The kinetic energy of a particle is `1/2momega^(2)(A^(2)-x^(2))`, where m, `omega` and A are constants. Prove that the motion of the particle is simple harmonic.

Explanation

Kinetic energy, `K=1/2mv^(2)=1/2momega^(2)(A^(2)-x^(2))`.
So, the velocity of the particle is,
`v=pmomegasqrt(A^(2)-x^(2))=pmomega(A^(2)-x^(2))^(1//2)`
`therefore" "(dv)/(dx)=pmomega*1/2(A^(2)-x^(2))^(-1/2)*(-2x)=pm(omegax)/(sqrt(A^(2)-x^(2)))`.
Now, acceleration,
`a=(dv)/(dt)=(dx)/(dt)(dv)/(dx)=v(dv)/(dx)`
= `pmomegasqrt(A^(2)-x^(2))*(pm)(omegax)/(sqrt(A^(2)-x^(2)))=-omega^(2)x`.
So, `a=-omega^(2)x`, which signifies that the motion is simple harmonic.