A) Simple harmonic function
B) Periodic function
C) Cylindrical functions
D) None of the above.
Correct Answer: A
Solution :
Here : function is \[F(t)=(\sin \omega t+\cos \omega t)\] It can be written as follows: \[F(t)=\sqrt{2}\,\,\left( \frac{1}{\sqrt{2}}\sin \omega t+\frac{1}{\sqrt{2}}\cos \omega t \right)\] \[\Rightarrow \] \[F(t)=\sqrt{2}\left( \cos \frac{\pi }{4}\sin \omega t+\sin \frac{\pi }{4}\cos \omega t \right)\] \[\left( \because \,\,\sin \frac{\pi }{4}=\cos \frac{\pi }{4}=\frac{1}{\sqrt{2}} \right)\] \[\Rightarrow \] \[F(t)=\sqrt{2}\sin \left( \omega t+\frac{\pi }{4} \right)\] \[[\because \,\,\sin A\,\cos B+\cos A\operatorname{sinB}=sin(A+B)]\] \[=\sqrt{2}\sin \left( \omega t+\frac{\pi }{4}+2\pi \right)\] \[(\because \,\,\sin \theta =\sin (\theta +2\pi ))\] \[=\sqrt{2}\,\sin \,\left[ \omega \left( t+\frac{2\pi }{\omega } \right)+\frac{\pi }{4} \right]\] This represent a simple harmonic motion with time period\[\frac{2\pi }{\omega }\].
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