Q178 / 225 · JEE Main 2004 · Physics
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 $\cos \omega t\left(\omega \neq \omega_0\right)$ is applied to the oscillator. The time displacement of the oscillator will be proportional to
- A$\frac{\mathrm{m}}{\omega_0^2-\omega^2}$
- Correct answerB$\frac{1}{m\left(\omega_0^2-\omega^2\right)}$
- C$\frac{1}{m\left(\omega_0^2+\omega^2\right)}$
- D$\frac{m}{\omega_0^2+\omega^2}$
Solution
For forced oscillations, the displacement is given by
$$
\mathrm{x}=\mathrm{A} \sin (\omega \mathrm{t}+\phi) \text { with } \mathrm{A}=\frac{\mathrm{F}_0 / \mathrm{m}}{\omega_0^2-\omega^2}
$$
Hence, the correct answer is option: 'B'.