A) \[180{}^\circ \]and \[90{}^\circ \]
B) \[90{}^\circ \]and\[180{}^\circ \]
C) \[0{}^\circ \]and \[90{}^\circ \]
D) \[90{}^\circ \]and\[0{}^\circ \]
Correct Answer: B
Solution :
\[Let\,x=A\sin \omega t\] \[x=A\omega \cos \omega t\] \[\ddot{x}=-A{{\omega }^{2}}\sin \omega \,t=A{{\omega }^{2}}\sin (180{}^\circ +\omega \,t)\]
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