A) the motion is oscillatory but not SHM
B) the motion is SHM with amplitude a
C) the motion is SHM with amplitude\[a\sqrt{2}\]
D) the motion is SHM with amplitude 2a
Correct Answer: C
Solution :
The equation of particle varying with time is \[y=a(\sin \omega t+\cos \omega t)\] Or \[y=a\sqrt{2}\left( \frac{1}{\sqrt{2}}\sin \omega t+\frac{1}{\sqrt{2}}\cos \omega t \right)\] or \[y=a\sqrt{2}\left( \cos \frac{\pi }{4}\sin \omega t+\sin \frac{\pi }{4}\cos \omega t \right)\] or \[y=a\sqrt{2}\sin \left( \omega t+\frac{\pi }{4} \right)\] ??.(i) This is the equation of simple harmonic motion with amplitude\[a\sqrt{2}\]. Note: We can represent the resultant Eq. (i) in angular SHM as \[\theta ={{\theta }_{0}}\sin \left( \omega t+\frac{\pi }{4} \right)\] where\[{{\theta }_{0}}\]is amplitude of angular SHM of particle.
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