Damped oscillation — explain with equation and graph of amplitude decay

Question

Write the equation of motion for a damped harmonic oscillator. Derive the expression for displacement as a function of time and explain how the amplitude decays. Under what condition does the system become overdamped?

(JEE Main 2022, similar pattern)

Solution — Step by Step

A damped oscillator experiences a restoring force $-kx$ and a damping force $-b\dot{x}$ (proportional to velocity, opposing motion):

m\ddot{x} + b\dot{x} + kx = 0

Dividing by $m$ and defining $2\beta = b/m$ (damping coefficient) and $\omega_0^2 = k/m$ (natural frequency):

\ddot{x} + 2\beta\dot{x} + \omega_0^2 x = 0

When $\beta < \omega_0$ (light damping), the solution is:

x(t) = A_0 e^{-\beta t}\cos(\omega_d t + \phi)

where the damped frequency is:

\omega_d = \sqrt{\omega_0^2 - \beta^2}

The amplitude envelope $A_0 e^{-\beta t}$ decays exponentially with time.

The amplitude at time $t$ is:

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

After one time constant $\tau = 2m/b$ , the amplitude drops to $A_0/e \approx 0.37 A_0$ . The energy, being proportional to $A^2$ , decays as $E(t) = E_0 e^{-2\beta t} = E_0 e^{-bt/m}$ .

Critically damped ( $\beta = \omega_0$ ): The system returns to equilibrium fastest without oscillating. $x(t) = (A + Bt)e^{-\beta t}$ .
Overdamped ( $\beta > \omega_0$ ): The system returns to equilibrium slowly without oscillating. $x(t) = C_1 e^{-(\beta - \gamma)t} + C_2 e^{-(\beta + \gamma)t}$ where $\gamma = \sqrt{\beta^2 - \omega_0^2}$ .

Why This Works

In any real oscillator — a pendulum in air, a spring in oil, a car suspension — energy is continuously lost to friction or viscous drag. The damping force $-b\dot{x}$ removes energy proportional to velocity. The exponential decay arises because the rate of energy loss is proportional to the energy itself (since both velocity and amplitude decrease together).

The damped frequency $\omega_d$ is always less than the natural frequency $\omega_0$ . The stronger the damping, the slower the oscillation, until at critical damping the oscillation disappears entirely.

Alternative Method

Using the quality factor $Q = \omega_0/(2\beta)$ : the number of oscillations before amplitude drops to $1/e$ is approximately $Q/\pi$ . High $Q$ means low damping (a tuning fork has $Q \approx 1000$ ). This approach is useful when JEE problems ask “how many oscillations until amplitude halves.”

For JEE, the most commonly tested formula is $A(t) = A_0 e^{-bt/(2m)}$ . If the question says “amplitude reduces to half after 10 oscillations,” set $A_0/2 = A_0 e^{-b \cdot 10T/(2m)}$ and solve for $b$ . The exponential nature of decay is the key — amplitude does not decrease linearly.

Common Mistake

Students often confuse the damping of amplitude with the damping of energy. Amplitude decays as $e^{-\beta t}$ , but energy decays as $e^{-2\beta t}$ (twice as fast, since $E \propto A^2$ ). If a question asks “after what time does energy reduce to 1/e,” the answer is $m/b$ , not $2m/b$ . Always check whether the question asks about amplitude or energy.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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