Damped Oscillation

A damped oscillation is a wave whose amplitude shrinks over time due to friction or some other energy-removing force. Mathematically, it's a sine wave multiplied by a decaying exponential:

$y(t) = A_0\,e^{-\gamma t}\sin(\omega t)$

The exponential factor $e^{-\gamma t}$ is the envelope that defines the wave's "container," shrinking with time at the rate $\gamma$ (the damping coefficient). The sine factor inside oscillates at angular frequency $\omega$.

This describes a guitar string after you pluck it, a pendulum swinging through air, a child losing momentum on a swing — almost any real oscillator.

Step-by-Step Solution

Given: $y(t) = e^{-0.2 t}\sin(4 t)$.

Find: The amplitude envelope, the period of oscillation, and the time it takes to lose half its amplitude.

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Step 1 — Identify the parameters.

Compare to the general form $A_0 e^{-\gamma t}\sin(\omega t)$:

• Initial amplitude: $A_0 = 1$
• Damping coefficient: $\gamma = 0.2\;\text{s}^{-1}$
• Angular frequency: $\omega = 4\;\text{rad/s}$

Step 2 — Find the period of oscillation.

The sine factor has period:

$T = \dfrac{2\pi}{\omega} = \dfrac{2\pi}{4} = \dfrac{\pi}{2} \approx 1.5708\;\text{s}$

The oscillator completes one full swing every $\approx 1.57$ seconds.

Step 3 — Find the half-life of the amplitude.

The envelope $e^{-\gamma t}$ shrinks to half its initial value when:

$e^{-\gamma t_{1/2}} = \dfrac{1}{2}$

Take the natural log of both sides:

$-\gamma t_{1/2} = \ln(1/2) = -\ln 2$

$t_{1/2} = \dfrac{\ln 2}{\gamma} = \dfrac{0.6931}{0.2} \approx 3.466\;\text{s}$

So after about 3.47 seconds, the amplitude has dropped to 50% of where it started.

Step 4 — Compute amplitude at several time points.

• $t = 0$: $A(0) = e^{0} = 1.000$
• $t = 1$: $A(1) = e^{-0.2} \approx 0.819$ (about 82% of initial)
• $t = 5$: $A(5) = e^{-1.0} \approx 0.368$ (37%)
• $t = 10$: $A(10) = e^{-2.0} \approx 0.135$ (13.5%)
• $t = 20$: $A(20) = e^{-4.0} \approx 0.018$ (less than 2%)

After about $5/\gamma = 25$ seconds, the wave is for all practical purposes gone.

Step 5 — Find the energy decay.

Energy is proportional to amplitude squared, so it decays at twice the rate:

$E(t) \propto A(t)^{2} = e^{-2\gamma t}$

The energy half-life is $t_{E,1/2} = \ln 2/(2\gamma) = 0.693/0.4 \approx 1.733\;\text{s}$ — half as long as the amplitude half-life.

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Answer: The damped oscillation $y(t) = e^{-0.2 t}\sin(4 t)$ has:

• Period of oscillation: $T = \pi/2 \approx 1.571\;\text{s}$
• Amplitude envelope: $\pm e^{-0.2 t}$ (decaying exponential)
• Amplitude half-life: $t_{1/2} = \ln 2 / 0.2 \approx 3.466\;\text{s}$
• Energy half-life: $\approx 1.733\;\text{s}$

The motion is a sine wave squeezed between two mirror-image exponential curves, decaying smoothly toward zero.

Try It

• Adjust the damping coefficient $\gamma$ — see the envelope decay faster or slower.
• Adjust the angular frequency $\omega$ — see the wave oscillate faster or slower inside the envelope.
• The dashed orange curves show the envelope $\pm e^{-\gamma t}$ — the wave is always tangent to them at its peaks.
• At $\gamma = 0$ (no damping), the envelope is constant and the wave is just a pure sine.