Question
The displacement vs time of a particle executing $$SHM$$ is shown in figure.
The initial phase $$\phi $$ is
A.
$$ - \pi < \phi < - \frac{\pi }{2}$$
B.
$$\pi < \phi < \frac{{3\pi }}{2}$$
C.
$$ - \frac{{3\pi }}{2} < \phi < - \pi $$
D.
$$\frac{\pi }{2} < \phi < \pi $$
Answer :
$$ - \pi < \phi < - \frac{\pi }{2}$$
Solution :
$$\eqalign{
& {\text{For}}\,\,x = \left( { - A} \right),{\text{we}}\,{\text{have}} \cr
& - A = A\sin \left( {\omega \times 0 + {\phi _0}} \right)\,\,{\text{or}}\,\,{\phi _0} = - \frac{\pi }{2}. \cr
& {\text{So}}\,{\text{for}}\,\,x < \left( { - A} \right),{\phi _0} < \left( { - \frac{\pi }{2}} \right). \cr} $$