A plucked guitar string or a car bouncing on its springs oscillates, but each swing is smaller than the last. To model that, keep a sinusoid for the back-and-forth and multiply it by a decaying exponential for the shrinking size.
Key idea
$$y = Ae^{-kt}\cos(\omega t) \quad (k > 0)$$ oscillates with period $\frac{2\pi}{\omega}$ between the curves $y = Ae^{-kt}$ and $y = -Ae^{-kt}$, called the envelope. Any decay factor works the same way, such as $A\left(\frac{1}{2}\right)^t\cos(\omega t)$.
Why it works
Because $-1 \le \cos(\omega t) \le 1$, multiplying by $Ae^{-kt}$ gives $-Ae^{-kt} \le y \le Ae^{-kt}$. The graph touches the top of the envelope wherever $\cos(\omega t) = 1$ and the bottom wherever it is $-1$, and crosses zero wherever cosine does, so the timing of the swings does not change. Only their size shrinks.
Worked examples
Worked example: A value
For $y = 8\left(\frac{1}{2}\right)^t\cos(2\pi t)$, find $y$ at $t = 3$.
- $\cos(6\pi) = 1$, so the point is on the top of the envelope.
- $\left(\frac{1}{2}\right)^3 = \frac{1}{8}$, so $y = 8 \cdot \frac{1}{8} \cdot 1 = 1$.
Answer: $y = 1$
Worked example: Writing a model
A mass starts $10$ cm above its rest position, oscillates with period $2$ seconds, and its swings decay like $e^{-0.3t}$. Write $y(t)$.
- At $t = 0$ it is at its highest, so use cosine with $A = 10$.
- $\omega = \frac{2\pi}{2} = \pi$.
Answer: $y(t) = 10e^{-0.3t}\cos(\pi t)$
Worked example: When is it small?
For $y = 16\left(\frac{1}{2}\right)^t\cos(\pi t)$, when has the envelope shrunk to $1$?
- Solve $16\left(\frac{1}{2}\right)^t = 1$, that is $\left(\frac{1}{2}\right)^t = \frac{1}{16} = \left(\frac{1}{2}\right)^4$.
- So $t = 4$. After that every swing is less than $1$ in size.
Answer: $t = 4$
Common mistake
Adding the exponential instead of multiplying. $y = \cos t + e^{-t}$ just slides the midline as the exponential fades; the swings stay the same size. Check: in a damped model, the distance between a peak and the next trough must shrink.