A particle is executing Simple Harmonic Motio | vedclass

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

Question

(D) The potential energy $(P.E.)$ of a particle in $SHM$ at displacement $x$ is given by $P.E. = \frac{1}{2} kx^2$.The kinetic energy $(K.E.)$ of the

Vedclass · Accepted Answer

$1: 3$

Vedclass · Answer

(D) The potential energy $(P.E.)$ of a particle in $SHM$ at displacement $x$ is given by $P.E. = \frac{1}{2} kx^2$.The kinetic energy $(K.E.)$ of the particle at displacement $x$ is given by $K.E. = \frac{1}{2} k(A^2 - x^2)$,where $A$ is the amplitude.Given that the displacement $x = \frac{A}{2}$.Substituting $x$ into the potential energy formula: $P.E. = \frac{1}{2} k(\frac{A}{2})^2 = \frac{1}{2} k(\frac{A^2}{4}) = \frac{1}{8} kA^2$.Substituting $x$ into the kinetic energy formula: $K.E. = \frac{1}{2} k(A^2 - (\frac{A}{2})^2) = \frac{1}{2} k(A^2 - \frac{A^2}{4}) = \frac{1}{2} k(\frac{3A^2}{4}) = \frac{3}{8} kA^2$.The ratio of potential energy to kinetic energy is $\frac{P.E.}{K.E.} = \frac{\frac{1}{8} kA^2}{\frac{3}{8} kA^2} = \frac{1}{3}$.

$A$ particle is executing Simple Harmonic Motion $(SHM)$. The ratio of potential energy and kinetic energy of the particle when its displacement is half of its amplitude will be

Similar Questions

$A$ particle is executing $S.H.M.$ with total mechanical energy $90 \,J$ and amplitude $6 \,cm$. If its energy is somehow decreased to $40 \,J$,then its amplitude will become ........ $cm$.

$A$ particle of mass $m$ oscillates with simple harmonic motion between points $x_1$ and $x_2$,the equilibrium position being $O$. Its potential energy is plotted. It will be as given below in the graph:

The maximum potential energy of a block executing $S.H.M.$ is $E$ and amplitude of oscillation is $A$. The kinetic energy of the block at amplitude $A/2$ is

$A$ 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: $\sin(\frac{\pi}{6}) = \frac{1}{2}$)

$A$ particle oscillates along the $x$-axis according to the law,$x(t) = x_0 \sin^2\left(\frac{t}{2}\right)$,where $x_0 = 1 \text{ m}$. The kinetic energy $(K)$ of the particle as a function of $x$ is correctly represented by the graph.

Vedclass Test Series

Exam Paper Generator

Online Exam Module