A particle of mass m is attached to a spring (of spring constant k) and has a natural angular frequency ω₀. An external force F(t) proportional to cos ω t (ω ≠ ω₀) is applied to the oscillator. The time displacement of the oscillator will be proportional to (a) (1)/(m(ω₀²+ω²)) (b) (1)/(m(ω₀²-ω²)) (c) (m)/(ω₀²-ω²) (d) (m)/(ω₀²+ω²)

Answer: (b). Steady state: mx + mω₀² x = F₀cos ω t, try x = Acos ω t. -mω² A + mω₀² A = F₀ ⇒ A = (F₀)/(m(ω₀² - ω²)). So the answer is (b) (1)/(m(ω₀² - ω²)).

A particle of mass m is attached to a spring (of spring constant k)… | JEE Main 2004 (AIEEE)

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Damped, Forced Oscillations & Superposition

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 $\cos\omega t\ (\omega \neq \omega_0)$ is applied to the oscillator. The time displacement of the oscillator will be proportional to

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Answer: (b)

Steady state: $m\ddot{x} + m\omega_0^2 x = F_0\cos\omega t$ , try $x = A\cos\omega t$ .
$-m\omega^2 A + m\omega_0^2 A = F_0 \Rightarrow A = \frac{F_0}{m(\omega_0^2 - \omega^2)}$ .
So the answer is (b) $\frac{1}{m(\omega_0^2 - \omega^2)}$ .

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