A body is executing a linear S.H.M. Its poten | vedclass

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

Question

(B) We know that the potential energy $E_P$ of a body in $S.H.M.$ is given by $E_P = \frac{1}{2} Kx^2$,where $K$ is the force constant.For displacemen

Vedclass · Accepted Answer

$(\sqrt{E_1}+\sqrt{E_2})^2$

Vedclass · Answer

(B) We know that the potential energy $E_P$ of a body in $S.H.M.$ is given by $E_P = \frac{1}{2} Kx^2$,where $K$ is the force constant.For displacement $x$,$E_1 = \frac{1}{2} Kx^2 \Rightarrow x = \sqrt{\frac{2E_1}{K}}$.For displacement $y$,$E_2 = \frac{1}{2} Ky^2 \Rightarrow y = \sqrt{\frac{2E_2}{K}}$.The potential energy $E$ at displacement $(x+y)$ is given by $E = \frac{1}{2} K(x+y)^2$.Substituting the values of $x$ and $y$:$E = \frac{1}{2} K \left( \sqrt{\frac{2E_1}{K}} + \sqrt{\frac{2E_2}{K}} \right)^2$$E = \frac{1}{2} K \left( \sqrt{\frac{2}{K}} \right)^2 (\sqrt{E_1} + \sqrt{E_2})^2$$E = \frac{1}{2} K \cdot \frac{2}{K} (\sqrt{E_1} + \sqrt{E_2})^2$$E = (\sqrt{E_1} + \sqrt{E_2})^2$.

$A$ body is executing a linear $S.H.M.$ Its potential energies at the displacements $x$ and $y$ are $E_1$ and $E_2$ respectively. Its potential energy at displacement $(x+y)$ will be

Similar Questions

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

If $T$ is the period of $SHM$,then write the period of kinetic and potential energy.

$A$ mass $0.4 \,kg$ performs $S.H.M.$ with a frequency $\frac{16}{\pi} \,Hz$. At a certain displacement, it has kinetic energy $2 \,J$ and potential energy $1.2 \,J$. The amplitude of oscillation is (in $m$)

The potential energy of a simple harmonic oscillator at the mean position is $2\,J$. If its mean kinetic energy $(K.E.)$ is $4\,J$,its total energy will be .... $J$

$A$ 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