Question

determine whether sin square omega t represent SHM,periodic but not SH.give the period.

Answer 1

condition of SHM is
d²y/dt² = -w²y
where w is omega ,

y = sin²wt

differentiate wrt t

dy/dt =2w.sinwt.coswt

again differentiate

d²y/dt² = 2w²{ cos²wt -sin²wt } =2w²cos2wt


here we see that ,

d²y/dt² ≠ -w²y

so, this y = sin²wt doesn't perform with SHM .

but this is perpidic ,

becoz f(x ) = f( x + T) valid here .

see this

f(x ) = sin²wt

f(x + π) = sin²( π + wt) = sin²wt

here , f(x ) = f(x + π)

hence , sin²wt is peroidic ,
and peroid = π

Answer 2

Answer:Yes , sin^2(wt) is in SHM . See the proof in the image .

Explanation: