The maximum potential energy of a block execu | vedclass

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

Question

(C) The total energy of a particle in simple harmonic motion is equal to its maximum potential energy,which is given by $E = \frac{1}{2} k A^2 = 25 \

Vedclass · Accepted Answer

$18.75$

Vedclass · Answer

(C) The total energy of a particle in simple harmonic motion is equal to its maximum potential energy,which is given by $E = \frac{1}{2} k A^2 = 25 \ J$.The potential energy at any displacement $x$ is given by $U = \frac{1}{2} k x^2$.At $x = A / 2$,the potential energy is $U = \frac{1}{2} k (A / 2)^2 = \frac{1}{4} (\frac{1}{2} k A^2) = \frac{1}{4} 	imes 25 \ J = 6.25 \ J$.The kinetic energy $K$ at any position is given by $K = E - U$.Substituting the values,$K = 25 \ J - 6.25 \ J = 18.75 \ J$.

The maximum potential energy of a block executing simple harmonic motion is $25 \ J$. $A$ is the amplitude of oscillation. At $x = A / 2$,the kinetic energy of the block is $...............$ (in $J$)

Similar Questions

In $SHM$,the restoring force is $F = -kx$,where $k$ is the force constant,$x$ is the displacement,and $A$ is the amplitude of motion. Then,the total energy depends upon:

$U$ is the potential energy $(PE)$ of an oscillating particle and $F$ is the force acting on it at a given instant. Which of the following is true?

For a particle executing simple harmonic motion,the ratio of kinetic and potential energies at a point where displacement is one half of the amplitude is

The potential energy of a simple harmonic oscillator of mass $2\, kg$ in its mean position is $5\, J.$ If its total energy is $9\, J$ and its amplitude is $0.01\, m,$ its time period would be

The total energy of a particle,executing simple harmonic motion is

Vedclass Test Series

Exam Paper Generator

Online Exam Module