A) \[\frac{a}{3}\]
B) \[\frac{a}{2}\]
C) \[\frac{a}{\sqrt{2}}\]
D) \[\frac{a}{2\sqrt{2}}\]
Correct Answer: C
Solution :
For a particle executing SHM with angular velocity g), displacement y, and amplitude a, the potential energy of particle is given by \[PE=\frac{1}{2}m{{\omega }^{2}}{{y}^{2}}\] ... (i) Total energy \[TE=\frac{1}{2}m{{\omega }^{2}}{{a}^{2}}\] ... (ii) Given, \[PE=\frac{1}{2}TE\] \[\therefore \] From Eqs. (i) and (ii), we get \[\frac{1}{2}m{{\omega }^{2}}{{y}^{2}}=\frac{1}{4}m{{\omega }^{2}}{{a}^{2}}\] \[\Rightarrow \] \[y=\frac{a}{\sqrt{2}}\]
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