The K.E. and P.E. of a particle executing SHM | vedclass

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

Question

(C) For a particle executing $SHM$,the kinetic energy $(KE)$ at displacement $y$ is given by $KE = \frac{1}{2} m \omega^{2} (A^{2} - y^{2})$.The poten

Vedclass · Accepted Answer

$A/\sqrt{2}$

Vedclass · Answer

(C) For a particle executing $SHM$,the kinetic energy $(KE)$ at displacement $y$ is given by $KE = \frac{1}{2} m \omega^{2} (A^{2} - y^{2})$.The potential energy $(PE)$ at displacement $y$ is given by $PE = \frac{1}{2} m \omega^{2} y^{2}$.Given that $KE = PE$,we equate the two expressions:$\frac{1}{2} m \omega^{2} (A^{2} - y^{2}) = \frac{1}{2} m \omega^{2} y^{2}$.Canceling the common terms $\frac{1}{2} m \omega^{2}$ from both sides,we get:$A^{2} - y^{2} = y^{2}$.Rearranging the terms:$A^{2} = 2y^{2}$.Solving for $y$:$y^{2} = \frac{A^{2}}{2} \Rightarrow y = \frac{A}{\sqrt{2}}$.

The $K.E.$ and $P.E.$ of a particle executing $SHM$ with amplitude $A$ will be equal when its displacement is

Similar Questions

$A$ body is executing $S$.$H$.$M$. Its potential energy is '$P_1$' and '$P_2$' at displacements '$x$' and '$y$' respectively. The potential energy at displacement $(x+y)$ is

$A$ simple pendulum of mass $m$ executes $S.H.M.$ with total energy $E$. If at an instant it is at one of the extreme positions,then its linear momentum after a phase shift of $\frac{\pi}{3} \, rad$ will be

In a simple harmonic oscillation,what fraction of total mechanical energy is in the form of kinetic energy,when the particle is midway between mean and extreme position?

The kinetic energy and potential energy of a particle executing simple harmonic motion will be equal,when displacement (amplitude = $a$) is

The potential energy of a particle with displacement $X$ is $U(X)$. The motion is simple harmonic,when ($K$ is a positive constant)

Vedclass Test Series

Exam Paper Generator

Online Exam Module