Free Vibrations

Vibrations or oscillations are very important in physics and engineering. You encounter mechanical vibrations when your car goes over a bump in the road. Electrical oscillations are also important: e.g. in a radio receiver. A simple system where we can study these phenomena consists of a mass $m$ suspended by a spring from a rigid support.

Gravity acts with a force $mg$ downwards on the mass. Moving the mass down stretches the spring, and according to Hooke's Law the spring exerts a force in the opposite direction, proportional to the amount of stretching. The spring constant $k$ is the force per unit of stretching (a positive number). The mass can hang motionless at a certain equilibrium position, at which the forces of gravity and the spring are in balance: $ks - mg= 0$, where $s$ is the amount of stretching. We take this equilibrium position as the origin of our coordinate system: our dependent variable $y$ is the displacement of the mass away from the equilibrium position. Boyce and DiPrima take a downward displacement as positive, but I always prefer to take up as positive.

Now if the mass is displaced by $y$, the spring's force is $k(s-y)$ while gravity is still $-mg$, and the net force is $-ky$. If there are no other forces acting on the mass, by Newton's second law we get the differential equation

$$m y'' + k y = 0$$

(where the independent variable is the time $t$). This equation describes free undamped oscillations. It is a linear constant-coefficient homogeneous equation. As we'll see, the motion is periodic: it goes on forever, repeating at regular intervals.

In most real situations, there are other forces acting on the mass that tend to resist the motion and would cause it to die out eventually. For example, there probably is some air resistance. Or some damping mechanism might be deliberately introduced (e.g. the shock absorbers on your car). Whatever produces this damping force, we assume that it is proportional to the velocity and in the opposite direction, with a positive constant $\gamma$ measuring the force per unit of velocity. Then the differential equation becomes

$$m y'' + \gamma y' + k y = 0$$

This equation describes free damped oscillations.

There may also be an external force that can depend on time but not on $y$ or its derivatives. If this force is $f(t)$, the equation is

$$m y'' + \gamma y' + k y = f(t)$$

This is the equation of forced damped oscillations (or undamped if we make $\gamma=0$). For example, the external force may be applied by moving the support at the top of the spring. If $x(t)$ is the amount of displacement of the top of the spring, this produces a force $f(t) = k x(t)$ on the mass.

Similar equations govern oscillations in electrical circuits. Consider a voltage source, resistor, inductor and capacitor in series.

If $I$ is the current in the circuit, the voltage drop across the resistor is $RI$ and that across the inductor is $L I'$, where $R$ is the resistance and $L$ the inductance. The voltage drop across a capacitor of capacitance $C$ is $Q/C$ where $Q$ is the charge on the capacitor. Now $I = Q'$ (current is the rate of flow of charge) so the total voltage drop across resistor, inductor and capacitor is $L Q'' + R Q' + Q/C$. This must be equal to the voltage across the source, say $v_0(t)$. We have the differential equation

$$L Q'' + R Q' + Q/C = v_0(t)$$

which is of the same form as for a mechanical forced, damped oscillation ($L$ corresponds to $m$, $R$ to $r$, $1/C$ to $k$, $v_0$ to $f$ and $Q$ to $y$). We'll concentrate on mechanical systems, but everything is applicable to electrical ones.

Free Undamped Vibrations

$$m y'' + k y = 0$$

The characteristic equation has imaginary roots $\pm \omega_0 i$ where $\omega_0 = \sqrt{k/m}$. A fundamental set of solutions is $\cos (\omega_0 t)$ and $\sin(\omega_0 t)$, which are the real and imaginary parts of $\exp(i \omega_0 t)$. The general solution can be written as $y = c_1 \cos(\omega_0 t) + c_2 \sin(\omega_0 t)$. The solution is periodic, with period $T = 2 \pi/\omega_0$: this means that $y(t+T)=y(t)$ for all $t$. We call $\omega_0$ the angular or circular frequency.

Another way to write the solutions comes from the complex solution $z = C \exp(i \omega_0 t)$ (where $C$ is any complex constant). Write $C$ in its polar representation as $C = R \exp(-i \delta)$ where $R \ge 0$ and $\delta$ is real. Then $z = R \exp(i \omega_0 t - i \delta)$. The real part of $z$, which is also a solution, is $y = R \cos(\omega_0 t - \delta)$. $R$ is called the amplitude of the solution and $\delta$ the phase. Note that $y$ varies from $-R$ to $R$, with $y(t) = R$ when $t = \delta/\omega_0 + nT$ for any integer $n$.

Of course, this is just another way to write the same general solutions. Using the trigonometric identity for cosine of a difference, $y = R \cos (\delta) \cos(\omega_0 t) + R \sin (\delta) \sin(\omega_0 t)$ so $c_1 = R \cos \delta$ and $c_2 = R \sin \delta$.

Free Damped Vibrations

$$m y'' + \gamma y' + k y = 0$$

The characteristic equation has roots

$$\frac{-\gamma \pm \sqrt{\gamma^2 - 4 k m}}{2 m}$$

There are three cases, depending on the sign of $\gamma^2 - 4 k m$.

• (overdamped case) If $\gamma^2 - 4 k m > 0$, the roots are both real. They are negative, since $$\sqrt{\gamma^2 - 4 k m} < \gamma$$. If we call the roots $-\alpha_1$ and $-\alpha_2$, the general solution is
  
  $$y = c_1 \exp(-\alpha_1 t) + c_2 \exp(-\alpha_2 t)$$
  
  
  There is no oscillation: $y(t)$ can change sign at most once. The picture below shows one solution with $m = 40$, $\gamma=24$, $k=2.7$.

  

  Click on the picture for an animation.

• (critically damped case) If $\gamma^2 - 4 k m = 0$, there is a double real root at $-\gamma/(2 m)$. The general solution is
  
  $$y = c_1 \exp(-\gamma t/(2 m)) + c_2 t \exp(-\gamma t/(2 m))$$
  
  
  Again there is no oscillation. I didn't provide a picture for this case, because it looks very much like the last one.
• (underdamped case) If $\gamma^2 - 4 k m < 0$, there is a pair of complex roots at $$(-\gamma \pm i \sqrt{4km - \gamma^2})/(2 m) = -\lambda \pm i \mu$$. Thus
  
  $$\lambda = -\frac{\gamma}{2m} \qquad \qquad\mbox{and}\qquad \... ...\frac{\sqrt{4km-\gamma^2}}{2m} = \sqrt{\frac{k}{m} - \lambda^2}$$
  
  
  A fundamental set of solutions is $\exp(-\lambda t) \cos(\mu t)$ and $\exp(-\lambda t) \sin(\mu t)$. We can also write the general solution as $R \exp(-\lambda t) \cos(\mu t - \delta)$. This oscillates between $R \exp(-\lambda t)$ and $-R \exp(-\lambda t)$. You can think of it as a cosine with an amplitude that decays exponentially. $\mu$ is called the quasi frequency. It's not an actual angular frequency because the motion doesn't repeat exactly. Instead, it repeats at a smaller amplitude:
  
  $$y(t+T) = \exp(-\lambda T) y(t)$$
  
  
  where $T = 2\pi/\mu$ is the quasi-period. Note that $\mu$ is less than the angular frequency $\omega_0 = \sqrt{k/m}$ that the system would have with no damping.

  The picture below shows a solution with $m = 40$, $\gamma=12$, $k=7.3$, resulting in $\lambda = .15$ and $\mu = .4$. The quasi-period is $T= 5\pi \approx 15.73$.

  

  Click on the picture for an animation.