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

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

Question

(C) Potential energy is given by $U = \frac{1}{2} m \omega^2 x^2$ and kinetic energy is given by $K = \frac{1}{2} m \omega^2 (A^2 - x^2)$.Given that t

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Potential energy is given by $U = \frac{1}{2} m \omega^2 x^2$ and kinetic energy is given by $K = \frac{1}{2} m \omega^2 (A^2 - x^2)$.Given that the potential energy is $8$ times the kinetic energy at displacement $x$:$U = 8K$Substituting the expressions:$\frac{1}{2} m \omega^2 x^2 = 8 	imes \frac{1}{2} m \omega^2 (A^2 - x^2)$$x^2 = 8(A^2 - x^2)$$x^2 = 8A^2 - 8x^2$$9x^2 = 8A^2$$x^2 = \frac{8A^2}{9}$Taking the square root on both sides:$x = \frac{\sqrt{8} A}{3} = \frac{2\sqrt{2} A}{3}$

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

$A$ 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

If the frequency of a simple harmonic motion is $f$,what is the frequency of its kinetic energy?

$A$ particle executes $SHM$ of amplitude $A$. The distance from the mean position when its kinetic energy becomes equal to its potential energy 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 $160\, J$. Which of the following statements is correct?

For a particle executing simple harmonic motion, the kinetic energy of the particle at a distance of $4 \,cm$ from the mean position is $\frac{1}{3}$ of the maximum kinetic energy. The amplitude of the motion is

Vedclass Test Series

Exam Paper Generator

Online Exam Module