A) \[\sin \omega t-\cos \omega t\]
B) \[{{\sin }^{2}}\omega t\]
C) \[\sin \omega t+\sin 2\,\omega t\]
D) \[\sin \omega t-\sin 2\,\omega t\]
Correct Answer: A
Solution :
Here, \[F|t|=\sin \omega t-cos\omega t\] \[\Rightarrow \] \[F(t)=\sqrt{2}\left[ \frac{1}{\sqrt{2}}\operatorname{sinwt}-\frac{1}{\sqrt{2}}\operatorname{coswt} \right]\] \[\Rightarrow \] \[F(t)=\sqrt{2}\left[ \operatorname{sinw}t\cos \frac{\pi }{4}-\operatorname{cosw}t\sin \frac{\pi }{4} \right]\] \[\left( \because \,\,\cos \frac{\pi }{4}=\sin \frac{\pi }{4}=\frac{1}{\sqrt{2}} \right)\] \[F(t)=\sqrt{2}\,\,\sin \,\,\left( wt-\frac{\pi }{4} \right)\] It is clear, this function represent SHM having period. \[T=\frac{2\pi }{w}\] With initial phase \[=-\frac{\pi }{4}\] rad.
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