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

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

Question

(A) The potential energy $U$ of a particle in $S.H.M.$ is given by the formula $U = \frac{1}{2} k x^2$,where $k$ is the restoring force constant and $

Vedclass · Accepted Answer

$12$

Vedclass · Answer

(A) The potential energy $U$ of a particle in $S.H.M.$ is given by the formula $U = \frac{1}{2} k x^2$,where $k$ is the restoring force constant and $x$ is the displacement.From the given graph,at the extreme position $x = 2 \ m$,the potential energy is $U = 24 \ J$.Substituting these values into the formula:$24 = \frac{1}{2} 	imes k 	imes (2)^2$$24 = \frac{1}{2} 	imes k 	imes 4$$24 = 2k$$k = \frac{24}{2} = 12 \ N/m$.Therefore,the value of the restoring force constant is $12 \ N/m$.

For a particle executing $S.H.M.$,the potential energy $V$ versus displacement $x$ graph is given. The value of the restoring force constant is ..... $N/m$.

Similar Questions

$A$ particle is executing simple harmonic motion starting from its mean position. If the time period of the particle is $1.5 \ s$,then the minimum time at which the ratio of the kinetic and total energies of the particle becomes $3: 4$ is

The kinetic energy and potential energy of a particle executing simple harmonic motion will be equal,when displacement (amplitude = $a$) is

If a particle repeats its motion after a fixed time interval of $8 \,s$,then after how much time will its maximum value of $PE$ be attained after attaining its minimum value?

$A$ 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?

For a particle executing simple harmonic motion,the kinetic energy $K$ is given by $K = K_0 \cos^2 \omega t$. The maximum value of potential energy is

Vedclass Test Series

Exam Paper Generator

Online Exam Module