A particle is executing SHM with an amplitude oif 0.2 m. At what distance from themean position the potential energy of the particle will be equal to its kinetic energy?

A particle is executing SHM with an amplitude oif 0.2 m. At what dista

1.	A particle is executing SHM with an amplitude oif 0.2 m. At what distance from themean position the potential energy of the particle will be equal to its kinetic energy?
Answer» Solution :P.E at the DISPLACEMENT x `P.E.=1/2m^(2)OMEGA^(2)x^(2)`………..1 K.E at the displacement x is given by `K.E. =1/2m^(2)omega^(2)(A^(2)-x^(2))`………2 But `PE=KE` i.e. `1/2m^(2)omega^(2)x^(2)=1/2 m omega^(2)(A^(2)-x^(2))` `impliesx^(2)=A^(2)-x^(2)` or `implies2x^(2)=A^(2) x^(2)=(A^(2))/2` `:. x=A/(SQRT(2))` But`A=0.2m` `:.x=(0.2)/(sqrt(2))=0.1xxsqrt(2)=0.1414m`