1.

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

Answer»

`(a)/(2)`
`(a)/(sqrt(2))`
`a sqrt(2)`
`(a sqrt(2))/(3)`

SOLUTION :K.E. = P.E.
`(1)/(2)momega^(2)(r^(2)-y^(2))=(1)/(2)m omega^(2)y^(2)`.
`implies""y=( r )/(sqrt(2))=(a)/(sqrt(2))`.
So correct CHOICE is (B).


Discussion

No Comment Found

Related InterviewSolutions