# 17.3 Forced Oscillation

When the object on the spring experiences an external force, that is some function, $$g(t)$$, the model we have been considering becomes

$mx'' = -kx - fx' + g(t)$

The forcing can be of any kind. We handle that by looking at the response to forcing that is sinusoidal with any given frequency. We do this for three separate reasons.

First, we can solve the resulting equations, and the solutions have properties that are interesting in their own right.

Second, the very same equations arise in many other contexts, such as in the study of electric circuits, and these properties are very important.

Third, the solutions can be used to solve the general problem. Any stimulus can be written as a sum or integral of sinusoidal functions, and these solutions then can be used to obtain a corresponding sum or integral that describes the solution.

Our model is then described by the equation

$mx'' + kx + fx' = A\sin\omega t$

Given any solution to this equation we can add any solution to the equation with $$0$$ as the right hand side term, and we will still have a solution. As we saw in the previous section, such solutions decay exponentially in $$t$$ as long as $$f$$ is non-zero. Solutions to the ‘homogeneous’ equation (with right hand side zero) are called transient solutions because of this decay. Thus we concentrate our attention on steady state solutions, which persist in time because the forcing function persists.

These solutions will have the same frequency and periodicity as the forcing function, and so we look at solutions of the form $$B\sin\omega t + C\cos\omega t$$. We find

$(B(-m\omega^2 + k) - fC\omega - A)\sin\omega t +(C(-\omega^2 + k) + fB\omega)\cos\omega t = 0$

From these we deduce that both coefficients here must vanish which tell us:

$C = \frac{-B\omega f}{k - m\omega^2}$

and

$B = \frac{A}{k - m\omega^2 + \frac{(ωf)^2}{k-m\omega^2}}$

$B = \frac{A(k -m\omega^2)}{(k -m\omega^2)^2 + (\omega f)^2}$

and

$C = \frac{-\omega fA}{(k-m\omega^2)^2 + (\omega f)^2}$

The magnitude of the response to the forcing is $$(B^2 + C^2)^{\frac{1}{2}}$$ and that becomes

$\frac{A}{(k -m\omega^2)^2 + (\omega f)^2}$

The unforced and undamped spring has its ‘natural frequency’ $$\omega_0$$ given by $$\omega_0^2 = \frac{k}{m}$$. The magnitude just described can be written in terms of ω0 as

$\frac{A}{m^2(\omega_0^2 - \omega^2) +(\omega f)^2}$

When $$f$$ is reasonably small compared to $$m$$, this response exhibits a phenomenon called resonance. That is, when $$\omega_0^2 - \omega^2$$ is very small, and $$f^2$$ is small compared to $$m$$, the denominator becomes very small and the response gets very large.