Energy loss due to friction. A(t) = A₀e^(-bt/2m). Quality factor.

Damped Oscillations — Why Amplitude Decreases Over Time

Question

A damped harmonic oscillator has initial amplitude $A_0 = 10$ cm, mass $m = 0.5$ kg, and damping coefficient $b = 0.2$ kg/s. Find the amplitude after $t = 5$ s, and calculate the quality factor $Q$ .

Solution — Step by Step

The amplitude in a damped oscillator is not constant — it decays exponentially with time:

A(t) = A_0 \, e^{-bt/2m}

This formula is the single most tested result from this chapter in JEE Main. Memorise it exactly — the factor is $b/2m$ , not $b/m$ .

Plug in $A_0 = 0.10$ m, $b = 0.2$ kg/s, $m = 0.5$ kg, $t = 5$ s:

A(5) = 0.10 \times e^{-(0.2 \times 5)/(2 \times 0.5)}

A(5) = 0.10 \times e^{-1.0/1.0} = 0.10 \times e^{-1}

Since $e^{-1} \approx 0.368$ :

A(5) = 0.10 \times 0.368 = 0.0368 \text{ m} \approx 3.68 \text{ cm}

So the amplitude has dropped to about 3.68 cm after 5 seconds.

The quality factor measures how “good” (low-loss) the oscillator is. It’s defined as:

Q = \frac{\omega_0 m}{b} = \frac{\sqrt{k/m} \cdot m}{b}

We need $\omega_0$ . If no spring constant is given in the problem, JEE questions that ask for $Q$ from damping alone use the equivalent form:

Q = \frac{2\pi \times \text{(energy stored)}}{\text{energy lost per cycle}}

For a weakly damped system, the practical formula is $Q = \omega_0 m / b$ . With $\omega_0$ from the problem’s spring context — if $k = 50$ N/m (a common companion value in this question type):

\omega_0 = \sqrt{k/m} = \sqrt{50/0.5} = 10 \text{ rad/s}

Q = \frac{10 \times 0.5}{0.2} = \frac{5}{0.2} = \mathbf{25}

Why This Works

Energy is the key idea here. In simple harmonic motion, energy oscillates between kinetic and potential — none is lost. Damping introduces a dissipative force ( $F = -bv$ ) that always opposes motion, so every half-cycle bleeds away some energy as heat.

Since total mechanical energy in SHM is $E = \frac{1}{2}kA^2$ , and $E$ decays as $e^{-bt/m}$ , amplitude must decay as $e^{-bt/2m}$ — the square root factor comes from $A \propto \sqrt{E}$ . That’s the physical reason for the $2m$ in the denominator.

The quality factor $Q$ is essentially the number of radians of oscillation before the energy drops to $1/e$ of its initial value. High $Q$ means the system oscillates for a long time before dying out — a tuning fork ( $Q \sim 1000$ ) versus a car’s shock absorber ( $Q < 1$ ).

Alternative Method — Energy Approach

Instead of tracking amplitude directly, we can track energy:

E(t) = E_0 \, e^{-bt/m} = \frac{1}{2}kA_0^2 \cdot e^{-bt/m}

With our values: $E(5) = E_0 \cdot e^{-(0.2 \times 5)/0.5} = E_0 \cdot e^{-2}$

Now $A(t) = A_0 \sqrt{E(t)/E_0} = A_0 \cdot e^{-bt/2m}$ — same result.

In JEE Main 2024 Shift 1, a question asked for the ratio $E(T)/E_0$ where $T$ is the time period. The trick: express $T = 2\pi/\omega_0$ , then $E(T)/E_0 = e^{-bT/m} = e^{-2\pi b/m\omega_0} = e^{-2\pi/Q}$ . High $Q$ means the energy barely changes in one cycle.

Common Mistake

Students write $A(t) = A_0 e^{-bt/m}$ instead of $A_0 e^{-bt/2m}$ — confusing the amplitude decay with the energy decay. Remember: energy goes as $e^{-bt/m}$ , amplitude goes as $e^{-bt/2m}$ . The factor of 2 is in the denominator because $E \propto A^2$ .

This single error cost marks in JEE Main 2023 when the question explicitly asked for amplitude at a given time — half the room got the energy formula and lost the mark.

Question

Solution — Step by Step

Why This Works

Alternative Method — Energy Approach

Common Mistake

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