In many of the books talking about damped simple harmonic motion, the underdamped oscillator is treated as follows:

Newton's second law says $$m\ddot{x} + r\dot{x} + sx = 0 $$where s is stiffness constant, r is viscosity constant and m is mass of the oscillating object. The solution of the differential equation is $$x=Ce^{\alpha t} \, .$$ Putting the above solution in the differential equation, we get $$\alpha = \frac{r}{2m} \pm \sqrt{\frac{r^2}{4m^2} - \frac{s}{m}} \, .$$ In the underdamped case the term in the square root is less than 0 and we have two solutions: \begin{align} x =& e^{-rt/2m}e^{\pm i w't} \\ \text{where} \qquad w' =& \sqrt{\frac{s}{m} - \frac{r^2}{4m^2} } \, . \end{align}

After that, I have two sources where they take the solution to be $$x= e^{-rt/2m}[A \cos(\omega t)+ B \sin(\omega t)] \, .$$

In one of the sources, R.D. Gregory, the author says

The real and imaginary parts of the first complex solution are $$x=\begin{cases} e^{-rt/2m} \cos(\omega t) \\ e^{-rt/2m} \sin(\omega t) \end{cases}$$ and these functions form a basis for the space of real solutions. The general real solution of the damped SHM equation, in this case, is therefore $$x= e^{-rt/2m}[A \cos(\omega t)+ B \sin(\omega t)] \, .$$

Here is one other source where I found the above to be a solution.

Now my confusion lies where the author takes both the real and imaginary parts as the basis for the "real" solution. Can anyone explain to me how the imaginary part is in the basis?


2 Answers 2


As the equation for $x$ is linear and of a second order, there are two independent solutions to it. You may choose two linearly independent complex solutions like $A\cdot exp(\text{i}\omega t)$ and $B\cdot exp(-\text{i}\omega t)$ with any coefficients (complex in general) or two real solutions, especially if your variable is real-valued. $\sin(\omega t)$ and $\cos(\omega t)$ are just linear combinations of your complex exponentials. When you apply the initial conditions $x(0)=x_0,\; \dot{x}(0)=v_0$, you will fix the coefficients $A$ and $B$, which are arbitrary otherwise. For $\sin(\omega t)$ and $\cos(\omega t)$ the resulting coefficients will be real-valued too. For complex exponentials the resulting coefficients will be complex, the unique solution being the same numerically.


\begin{align} c1 \exp(i \omega t) + c2 \exp(-i \omega t) =& c1 (\cos(\omega t)+ i \sin(\omega t)) + c2 (\cos(\omega t) - i \sin(\omega t)) \\ =& (c1+c2) \cos(w t)+i (c1 - c2) \sin(w t) \\ =& A \cos(\omega t) + B \sin(\omega t) \end{align} where $A \equiv c1 + c2$ and $B \equiv i(c1 - c2)$.

  • $\begingroup$ Since the solution is real-valued, then complex $c_2^*=c_1$. So $A=c_1+c_1^*=2\cdot \text{Re}(c_1)$ and $B=-2\cdot \text{Im}(c_1)$ are real coefficients at the real independent solutions. $\endgroup$ Commented Jan 15, 2019 at 7:23
  • $\begingroup$ @VladimirKalitvianski If you solve my equations for c1 and c2 you get c1 = (A - i B)/2 and c2 = (A + i B)/2 which do match your equations, although I'm not sure what you are getting at, since mine work too. Sorry for the lack of LATEX, but I've never used it before today. Maybe I'll get better. $\endgroup$
    – Bill Watts
    Commented Jan 15, 2019 at 8:21

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