Question

At what distance from the mean position is the kinetic energy of particular performing SHM of amplitude 8cm three times it's potential energy?

Answer 1

Given :

A particle is performing SHM of amplitude $\tt\blue{8\:cm.}$

To Find :

We have to find distance from mean position at where kinetic energy of particle becomes three times it's potential energy$.$

SoluTioN :

$\star\tt\:P.E.\:at\:distance\:x\:from\:mean\:position,$

• $\boxed{\tt\green{P.E.=\dfrac{1}{2}\times kx^2}}$

$\star\:\tt\:K.E.\:at\:distance\:x\:from\:mean\:position,$

• $\boxed{\tt\orange{K.E.=\dfrac{1}{2}\times k(A^2-x^2)}}$

ATQ,

$:\implies\tt\:K.E.=3(P.E.)$

$:\implies\tt\:\dfrac{1}{2}\times k(A^2-x^2)=3\times \dfrac{1}{2}\times kx^2$

$:\implies\tt\:A^2-x^2=3x^2$

$:\implies\tt\:4x^2=A^2$

$:\implies\tt\:x=\dfrac{A}{2}$

$:\implies\tt\:x=\dfrac{8}{2}$

$:\implies\underline{\boxed{\bf{\pink{x=4\:cm}}}}$