If x, v and a denote the displacement, the velocity and the acceleration of a particle executing simple harmonic motion of time period T, show that the expressions `(aT)/xanda^(2)T^(2)+4pi^(2)v^(2)` do not change with time.

Explanation

If A is the amplitude, `omega` is the angular frequency and x is the displacement of a particle executing SHM then,
velocity `v=pmomegasqrt(A^(2)-x^(2))`
acceleration `a=-omega^(2)x`
time period `T=(2pi)/omega`
`(aT)/x=(-omega^(2)x*(2pi)/omega)/x=-2piomega` = constant
`a^(2)T^(2)+4pi^(2)v^(2)=(-omega^(2)x)^(2)((2pi)/omega)^(2)+4pi^(2)omega^(2)(A^(2)-x^(2))`
`=omega^(4)x^(2)(4pi^(2))/omega^(2)+4pi^(2)omega^(2)A^(2)-4pi^(2)omega^(2)x^(2)`
`=4pi^(2)omega^(2)A^(2)` = constant