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

$A$ 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

$A$ particle of mass $1 \ kg$ is moving in $SHM$ with a path length of $0.01 \ m$ and frequency of $50 \ Hz$. The maximum force in newton,acting on the particle is (in $\pi^2$)

Two simple harmonic motions of angular frequency $100 \, rad \, s^{-1}$ and $1000 \, rad \, s^{-1}$ have the same displacement amplitude. The ratio of their maximum acceleration is

$A$ particle is executing a simple harmonic motion. Its maximum acceleration is $\alpha$ and maximum velocity is $\beta$. Then its frequency of vibration will be

What is the maximum acceleration of a particle performing $SHM$ given by the equation $y = 2\sin \left( {\frac{{\pi t}}{2} + \phi } \right)$,where $y$ is in $cm$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module