Question

Differentiating velocity for acceleration in SHM

Answer 1

d/dt[\[\omega \cos (\omega t+ \phi)\] ] on differentiating becomes \[-\omega ^{2} a \sin (\omega t + \phi)\]. why do we change \[\omega \] to \[\omega ^{2}\] ?

Answer 2

Thats how differentiation works.. we are not changing omega to omega squared
\[\frac{ d }{ dt }(\sin(Ax+B)) = \cos(Ax+B)\frac{ d }{ dt}(Ax+B)= \cos(Ax+B)\times A = Acos(Ax+B)\]

hope now you get it!