Question

A particle of mass m is attached to a spring (of spring constant k) and has a natural angular frequency $ {{\omega }_{0}} $ . An external force F(t) proportional to $ \cos \omega t(\omega \ne {{\omega }_{0}}) $ is applied to the oscillator. The time displacement of the oscillator will be proportional to

• 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}}} $

Answer 1

Answer: (B)

Initial angular velocity of particle $ ={{\omega }_{0}} $ and at any instant t, angular velocity $ =\omega $ Therefore, for a displacement $ x, $ the resultant acceleration $ f=(\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) $ \therefore $ 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 ) $ at $ t=0;x=A $ $ \therefore $ $ A=A\sin (0+\phi ) $ $ \Rightarrow $ $ \phi =\frac{\pi }{2} $ $ \therefore $ $ x=A\sin \left( \omega t+\frac{\pi }{2} \right)=A\cos \omega t $ 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}})} $