A) simple harmonic motion of frequency \[\omega /\pi \]
B) simple harmonic motion of frequency\[3\omega /2\pi \]
C) non simple harmonic motion
D) simple harmonic motion of frequency \[\omega /2\pi \]
Correct Answer: C
Solution :
For a particle executing SHM acceleration (a) \[\propto -{{\omega }^{2}}\] displacement (x) ... (i) Given \[x=a{{\sin }^{2}}\omega t\] ...(ii) Differentiating the above equation w.r.t, we get\[\frac{dx}{dt}=2a\omega (\sin \omega t)(\cos \omega t)\] Again differentiating, we get \[\frac{{{d}^{2}}x}{d{{t}^{2}}}=a=2a{{\omega }^{2}}[{{\cos }^{2}}\omega t-{{\sin }^{2}}\omega t]\] \[=2a{{\omega }^{2}}{{\cos }^{2}}\omega t\] The given equation does not satisfy the condition for SHM [Eq. (i)]. Therefore, motion is not simple harmonic.
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