State an expression for K.E. (kinetic energy) and P.E. (potential energy) at displacement 'x' for a particle performing linear S.H.M. Represent them graphically. Find the displacement at which K.E. is equal to P.E.

Explanation

`K.E.=(1)/(2)momega^(2)(a^(2)-y^(2))`
where, y is the displacement of the particle and a is the amplitude.
and `" "P.E.=(1)/(2)momega^(2)y^(2)`
Now, `" "K.E.=P.E.`

then, `(1)/(2)momega^(2)(a^(2)-y^(2))=(1)/(2)momega^(2)y^(2)`
`implies" "(1)/(2)momega^(2)a^(2)-(1)/(2)momega^(2)y^(2)=(1)/(2)momega^(2)y^(2)`
`implies" "(1)/(2)momega^(2)a^(2)=(1)/(2)momega^(2)y^(2)+(1)/(2)momega^(2)y^(2)`
`implies" "(1)/(2)momega^(2)a^(2)=momega^(2)y^(2)`
`implies" "y=+-(a)/(sqrt(2))`