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

What is the ratio of the kinetic energy at the mean position to the potential energy at $y = A / 2$ for a particle performing $SHM$ (in $: 1$)?

$A$ particle is vibrating in a simple harmonic motion with an amplitude of $4\, cm$. At what displacement from the equilibrium position is its energy half potential and half kinetic?

For a simple pendulum,a graph is plotted between its kinetic energy $(KE)$ and potential energy $(PE)$ against its displacement $d.$ Which one of the following represents these correctly? (graphs are schematic and not drawn to scale)

For a particle executing $S.H.M.$,the displacement $x$ is given by $x = A \cos \omega t$. Identify the graphs which represent the variation of potential energy $(P.E.)$ as a function of time $t$ and displacement $x$.

$A$ particle is executing simple harmonic motion starting from its mean position. If the time period of the particle is $1.5 \ s$,then the minimum time at which the ratio of the kinetic and total energies of the particle becomes $3: 4$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module