A) The motion is oscillatory but not SHM
B) The motion is SHM with amplitude \[a\]
C) The motion is SHM with amplitude \[\alpha \sqrt{2}\]
D) The motion is SHM with amplitude \[2\alpha \]
Correct Answer: C
Solution :
Key Idea: Rearrange the given equation of particle motion. The given equation is written as \[y=a(sin\omega t+cos\omega t)\] 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 \[y=a\sqrt{2}\sin (\omega t+\pi /4)\] Thus, we have seen that the particles motion is simple harmonic with amplitude \[a\sqrt{2}.\] Note: We can represent the resultant equation in angular from as \[\theta ={{\theta }_{0}}sin(\omega t+\pi /4)\] where\[{{\theta }_{0}}\]is amplitude of angular SHM of particle.
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