A body is oscillating in simple harmonic moti | vedclass

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

Question

(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

$12 \pi^2$

Vedclass · Answer

(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 $|a| = 12 \pi^2 \ m/s^2$.

$A \ (mm \ s^{-2})$	$16$	$8$	$0$	$-8$	$-16$
$x \ (mm)$	$-4$	$-2$	$0$	$2$	$4$

$A$ 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

$A$ man of mass $M$ is standing on a platform. The platform is executing $S.H.M.$ of frequency $f$ in the vertical direction. The span of oscillation is $L$. What is the acceleration of the platform at the top of the oscillation?

For a particle in $SHM$,if the amplitude of the displacement is $a$ and the amplitude of velocity is $v$,the amplitude of acceleration is

$A$ point mass oscillates along the $x$-axis according to the law $x = x_0 \cos(\omega t + \pi/4)$. If the acceleration of the particle is written as $a = A \cos(\omega t + \delta)$,then:

$A$ point mass oscillates along the $x$-axis according to the law $x=x_0 \cos(\omega t - \frac{\pi}{4})$. If the acceleration of the particle is written as $a=A \cos(\omega t + \delta)$,then:

Values of the acceleration $A$ of a particle moving in simple harmonic motion as a function of its displacement $x$ are given in the table below:
$A \ (mm \ s^{-2})$ $16$ $8$ $0$ $-8$ $-16$
$x \ (mm)$ $-4$ $-2$ $0$ $2$ $4$

The period of the motion is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module