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

A particle is executing SIIM with an amplitude of 0.2 m. At what dista

1.	A particle is executing SIIM with an amplitude of 0.2 m. At what distance from the mean 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)/(2)m^(2)omega^(2)x^(2)`…….(1) K.E. at the displacement x is given by `K.E.= (1)/(2)m^(2)omega^(2)(A^(2)-x^(2))` …………(2) But PE=KE i.e., `(1)/(2)m^(2)omega^(2)x^(2)= (1)/(2)momega^(2)(A^(2)-x^(2))` `IMPLIES x^(2)= A^(2)-x^(2)` or `implies 2X^(2)=A^(2) x^(2)= (A^(2))/(2)` `:. x= (A)/(sqrt(2)).` But A=0.2m `:. x= (0.2)/(sqrt(2))= 0.1 xx sqrt(2)= 0.1414m`