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 $x=A\sin (\omega t+\varphi )$ at $t=0;x=A$ $\therefore$ $A=A\sin (0+\varphi )$ $\Rightarrow$ $\varphi =\frac{\pi }{2}$ $\therefore$ $x=A\sin \left(\omega t+\frac{\pi }{2}\right)=A\cos \omega t$ Hence, from Eqs. (iii) and (v), we finally get $m({\omega }_0^2-{\omega }^2)A\cos \omega t\propto \cos \omega t$ $\Rightarrow$ $A\propto \frac{1}{m({\omega }_0^2-{\omega }^2)}$