The oscillation of a body on a smooth horizon | vedclass

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

Question

(C) Given the displacement equation: $x = A \cos \omega t$.The velocity $v$ is the first derivative of displacement with respect to time:$v = \frac{dx

Vedclass · Accepted Answer

Vedclass · Answer

(C) Given the displacement equation: $x = A \cos \omega t$.The velocity $v$ is the first derivative of displacement with respect to time:$v = \frac{dx}{dt} = \frac{d}{dt}(A \cos \omega t) = -A \omega \sin \omega t$.The acceleration $a$ is the derivative of velocity with respect to time:$a = \frac{dv}{dt} = \frac{d}{dt}(-A \omega \sin \omega t) = -A \omega^2 \cos \omega t$.Comparing this with the original displacement equation $x = A \cos \omega t$,we see that $a = -\omega^2 x$. At $t = 0$,$x = A$,so $a = -A \omega^2$. This means the acceleration starts at a negative maximum value and follows a negative cosine curve. Graph $C$ represents this variation correctly.

Similar Questions

$A$ particle of mass $5 \,g$ is executing $S.H.M.$ with an amplitude $0.3 \,m$ and period $\frac{\pi}{5} \,s$. The maximum value of the force acting on the particle is (in $\,N$)

Obtain instantaneous acceleration of a $SHM$ particle with the help of reference circle.

Write the force law for $SHM$.

The acceleration of a particle in $S.H.M.$ is

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module