A) \[\frac{m}{\omega _{0}^{2}-{{\omega }^{2}}}\]
B) \[\frac{1}{m(\omega _{0}^{2}-{{\omega }^{2}})}\]
C) \[\frac{1}{m(\omega _{0}^{2}+{{\omega }^{2}})}\]
D) \[\frac{m}{\omega _{0}^{2}+{{\omega }^{2}}}\]
Correct Answer: B
Solution :
Initial angular velocity of particle\[={{\omega }_{0}}\] and at any instant t, angular velocity\[=\omega \] Therefore, for a displacement\[x,\]the resultant acceleration \[f=m(\omega _{0}^{2}-{{\omega }^{2}})x\] ...(i) External force, \[F=m(\omega _{0}^{2}-{{\omega }^{2}})x\] ...(ii) Since, \[F\propto \cos \omega t\] (given) From Eq. (ii), \[m(\omega _{0}^{2}-{{\omega }^{2}})x\propto \cos \omega t\] ...(iii) Now, equation of simple harmonic motion \[x=A\sin (\omega t+\phi )\] ...(iv) At \[t=0;x=A\] \[\therefore \]\[A=A\sin (0+\phi )\] \[\Rightarrow \]\[\phi =\frac{\pi }{2}\] (as\[\sin \phi =1\]) \[\therefore \] \[x=A\sin \left( \omega t+\frac{\pi }{2} \right)=A\cos \omega t\] ...(v) Hence, from Eqs. (iii) and (v). we finally get \[m(\omega _{0}^{2}-{{\omega }^{2}})A\cos \omega t\propto \cos \omega t\] \[\Rightarrow \]\[A\propto \frac{1}{m(\omega _{0}^{2}-{{\omega }^{2}})}\]
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