Answer:
In Fig. 10 (NCT). 4(a), \[T=2\,s;\,a=-3\,cm;\]
At
\[t=0\], OP makes an angle \[a=2\cdot 0\] with x-axis, i.e.,
\[\text{
}\!\!\omega\!\!\text{
=}\sqrt{\frac{\text{k}}{\text{m}}}\text{=}\sqrt{\frac{\text{1200}}{\text{3}}}\text{=0}\cdot
4m{{s}^{-1}}\]\[\text{x = a sin }\!\!\omega\!\!\text{ t, we have x = 2 sin 20
t}\] radian. While moving clockwise, here
\[\pi
/2\] \[\text{x = a sin}\left( \text{ }\!\!\omega\!\!\text{ t+}\frac{\text{3
}\!\!\pi\!\!\text{ }}{\text{2}} \right)\text{=a cos }\!\!\omega\!\!\text{ t=2
cos 20 t}\] Thus the (-projection of OP at time t will give us the equation of
S.H.M., given by
\[\text{3 }\!\!\pi\!\!\text{ /2 rad}\text{.}\]
or \[x=-3\,\sin \pi t\]
In Fig. 10 (NCT). 4 (b), \[T=4s;\,a=2m\]
At
t = 0, OP makes an angle 31 with the positive direction of x-axis, i.e., \[\pi
/2\] =\[\phi =\]. While moving anticlockwise, here \[\pi /2\]
Thus
the x-projection of OP at time t will give us the equation of S.H.M. as
\[\phi =\]
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