Forced Vibrations

We consider a spring-mass system to which an external force $F_0 \cos (\omega t)$ is applied, where $F_0$ and $\omega$ are constants. The equation of motion is then $m y'' + \gamma y' + k y = F_0 \cos (\omega t)$. One way of supplying such an external force is by moving the support of the spring up and down, with a displacement $(F_0/k) \cos(\omega t)$.

Undamped Forced Vibrations

We begin with the undamped case: $\gamma=0$. As we have done in the Constant Coefficients: Complex Roots page, we look for a particular solution of the form $y = {\rm Re }z$ where $m z'' + k z = F_0 \exp(i\omega t)$. Moreover, for that equation we would take a trial solution $z = B \exp(i\omega t)$ (in the case where $i\omega$ is not a root of the characteristic equation, i.e. $\omega \ne \omega_0$). We get $B = F_0/(-m\omega^2 + k) = F_0/(m (\omega_0^2 - \omega^2))$ where $\omega_0 = \sqrt{k/m}$ is the natural angular frequency of the spring-mass system (i.e. the homogeneous system has solutions with angular frequency $\omega_0$). Thus our particular solution is

$$y = \frac{F_0}{m(\omega^2-\omega_0^2)} \cos(\omega t)$$

When $\omega_0 > \omega$, i.e. the natural frequency of the system is higher than the frequency of the external force, we find that $y$ is in phase with the external force. Here is a picture of this: the external force is provided by moving the support of the spring up and down (the top curve), and the response $y$ is the bottom curve. In this case $m=k=\omega_0=1$ and $\omega = 2/3$. Note that the mass is up when the support is up, and down when the support is down.

Click on picture for animation.

On the other hand, when $\omega_0 < \omega$, i.e. the natural frequency is lower than the frequency of the external force, the coefficient of $\cos(\omega t)$ is negative, which means that $y$ is 180 degrees out of phase with the external force. Here is a picture of this, with $\omega_0 = 1$ and $\omega = 4/3$. Note that the mass is down when the support is up, and up when the support is down.

Click on picture for animation.

Of course, this is only one particular solution. For the general solution, we must add the general solution of the homogeneous equation, obtaining

$$y = \frac{F_0}{m(\omega^2-\omega_0^2)} \cos(\omega t) + c \cos(\omega_0 t - \delta)$$

For example, consider the initial conditions $y(0) = 0$, $y'(0)=0$. A solution satisfying these initial conditions is $$y = \frac{F_0}{m(\omega^2-\omega_0^2)} (\cos(\omega t) - \cos(\omega_0 t))$$. We can rewrite this using a trigonometric identity:

$$\cos(\omega t) - \cos(\omega_0 t) = 2 \sin\left(\frac{\omega-... ...0}{2} t \right) \sin\left(\frac{\omega+\omega_0}{2} t \right)$$

Here is the case $\omega_0 = 1$, $\omega = 2/3$, the external force coming from moving the support of the spring up and down with amplitude $0.8$.

Click on picture for animation

The effect is that of an angular frequency that is the average of $\omega_0$ and $\omega$, with an amplitude that varies because of the low-frequency factor $$\sin\left(\frac{\omega-\omega_0}{2} t \right)$$. This phenomenon is known as ``beats'': in an audio signal you hear the signal grow louder and softer, dying out when $(\omega - \omega_0) t/2$ is a multiple of $\pi$ and sounding loudest when it is an odd multiple of $\pi/2$. This can be used to tune stringed instruments such as guitars.

Resonance

Our formula for the particular solution doesn't work in the case $\omega = \omega_0$, for which we would have $0$ in the denominator. Instead, our methods for ``Case 2'' tell us to take a trial solution for $m z'' + k z = F_0 \exp(i \omega_0 t)$ of the form $$z = B t \exp(i\omega_0 t)$$. We then get $B = F_0/(2 m \omega_0 i)$ so our particular solution is

$$y = {\rm Re }z = \frac{F_0 t}{2m\omega} \sin(\omega t)$$

This happens to be the solution satisfying initial conditions $y(0)=y'(0) = 0$. It describes an oscillation with an amplitude that grows steadily. This phenomenon is called resonance. Of course, in a real physical system the amplitude can't grow forever, and eventually the differential equation will not be valid. In many cases the reason the equation is no longer valid is that something breaks. This is why resonance is generally a bad thing.

The picture shows our spring-mass system, starting from rest, with the support moving up and down at the resonant frequency with amplitude 0.2. Eventually the mass collides with the support of the spring.

Click on picture for animation.

Damped Forced Vibrations

Now suppose the damping constant $\gamma$ is positive. Again we take $y = {\rm Re }z$ with $m z'' + \gamma z' + k z = F_0 \exp(i\omega t)$, and use a trial solution $z = B \exp(i\omega t)$. This should work since the roots of the characteristic equation will all have negative real parts. We get

$$B = \frac{F_0}{-m\omega^2 + i \gamma \omega + k} = \frac{F_0}{m(\omega_0^2 - \omega^2) + i \gamma \omega}$$

We want to write this in a polar representation $B = R \exp(-i\delta)$ where $R > 0$. Thus

$$R = \vert B\vert = \frac{F_0}{\left\vert m(\omega_0^2 - \omega^2) + i \gamma \omega\right\vert} = \frac{F_0}{\Delta}$$

where

$$\,\vcenter{\openup.7ex\mathsurround=0pt \ialign{\strut\hfil... ...2)/\Delta\cr \sin \delta &= \gamma \omega/\Delta\cr\crcr}}\,$$

For example, if $m = 40$, $k = 90$ and $\gamma=12.4$, we have $\omega_0 = \sqrt{k/m} = 1.5$. Suppose $\omega = 1.6$ and $F_0 = 0.5 k = 45$. We get $\Delta = 23.3963$ so $R = 1.92338$, $\cos \delta = -.529999$, $\sin \delta = .847998$ so $\delta = \cos^{-1} (-.529999) = \pi - \sin^{-1} (.847998) = 2.12940$.

Again, this is one particular solution. It is called the steady-state solution, because it represents a vibration with constant amplitude. A solution of the homogeneous equation, which we add to get any other solution, is called a transient: because of the damping, these contain a factor $\exp(-\lambda t)$ and go to 0 as $t \to \infty$. Thus whatever initial conditions you start with, if you wait long enough the solution will be very close to the steady state solution.

In our example, which is underdamped, $\lambda = \gamma/(2 m) = .155$ and $\mu = \sqrt{k/m -\lambda^2} = 1.49197$. Consider the initial conditions $y(0)=y'(0) = 0$. The steady-state solution has $y(0) = -1.01939$ and $y'(0) = 2.60964$, so the transient will have $y(0)=1.01939$ and $y'(0)=-2.60964$. The transient is $1.01939 \exp(-\lambda t) \cos(\mu t) - 1.64322 \exp(-\lambda t) \sin(\mu t)$. Here is the solution:

Click on picture for animation.

The above example had $\omega_0 < \omega$, which in an undamped system would result in $y$ being 180 degrees or $\pi$ radians out of phase with the external force. With damping, if $\omega_0 < \omega$ we will have $\cos \delta < 0$ and $\sin \delta > 0$, so that $\pi/2 < \delta < \pi$. If $\omega_0 > \omega$ we have $\cos \delta > 0$ and again $\sin \delta > 0$, so $0 < \delta < \pi/2$. Thus damping increases $\delta$ when $\omega_0 > \omega$ (so that the maximum of the response comes after the maximum of the force) and decreases it when $\omega_0 < \omega$ (so that the maximum of the response comes before the minimum of the force). When $\omega_0 = \omega$ we have $\cos \delta = 0$ and $\delta = \pi/2$.

For any given $\omega$ and $\omega_0$ the amplitude of the steady state solution will be less than the amplitude of the corresponding solution of the undamped system, and as the damping constant $r$ increases the amplitude will decrease (because $\Delta$ is an increasing function of $r$). If you keep $\omega$, $m$ and $\gamma$ fixed and allow $k$ (and thus $\omega_0$) to vary, it's easy to see that the maximum amplitude will occur when $\omega_0 = \omega$.

On the other hand, if you fix $m$, $k$ and $\gamma$ and allow $\omega$ to vary, it's not so easy to see where the maximum amplitude will occur. To maximize the amplitude we want to minimize $\Delta$. We may as well minimize $\Delta^2 = m^2 (\omega_0^2 - \omega^2)^2 + \gamma^2 \omega^2$. Letting $x = \omega^2$ for convenience, we have $\Delta^2 = m^2(\omega_0^2 - x)^2 + \gamma^2 x$ so

$$\frac{d}{dx} \Delta^2 = 2 m^2 (x - \omega_0^2) + \gamma^2$$

This is $0$ when $\omega^2 = x = \omega_0^2 - \gamma^2/(2 m^2)$. That is the value that yields the maximum amplitude.

Thus we have seen three effects of damping on forced vibrations:

1. It causes the solutions of the homogeneous equation to die away, so that after a long enough time all you see is the steady state solution.
2. It shifts the phase of the response.
3. It reduces the amplitude of the response.