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

(B) $\frac{1}{m(\omega_{0}^{2}-\omega^{2})}$

(A) $\frac{m}{\omega_{0}^{2}-\omega^{2}}$

(C) $\frac{1}{m(\omega_{0}^{2}+\omega^{2})}$

(D) $\frac{m}{\omega_{0}^{2}+\omega^{2}}$

JEE Main · 2004 · 24 Apr 2004 · Offline · Elasticity & SHM

A particle of mass

m

is attached to a spring (of spring constant

k

) and has a natural angular frequency

ω_{0}

. An external force

F (t)

proportional to

cos ω t

(

ω \neq = ω_{0}

) is applied to the oscillator. The time displacement of the oscillator will be proportional to

Answer: (B)

For an oscillator driven by

F_{0} cos ω t

, the steady displacement is

x = A sin (ω t + ϕ)

with amplitude

A = \frac{F _{0} / m}{ω _{0}^{2} - ω ^{2}}

. Treating

F_{0}

as a fixed constant, the amplitude varies as

\frac{1}{m ( ω _{0}^{2} - ω ^{2} )}

, so the displacement is inversely proportional both to the mass and to the gap between the squares of the natural and the driving frequencies.

JEE Main · 2004 · 24 Apr 2004 · Offline · Single correct option