A particle undergoing simple harmonic motion | vedclass

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

Question

(D) The displacement is given by $x(t) = A \sin(\omega t)$,where $\omega = \frac{\pi}{90} \ rad/s$.The velocity is $v(t) = \frac{dx}{dt} = A \omega \c

Vedclass · Accepted Answer

$1/3$

Vedclass · Answer

(D) The displacement is given by $x(t) = A \sin(\omega t)$,where $\omega = \frac{\pi}{90} \ rad/s$.The velocity is $v(t) = \frac{dx}{dt} = A \omega \cos(\omega t)$.Kinetic energy $K = \frac{1}{2} m v^2 = \frac{1}{2} m A^2 \omega^2 \cos^2(\omega t)$.Potential energy $U = \frac{1}{2} m \omega^2 x^2 = \frac{1}{2} m \omega^2 A^2 \sin^2(\omega t)$.The ratio of kinetic energy to potential energy is $\frac{K}{U} = \frac{\cos^2(\omega t)}{\sin^2(\omega t)} = \cot^2(\omega t)$.At $t = 210 \ s$,the phase is $\omega t = \frac{\pi}{90} 	imes 210 = \frac{21\pi}{9} = \frac{7\pi}{3} = 2\pi + \frac{\pi}{3}$.Therefore,$\frac{K}{U} = \cot^2\left( 2\pi + \frac{\pi}{3} ight) = \cot^2\left( \frac{\pi}{3} ight) = \left( \frac{1}{\sqrt{3}} ight)^2 = \frac{1}{3}$.

$A$ particle undergoing simple harmonic motion has time-dependent displacement given by $x(t) = A \sin \left( \frac{\pi t}{90} \right)$. The ratio of kinetic energy to potential energy of the particle at $t = 210 \ s$ will be:

Similar Questions

$A$ particle performs $S.H.M.$ of amplitude $A$ with angular frequency $\omega$ along a straight line. When it is at a distance $\frac{\sqrt{3}}{2}A$ from the mean position,its kinetic energy is increased by an amount $\frac{1}{2}m\omega^2A^2$ due to an impulsive force. What is its new amplitude?

$A$ body performs simple harmonic oscillations along the straight line $ABCDE$ with $C$ as the midpoint of $AE.$ Its kinetic energies at $B$ and $D$ are each one-fourth of its maximum value. If $AE = 2R,$ the distance between $B$ and $D$ is

The potential energy of a particle executing $S.H.M.$ is $2.5 \ J$,when its displacement is half of its amplitude. The total energy of the particle is .... $J$.

Which graph represents the difference between total energy and potential energy of a particle executing $SHM$ versus its distance from the mean position?

$A$ bob of a simple pendulum of mass $m$ performs $SHM$ with amplitude $A$ and period $T$. The kinetic energy of the pendulum at displacement $x = \frac{A}{2}$ will be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module