A particle starts oscillating simple harmonic | vedclass

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

Question

(A) For a particle in simple harmonic motion $(SHM)$ starting from the equilibrium position,the displacement is given by $x = A \sin(\omega t)$.Given

Vedclass · Accepted Answer

$1: 3$

Vedclass · Answer

(A) For a particle in simple harmonic motion $(SHM)$ starting from the equilibrium position,the displacement is given by $x = A \sin(\omega t)$.Given $t = \frac{T}{12}$ and $\omega = \frac{2\pi}{T}$,the displacement is $x = A \sin(\frac{2\pi}{T} \cdot \frac{T}{12}) = A \sin(\frac{\pi}{6}) = \frac{A}{2}$.The potential energy ($P$.$E$.) is given by $U = \frac{1}{2} k x^2 = \frac{1}{2} m \omega^2 x^2$.Substituting $x = \frac{A}{2}$,we get $U = \frac{1}{2} m \omega^2 (\frac{A^2}{4}) = \frac{1}{8} m \omega^2 A^2$.The kinetic energy ($K$.$E$.) is given by $K = \frac{1}{2} m \omega^2 (A^2 - x^2)$.Substituting $x = \frac{A}{2}$,we get $K = \frac{1}{2} m \omega^2 (A^2 - \frac{A^2}{4}) = \frac{1}{2} m \omega^2 (\frac{3A^2}{4}) = \frac{3}{8} m \omega^2 A^2$.The ratio of potential energy to kinetic energy is $\frac{U}{K} = \frac{\frac{1}{8} m \omega^2 A^2}{\frac{3}{8} m \omega^2 A^2} = \frac{1}{3}$.

$A$ particle starts oscillating simple harmonically from its equilibrium position with time period $T$. What is the ratio of potential energy to kinetic energy of the particle at time $t = \frac{T}{12}$? (Given: $\sin(\frac{\pi}{6}) = \frac{1}{2}$)

Similar Questions

Amplitude of a $SHO$ is $\sqrt{5} \ cm$. At what displacement from the mean position the ratio of kinetic energy to potential energy is $4$ (in $cm$)?

The potential energy of a simple harmonic oscillator of mass $2\, kg$ in its mean position is $5\, J.$ If its total energy is $9\, J$ and its amplitude is $0.01\, m,$ its time period would be

Assertion $(A)$: In $S.H.M$,kinetic and potential energy become equal when the distance is $1/\sqrt{2}$ times its amplitude. Reason $(R)$: The potential energy of a particle executing $S.H.M$ is periodic with time period being maximum at the extreme displacement.

If the amplitude of linear $S.H.M.$ is decreased,then:

$A$ particle is executing simple harmonic motion $(SHM)$ of amplitude $A$ along the $x$-axis about $x = 0$. When its potential energy $(PE)$ equals kinetic energy $(KE)$,the position of the particle will be

Vedclass Test Series

Exam Paper Generator

Online Exam Module