For a particle performing linear S.H.M. of am | vedclass

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

Question

(C) The potential energy $(P.E.)$ of a particle performing linear $S.H.M.$ is given by $P.E. = \frac{1}{2} k x^2$, where $k = m \omega^2$ is the force

Vedclass · Accepted Answer

$r\sqrt{\lambda}$

Vedclass · Answer

(C) The potential energy $(P.E.)$ of a particle performing linear $S.H.M.$ is given by $P.E. = \frac{1}{2} k x^2$, where $k = m \omega^2$ is the force constant and $x$ is the displacement.The total energy $(T.E.)$ of the particle is given by $T.E. = \frac{1}{2} k r^2$, where $r$ is the amplitude.According to the problem, $P.E. = \lambda 	imes T.E.$Substituting the expressions, we get $\frac{1}{2} k x^2 = \lambda 	imes (\frac{1}{2} k r^2)$.Canceling $\frac{1}{2} k$ from both sides, we get $x^2 = \lambda r^2$.Taking the square root of both sides, we find $x = r\sqrt{\lambda}$.

For a particle performing linear $S.H.M.$ of amplitude '$r$',the potential energy is '$\lambda$' times its total energy. The displacement of the particle is

Similar Questions

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

$A$ 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 ........ $cm$.

The potential energy of a simple harmonic oscillator,when the particle is halfway to its endpoint,is

The displacement of a particle, executing simple harmonic motion with time period $T$, is expressed as $x(t) = A \sin \omega t$, where $A$ is the amplitude. The maximum value of potential energy of this oscillator is found at $t = T / (2 \beta)$. The value of $\beta$ is . . . . . . .

Vedclass Test Series

Exam Paper Generator

Online Exam Module