(a) A particle executes simple harmonnic motion with an amplitude of `10cm`. At what distance from the mean position are the kinetic and potential energies equal?
(b) The maximum speed and acceleration of a particle executing simple harmonic motion are `10m//s` and `50cm//s`. Find the position(s) of the particle when the speed is `8cm//s`.

Explanation

(a) `K=U`
`1/2momega^2(A^2-x^2)=1/2momega^2x^2`
`A^2-x^2=x^2`
`x^2=A^2/2`
`x=+-A/sqrt2=+-(10)/(sqrt2)=+-5sqrt2cm`
(b) `v_(max)=Aomega=10` (i)
`a_(max)=Aomega^2=50` (ii)
`(ii)//(i)impliesomega=5rad//s, A=2cm`
`v=omegasqrt(A^2-x^2)`
`8=5sqrt((2)^2-x^2)`
`x^2=(2)^2-(1.6)^2=1.44`
`x=+-1.2cm`