[6 \cos \left(2 \pi t + \frac{\pi}{3}ight)ight] = -6 \sin \left(2 \pi t + \frac{\pi}{3}ight) 	imes 2 \pi = -12 \pi \sin \left(2 \pi t + \frac{\pi}{3}ight)$.The acceleration $a$ is the derivative of velocity with respect to time:$a = \frac{dv}{dt} = \frac{d}{dt} \left[-12 \pi \sin \left(2 \pi t + \frac{\pi}{3}ight)ight] = -12 \pi \cos \left(2 \pi t + \frac{\pi}{3}ight) 	imes 2 \pi = -24 \pi^2 \cos \left(2 \pi t + \frac{\pi}{3}ight)$.At $t = 1 \ s$, the acceleration is:$a = -24 \pi^2 \cos \left(2 \pi(1) + \frac{\pi}{3}ight) = -24 \pi^2 \cos \left(2 \pi + \frac{\pi}{3}ight)$.Since $\cos(2 \pi + 	heta) = \cos 	heta$, we have $\cos \left(2 \pi + \frac{\pi}{3}ight) = \cos \frac{\pi}{3} = \frac{1}{2}$.Therefore, $a = -24 \pi^2 	imes \frac{1}{2} = -12 \pi^2 \ m/s^2$.The magnitude of acceleration is $|a| = 12 \pi^2 \ m/s^2$.

$A$ body is oscillating in simple harmonic motion according to the equation $x = 6 \cos \left(2 \pi t + \frac{\pi}{3}\right) \ m$. The magnitude of the acceleration (in $m/s^2$) of the body at $t = 1 \ s$ is:

Similar Questions

$Assertion :$ In $SHM$,acceleration is always directed towards the mean position.
$Reason :$ In $SHM$,the body has to stop momentarily at the extreme position and move back to the mean position.

The ratio of maximum acceleration to maximum velocity in a simple harmonic motion is $10\,s^{-1}$. At $t = 0$,the displacement is $5\,m$. What is the maximum acceleration? The initial phase is $\frac{\pi}{4}$.

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

The amplitude of a particle executing $SHM$ is $2 \, cm$ and the force acting at the extreme position on the particle is $4 \, N$. What is the force at the midway point between the mean position and the extreme point (in $, N$)?

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