Underdamped Oscillations

We first consider the situation where $\beta < \omega_0$ which is referred to as underdamped. Defining

$$\omega= \sqrt{\omega_0^2-\beta^2}$$ (2.8)

the two roots which are both complex have values

$$\alpha_1=-\beta + i \omega \, \, \, {\rm and } \, \, \, \alpha_2=-\beta - i \omega$$ (2.9)

The resulting solution is a superposition of the two roots

$$x(t)=e^{-\beta t} [ A_1 e^{i \omega t} + A_2 e^{-i \omega t} ]$$ (2.10)

where $A_1$ and $A_2$ are constants which have to be determined from the initial conditions. The term $[ A_1 e^{i \omega t} + A_1 e^{i \omega t} ]$ is a superposition of $\sin$ and which can be written as

$$x(t)=A e^{-\beta t} \cos(\omega t + \phi)$$ (2.11)

This can also be expressed in the complex notation as

$$\tilde{x}(t)=\tilde{A} e^{(i \omega-\beta )t}$$ (2.12)

where $\tilde{A}= A e^{i \phi}$ is the complex amplitude which has both the amplitude and phase information. Figure 2.1 shows the underdamped motion $x(t)=e^{-t} \cos(2 \pi t)$.

Figure 2.1:

In all cases damping reduces the frequency of the oscillations ie. $\omega < \omega_0$. The main effect of damping is that it causes the amplitude of the oscillations to decay exponentially with time. It is often useful to quantify the decay in the amplitude during the time period of a single oscillation $T=2 \pi/\omega$. This is quantified by the logarithmic decrement which is defined as

$$\lambda=\ln \left[ \frac{x(t)}{x(t+T)} \right]=\frac{2 \pi \beta}{\omega}$$ (2.13)

Problem 1.: An under-damped oscillator with $\tilde{x}(t)=\tilde{A} e^{(i \omega -\beta) t }$ has initial displacement and velocity $x_0$ and $v_0$ respectively. Calculate $\tilde{A}$ and obtain $x(t)$ in terms of the initial conditions.
Solution: $\tilde{A}=x_0-i (v_0+\beta x_0)/\omega$ and $x(t)=e^{-\beta t} \left[ x_0 \cos \omega t + ((v_0 + \beta x_0)/\omega) \, \sin \omega t \right]$.