For a particle executing S.H.M., its potentia | vedclass

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

Question

(C) For a particle executing $S.H.M.$, the potential energy $(U)$ at displacement $x$ is given by $U = \frac{1}{2} k x^2$.The kinetic energy $(K)$ at

Vedclass · Accepted Answer

$\frac{2\sqrt{2}}{3} A$

Vedclass · Answer

(C) For a particle executing $S.H.M.$, the potential energy $(U)$ at displacement $x$ is given by $U = \frac{1}{2} k x^2$.The kinetic energy $(K)$ at displacement $x$ is given by $K = \frac{1}{2} k (A^2 - x^2)$, where $A$ is the amplitude.According to the problem, $U = 8K$.Substituting the expressions, we get $\frac{1}{2} k x^2 = 8 	imes \frac{1}{2} k (A^2 - x^2)$.Canceling $\frac{1}{2} k$ from both sides, we have $x^2 = 8(A^2 - x^2)$.$x^2 = 8A^2 - 8x^2$.$9x^2 = 8A^2$.$x^2 = \frac{8}{9} A^2$.Taking the square root of both sides, $x = \sqrt{\frac{8}{9}} A = \frac{2\sqrt{2}}{3} A$.

For a particle executing $S.H.M.$, its potential energy is $8$ times its kinetic energy at a certain displacement '$x$' from the mean position. If '$A$' is the amplitude of $S.H.M.$, the value of '$x$' is

Similar Questions

The amplitude of a particle executing $SHM$ is $3\,cm$. The displacement at which its kinetic energy will be $25\%$ more than the potential energy is: $.............cm$.

$A$ particle performs $S.H.M.$ Its potential energies are $U_{1}$ and $U_{2}$ at displacements $x_{1}$ and $x_{2}$ respectively. At displacement $(x_{1} + x_{2})$,its potential energy $U$ is:

$A$ linear harmonic oscillator of force constant $2 \times 10^6 \, N/m$ and amplitude $0.01 \, m$ has a total mechanical energy of $160 \, J$. Its

$A$ particle of mass $m$ performs $SHM$ along a straight line with frequency $f$ and amplitude $A.$

Obtain the expressions for kinetic energy,potential energy,and total energy in simple harmonic motion.

Vedclass Test Series

Exam Paper Generator

Online Exam Module