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

For a forced harmonic oscillator, the differential equation of motion is:

$$ m\frac{d^2x}{dt^2} + kx = F_0 \cos(\omega t) $$

We know that the natural angular frequency is $$\omega_0 = \sqrt{\frac{k}{m}}$$, so we can write the spring constant as $$k = m\omega_0^2$$.

The steady-state displacement $$x(t)$$ will oscillate at the driving frequency $$\omega$$, taking the form:

$$ x(t) = A \cos(\omega t) $$

To find the amplitude $$A$$, we need the second derivative:

$$ \frac{d^2x}{dt^2} = -A\omega^2 \cos(\omega t) $$

Substitute $$x(t)$$, its derivative, and $$k$$ back into the equation of motion:

$$ m(-A\omega^2 \cos(\omega t)) + (m\omega_0^2)(A \cos(\omega t)) = F_0 \cos(\omega t) $$

Divide by $$\cos(\omega t)$$:

$$ -mA\omega^2 + mA\omega_0^2 = F_0 $$

$$ A m(\omega_0^2 - \omega^2) = F_0 $$

$$ A = \frac{F_0}{m(\omega_0^2 - \omega^2)} $$

The time displacement magnitude $$A$$ is proportional to the term:

$$ \frac{1}{m(\omega_0^2 - \omega^2)} $$