A weight (orange box) is attached to a linear spring. If the weight is pulled down and then released, it starts swinging up and down.
Left: Oscillation with friction
The oscillation loses energy due to friction, thereby the weight oscillates ever closer around the equilibrium position and eventually stops oscillating.
Right: Oscillation without friction
The weight oscillates smoothly around equilibrium position.
We have described the oscillation without friction in the last section harmonic oscillator. Now the damped oscillation is described.
Physical systems always transfer energy to their surroundings e.g. by friction. They are therefore called damped. If such system is not added energy there won't be any motion at some time. As such perpetual motion machines are not possible (see conservation of energy).
Much of the energy of the pendulum spring is converted into thermal energy during the deformation of the spring. But the air friction can (depending on the section of the weight) also be important.
Most damped oscillations can be described with the help of a damping coefficient \( \delta \). It indicates how strongly the oscillation is dampened.
If you look at how the damping coefficient is incorporated into the oscillation equation, you observe that it does not affect the sine function, but only the amplitude.
\begin{aligned}
s_{harmonic}(t) & = \underset{\text{Amplitude}}{\underbrace{\hspace{1em} s_0 \hspace{1em}}} \cdot \sin (\omega t + \phi_0) \\
& \\
s_{damped}(t) & = \underset{\text{Amplitude}}{\underbrace{ s_0 \cdot e^{-\delta t} }} \cdot \sin (\omega t + \phi_0) \\
\end{aligned}
The first part of the oscillation equation is referred to as amplitude function: $$ \hat{s}(t) = s_0 \cdot e^{-\delta t} $$
Left:
The amplitude function for different \( \delta \) (in gray).
We observe that the amplitude decreases exponentially.
Special case \( \delta = 0 \):
The oscillation is not damped -> It is harmonic.
Example 1:
\( s_0 = 2 m \), \( f = \frac{1}{5} Hz \) und \( \phi_0 = 0 \) und \( \delta = 0,1 \)
If you have the graph of a oscillation or a table of values with the amplitudes, you can calculate the damping coefficient.
Table:
# | Zeit \( t \) | Amplitude \( \hat{s}(t) \) |
---|---|---|
1 | 1,25 \( s \) | 1,76 \( m \) |
2 | 6,25 \( s \) | 1,07 \( m \) |
3 | 11,25 \( s \) | 0,65 \( m \) |
4 | 16,25 \( s \) | 0,39 \( m \) |
5 | 21,25 \( s \) | 0,24 \( m \) | 6 | 26,25 \( s \) | 0,14 \( m \) |
Calculation with known initial amplitude \( s_0 \) and amplitude #2:
\( s_0 = 2 m \), \( t_2 = 6,25 s \) und \( \hat{s}_2 = 1,07 m \)
\begin{aligned}
\hat{s}(t) & = s_0 \cdot e^{-\delta t} \\
& \\
\hat{s}_2 & = s_0 \cdot e^{-\delta t_2} \\
& \\
\dfrac{\hat{s}_2}{s_0} & = e^{-\delta t_2} \\
& \\
\ln \left( \dfrac{\hat{s}_2}{s_0} \right) & = -\delta t_2
& \\
\ln \left( \dfrac{\hat{s}_2}{s_0} \right) / t_2 & = -\delta
& \\
-\ln \left( \dfrac{\hat{s}_2}{s_0} \right) / t_2 & = \delta
& \\
0.1 & = \delta
\end{aligned}
Calculation with two rows from the table:
\( t_2 = 6,25 s \), \( \hat{s}_2 = 1,07 m \), \( t_3 = 11,25 s \) und \( \hat{s}_3 = 0,65 m \)
\begin{eqnarray}
I & \hat{s}_2 & = s_0 \cdot e^{-\delta t_2} \\
II & \hat{s}_3 & = s_0 \cdot e^{-\delta t_3} \\
\end{eqnarray}