Background

Students should have taken the equivalent of a high school physics class before attempting this laboratory exercise. Knowledge of college-level freshman physics would be helpful but is not required.

The lightly-damped oscillator is a fundamental concept from freshman physics. It consists of a spring with constant $k$, a mass $m$, and a damper with a small damping coefficient $R$ (see Figure 1, left). It is a very simplified model of a plucked vibrating string (see Figure 1, right). Any string has a rest position. If it is pulled away from its rest position (i.e. plucked), then it will vibrate until friction and other losses cause it to cease its motion. Other analogies may be drawn as well.

Figure 1: Lightly-damped oscillator (left), vibrating string (right)

When the lightly-damped oscillator is given an initial condition, such as if the $1$kg mass is pulled away from its resting position to $x(0)=2$m and released, the mass will vibrate back and forth (see Figure 2, top). It is assumed that the force of gravity is negligible in comparison with the force due to the spring. The spring constant $k=(2\pi)^2$, and so the frequency of vibration is $f \approx \frac{1}{2\pi}\sqrt{\frac{k}{m}}=1$ Hz. The energy of the system is stored alternately in the potential energy of the displaced spring $E_{PE}(t)=kx(t)^2/2$ and the kinetic energy of the moving mass $E_{KE}(t)=mv(t)^2/2$. Recall that $v(t)=dx(t)/dt$ is the velocity of the mass. The total energy $E(t)$ is the sum of the potential and kinetic energies.

$$E(t) = E_{PE}(t) + E_{KE}(t)$$ (1)

Figure 2 (middle) shows the behavior of the energies. The damper causes the energy stored in the system to dissipate over time, and so eventually the mass will come to rest again. In fact, the faster the mass moves, the more quickly energy is removed from the system. This means that for small amounts of damping, the function of the energy in the system $E(t)$ decays exponentially over time. $\tau$ is defined to be the time constant for the exponential decay.

$$E(t) = E(0)e^{-t/\tau}$$ (2)

It is a little bit hard to tell what $\tau$ is from a plot of $E(t)$, so often times $10\log_{10}{E(t)}$ is plotted instead because it is a line (see Figure 2, middle).

$$10\log_{10}{E(t)} = 10\log_{10}{(E(0)e^{-t/\tau})} = 10\frac{\log_e{(E(0)e^{-t/\tau})}}{\log_e{10}}$$ (3)

$$10\log_{10}{E(t)} = 10\frac{\log_e{e^{-t/\tau}}}{\log_e{10}}... ...rac{-10t/\tau}{\log_e{10}} + 10\frac{\log_e{E(0)}}{\log_e{10}}$$ (4)

We see that $\tau$ is inversely proportional to the negative of the slope $s$ of $10\log_{10}{E(t)}$. Judging by the red portion of the line in Figure 2 (bottom), we estimate that $s=\frac{-1 -14}{9 - 2}=-2.15$dB/sec.

$$\frac{-10t/\tau}{\log_e{10}} = st$$ (5)

$$\tau = \frac{-10}{s\log_e{10}}$$ (6)

We find that $\tau=2$sec. This is the amount of time it takes for $E(t)$ to decay to $e^{-\tau/\tau}=e^{-1}\approx 0.37$ of its initial value $E(0)$ (see Figure 2 (middle)). Just above, in Figure 2 (top), the time constant for the exponential decay of the displacement $x(t)$ can be seen to be about $4$s. It is in fact exactly 4s. Consider where the extra factor of 2 may come from. (Hint: This has to do with the fact that sometimes $20\log_{10}$ is used instead of $10\log_{10}$ to convert exponential decays to linear decays.)

Figure 2: Displacement of a lightly-damped oscillator (top), energy stored in the oscillator (middle), and log-total energy stored in the oscillator (bottom).