Welcome to Lecture 107 of our comprehensive JEE Physics Master Series. In this lecture, we depart from idealized undamped motion to examine realistic oscillatory systems: Damped and Driven (Forced) Oscillations. We formulate the second-order differential equations governing viscous damping, classify the three damping regimes (underdamping, critical damping, and overdamping), analyze energy dissipation and the Quality Factor ($Q$), solve for steady-state forced oscillations, investigate amplitude and velocity resonance, and conclude with 4 graded solved examples followed by a 10-question practice worksheet (DPP) with complete solutions.
1. Damped Harmonic Oscillations
In physical media, oscillating bodies experience resistive forces that dissipate mechanical energy into thermal energy. For low-velocity laminar flow, the retarding damping force $F_d$ is directly proportional to velocity:
$$F_d = -b v = -b \frac{dx}{dt}$$
where $b$ is the damping coefficient ($\text{kg/s}$ or $\text{N}\cdot\text{s/m}$).
Summing restoring force $F_r = -kx$ and damping force $F_d$ via Newton’s second law:
$$m \frac{d^2 x}{dt^2} = -k x – b \frac{dx}{dt} \implies m \frac{d^2 x}{dt^2} + b \frac{dx}{dt} + k x = 0$$
Dividing by $m$, we define the undamped natural angular frequency $\omega_0 = \sqrt{\frac{k}{m}}$ and the damping factor $\gamma = \frac{b}{2m}$:
$$\frac{d^2 x}{dt^2} + 2\gamma \frac{dx}{dt} + \omega_0^2 x = 0$$
2. Classification of Damping Regimes
Using the trial solution $x(t) = e^{\alpha t}$, the auxiliary characteristic equation is $\alpha^2 + 2\gamma \alpha + \omega_0^2 = 0$, giving roots $\alpha = -\gamma \pm \sqrt{\gamma^2 – \omega_0^2}$. Three distinct physical regimes arise:
1. Underdamped Motion ($\gamma < \omega_0$, i.e., $b < 2\sqrt{k m}$):
The roots are complex conjugates: $\alpha = -\gamma \pm i \omega_d$, where $\omega_d = \sqrt{\omega_0^2 – \gamma^2} = \sqrt{\frac{k}{m} – \frac{b^2}{4m^2}}$.
The kinematic displacement is:
$$x(t) = A_0 e^{-\gamma t} \cos(\omega_d t + \phi) = A_0 e^{-\left(\frac{b}{2m}\right)t} \cos(\omega_d t + \phi)$$
The amplitude decays exponentially with time: $A(t) = A_0 e^{-\gamma t}$. The time period of damped oscillation is $T_d = \frac{2\pi}{\omega_d} > T_0$.
2. Critically Damped Motion ($\gamma = \omega_0$, i.e., $b = 2\sqrt{k m}$):
Repeated real roots $\alpha = -\gamma$. The displacement is $x(t) = (C_1 + C_2 t)e^{-\gamma t}$. The system returns to equilibrium in the shortest possible time without oscillating (essential for analog galvanometers and vehicle shock absorbers).
3. Overdamped Motion ($\gamma > \omega_0$, i.e., $b > 2\sqrt{k m}$):
Two distinct real negative roots. The displacement returns to equilibrium sluggishly without crossing the equilibrium point more than once.
3. Energetics, Relaxation Time & Quality Factor ($Q$)
The total mechanical energy of an underdamped oscillator is $E(t) \approx \frac{1}{2} k [A(t)]^2 = \frac{1}{2} k A_0^2 e^{-2\gamma t} = E_0 e^{-2\gamma t}$.
1. Energy Relaxation Time ($\tau$): Time required for total energy to drop to $1/e$ of its initial value:
$$\tau = \frac{1}{2\gamma} = \frac{m}{b}$$
2. Logarithmic Decrement ($\delta$): The natural logarithm of the ratio of two successive amplitudes separated by one full period $T_d$:
$$\delta = \ln\left(\frac{A(t)}{A(t + T_d)}\right) = \gamma T_d = \frac{b T_d}{2m}$$
3. Quality Factor ($Q$): A dimensionless metric defining the sharpness of resonance and energy retention:
$$Q = 2\pi \frac{\text{Energy Stored}}{\text{Energy Dissipated per Cycle}} = \frac{\omega_0}{2\gamma} = \frac{\omega_0 m}{b} = \omega_0 \tau$$
4. Driven (Forced) Oscillations & Resonance
When an external periodic force $F(t) = F_0 \cos(\omega t)$ drives the damped oscillator, the equation of motion is:
$$m \frac{d^2 x}{dt^2} + b \frac{dx}{dt} + k x = F_0 \cos(\omega t) \implies \frac{d^2 x}{dt^2} + 2\gamma \frac{dx}{dt} + \omega_0^2 x = f_0 \cos(\omega t)$$
where $f_0 = \frac{F_0}{m}$. In the steady-state regime (after initial transient terms decay to zero), the particle oscillates at the driving frequency $\omega$:
$$x(t) = A(\omega) \cos(\omega t – \phi)$$
1. Steady-State Amplitude ($A$):
$$A(\omega) = \frac{F_0}{\sqrt{m^2(\omega_0^2 – \omega^2)^2 + b^2 \omega^2}} = \frac{f_0}{\sqrt{(\omega_0^2 – \omega^2)^2 + 4\gamma^2 \omega^2}}$$
2. Phase Lag ($\phi$):
$$\tan\phi = \frac{b \omega}{m(\omega_0^2 – \omega^2)} = \frac{2\gamma \omega}{\omega_0^2 – \omega^2}$$
3. Amplitude Resonance ($\omega_r$): Minimizing the denominator in $A(\omega)$ yields the amplitude resonance frequency:
$$\omega_r = \sqrt{\omega_0^2 – 2\gamma^2} = \sqrt{\omega_0^2 – \frac{b^2}{2m^2}}$$
Notice $\omega_r < \omega_d < \omega_0$. At exact velocity resonance ($\omega = \omega_0$), the phase lag is $\phi = \pi/2$ ($90^\circ$), and power absorbed from the driver is maximized.
5. Solved Examples
Example 1 (Foundational – JEE Main):
An underdamped oscillator of mass $m = 200\text{ g}$ has a spring constant $k = 80\text{ N/m}$ and a damping coefficient $b = 0.04\text{ kg/s}$. Find: (a) the period of undamped oscillation $T_0$, (b) the period of damped oscillation $T_d$, and (c) the time taken for the amplitude to drop to half its initial value.
Solution:
(a) $\omega_0 = \sqrt{\frac{k}{m}} = \sqrt{\frac{80}{0.2}} = \sqrt{400} = 20\text{ rad/s}$.
$$T_0 = \frac{2\pi}{\omega_0} = \frac{2\pi}{20} = \frac{\pi}{10} \approx \mathbf{0.314\text{ s}}$$
(b) Damping factor: $\gamma = \frac{b}{2m} = \frac{0.04}{2 \times 0.2} = \frac{0.04}{0.4} = 0.1\text{ s}^{-1}$.
$\omega_d = \sqrt{\omega_0^2 – \gamma^2} = \sqrt{400 – 0.01} \approx 20.00\text{ rad/s} \implies T_d \approx T_0 = \mathbf{0.314\text{ s}}$.
(c) Amplitude decay: $A(t) = A_0 e^{-\gamma t}$. For $A(t) = \frac{1}{2} A_0$:
$$e^{-\gamma t_{1/2}} = \frac{1}{2} \implies t_{1/2} = \frac{\ln 2}{\gamma} = \frac{0.693}{0.1} = \mathbf{6.93\text{ s}}$$
Example 2 (Energy Dissipation & Quality Factor – JEE Main/Advanced):
A damped harmonic oscillator has an initial energy of $E_0 = 10\text{ J}$. After completing $50$ oscillations, its energy drops to $1\text{ J}$. If the natural frequency is $\omega_0 = 100\text{ rad/s}$, determine: (a) the damping factor $\gamma$, and (b) the Quality Factor $Q$ of the oscillator.
Solution:
(a) For light damping, $\omega_d \approx \omega_0 = 100\text{ rad/s}$.
Time for 50 oscillations: $t = 50 \times T \approx 50 \times \frac{2\pi}{\omega_0} = 50 \times \frac{2\pi}{100} = \pi\text{ s} \approx 3.1416\text{ s}$.
Energy decays as $E(t) = E_0 e^{-2\gamma t}$:
$$\frac{E(t)}{E_0} = e^{-2\gamma t} \implies \frac{1}{10} = e^{-2\gamma \pi} \implies 2\gamma \pi = \ln 10 \approx 2.3026$$
$$\gamma = \frac{2.3026}{2\pi} = \frac{2.3026}{6.2832} \approx \mathbf{0.366\text{ s}^{-1}}$$
(b) Quality factor:
$$Q = \frac{\omega_0}{2\gamma} = \frac{100}{2 \times 0.3665} = \frac{100}{0.733} \approx \mathbf{136.4}$$
Example 3 (Driven Amplitude Resonance – JEE Advanced):
A particle of mass $m = 1\text{ kg}$ is attached to a spring of constant $k = 100\text{ N/m}$ and subject to damping with $b = 2\text{ kg/s}$. It is driven by an external periodic force $F(t) = 10 \cos(\omega t)\text{ N}$. (a) Find the amplitude resonance angular frequency $\omega_r$. (b) Calculate the maximum steady-state amplitude $A_{\text{max}}$ at this resonant frequency.
Solution:
$\omega_0^2 = \frac{k}{m} = \frac{100}{1} = 100\text{ rad}^2/\text{s}^2$.
$\gamma = \frac{b}{2m} = \frac{2}{2(1)} = 1\text{ s}^{-1}$.
(a) Amplitude resonance occurs at:
$$\omega_r = \sqrt{\omega_0^2 – 2\gamma^2} = \sqrt{100 – 2(1)^2} = \sqrt{98} \approx \mathbf{9.899\text{ rad/s}}$$
(b) Evaluating the amplitude at $\omega = \omega_r$:
$$A(\omega_r) = \frac{F_0}{\sqrt{m^2(\omega_0^2 – \omega_r^2)^2 + b^2 \omega_r^2}}$$
Here $\omega_0^2 – \omega_r^2 = 2\gamma^2 = 2(1) = 2$.
Denominator term: $(\omega_0^2 – \omega_r^2)^2 + 4\gamma^2 \omega_r^2 = 2^2 + 4(1)(98) = 4 + 392 = 396$.
$$A_{\text{max}} = \frac{F_0/m}{\sqrt{396}} = \frac{10}{\sqrt{396}} = \frac{10}{19.8997} \approx \mathbf{0.503\text{ m}}$$
Example 4 (Power Dissipation in Steady-State – JEE Advanced):
A forced oscillator satisfies $m \ddot{x} + b \dot{x} + k x = F_0 \cos(\omega t)$. Prove that in the steady state, the average mechanical power supplied by the driving force over one full cycle equals the average power dissipated by the damping force, and determine this power at exact velocity resonance ($\omega = \omega_0$).
Solution:
In steady state, $x(t) = A \cos(\omega t – \phi) \implies v(t) = \dot{x} = -\omega A \sin(\omega t – \phi)$.
1. Instantaneous power delivered by driving force: $P_{\text{drive}}(t) = F(t) v(t) = [F_0 \cos(\omega t)][-\omega A \sin(\omega t – \phi)]$.
Expanding $\sin(\omega t – \phi) = \sin(\omega t)\cos\phi – \cos(\omega t)\sin\phi$:
$$\langle P_{\text{drive}} \rangle = -F_0 \omega A \left[ \langle \cos(\omega t)\sin(\omega t) \rangle \cos\phi – \langle \cos^2(\omega t) \rangle \sin\phi \right] = \frac{1}{2} F_0 \omega A \sin\phi$$
2. Damping force power: $P_{\text{damp}}(t) = F_d(t) v(t) = (-b v) v = -b v^2$.
Average power dissipated: $\langle P_{\text{diss}} \rangle = b \langle v^2 \rangle = b (\omega A)^2 \langle \sin^2(\omega t – \phi) \rangle = \frac{1}{2} b \omega^2 A^2$.
From the forced oscillator triangle, $\sin\phi = \frac{b \omega}{\sqrt{m^2(\omega_0^2 – \omega^2)^2 + b^2 \omega^2}} = \frac{b \omega A}{F_0}$.
Substituting $\sin\phi$ into $\langle P_{\text{drive}} \rangle$ gives $\frac{1}{2} F_0 \omega A \left(\frac{b \omega A}{F_0}\right) = \frac{1}{2} b \omega^2 A^2 = \langle P_{\text{diss}} \rangle$. (Energy balance proven).
3. At resonance $\omega = \omega_0$: $\phi = \frac{\pi}{2} \implies \sin\phi = 1$ and $A = \frac{F_0}{b \omega_0}$.
$$\langle P \rangle_{\text{res}} = \frac{1}{2} b \omega_0^2 \left(\frac{F_0}{b \omega_0}\right)^2 = \mathbf{\frac{F_0^2}{2b}}$$
6. 10-Question Practice Worksheet (DPP)
Q1 (Single Choice): For an underdamped oscillator, the amplitude of oscillation reduces to $1/e$ of its initial value in time $\tau$. The damping constant $b$ is equal to:
(A) $m/\tau$
(B) $2m/\tau$
(C) $m/(2\tau)$
(D) $\tau/(2m)$
Q2 (Single Choice): The condition for critical damping in an oscillator of mass $m$ and spring constant $k$ with damping constant $b$ is:
(A) $b = \sqrt{km}$
(B) $b = 2\sqrt{km}$
(C) $b = 4\sqrt{km}$
(D) $b = \frac{1}{2}\sqrt{km}$
Q3 (Single Choice): In a driven harmonic oscillator, at very low driving frequencies ($\omega \ll \omega_0$), the phase difference between the driving force and displacement approaches:
(A) $0$
(B) $\pi/2$
(C) $\pi$
(D) $3\pi/2$
Q4 (Single Choice): The amplitude of a damped oscillator becomes half in $10\text{ s}$. The time in seconds taken for the total energy to drop to $1/16$ of its initial value is:
(A) $10\text{ s}$
(B) $20\text{ s}$
(C) $30\text{ s}$
(D) $40\text{ s}$
Q5 (Single Choice): A pointer galvanometer is designed to come to rest rapidly at its deflected position without oscillating. The motion of the coil should be:
(A) Undamped
(B) Lightly underdamped
(C) Critically damped
(D) Heavily overdamped
Q6 (Single Choice): In forced resonance, the velocity of the oscillating particle and the driving force are:
(A) In the same phase
(B) Out of phase by $\pi/2$
(C) Out of phase by $\pi$
(D) Out of phase by $\pi/4$
Q7 (One or More than One Correct): Which of the following statements regarding damped and driven oscillations are strictly TRUE?
(A) In an underdamped oscillator, the frequency of oscillation is strictly smaller than the natural undamped frequency.
(B) The amplitude resonance frequency $\omega_r$ is always less than the velocity resonance frequency $\omega_0$.
(C) At exact velocity resonance ($\omega = \omega_0$), displacement lags behind the driving force by $\pi/2\text{ rad}$.
(D) The higher the Quality Factor $Q$, the broader and less sharp is the resonance peak.
Q8 (One or More than One Correct): A mass $m$ attached to a spring of stiffness $k$ is subjected to viscous resistance $F_d = -b v$ and a periodic force $F_0 \cos(\omega t)$. Then:
(A) In steady state, the system oscillates at driving frequency $\omega$ regardless of its initial conditions.
(B) For $\gamma \ge \omega_0$, no oscillatory motion occurs in free damping.
(C) Power absorbed from the driving force is maximum when $\omega = \omega_0$.
(D) The amplitude of steady-state oscillation is independent of mass $m$.
Q9 (Numerical Value): An underdamped oscillator completes $100$ full cycles while its amplitude decreases to $1/e$ of its initial value. The Quality Factor $Q$ of this oscillator (rounded to the nearest integer) is ________. (Take $\pi = 3.1416$)
Q10 (Numerical Value): An oscillator of mass $m = 0.5\text{ kg}$, spring constant $k = 200\text{ N/m}$, and damping constant $b = 1.0\text{ kg/s}$ is driven by a sinusoidal force of amplitude $F_0 = 10\text{ N}$ at velocity resonance ($\omega = \omega_0$). The average power dissipated in Watts is ________.
7. Answer Key & Detailed Step-by-Step Solutions
Answer Key:
Q1: (B) | Q2: (B) | Q3: (A) | Q4: (B) | Q5: (C) | Q6: (A) | Q7: (A, B, C) | Q8: (A, B, C) | Q9: 314 | Q10: 50
Detailed Solutions
Solution 1:
Amplitude decays as $A(t) = A_0 e^{-\gamma t} = A_0 e^{-\frac{b}{2m}t}$.
Given $A(\tau) = A_0/e \implies \frac{b}{2m}\tau = 1 \implies b = \frac{2m}{\tau}$. Hence, (B) is correct.
Solution 2:
Critical damping occurs when $\gamma = \omega_0 \implies \frac{b}{2m} = \sqrt{\frac{k}{m}} \implies b = 2\sqrt{k m}$. Hence, (B) is correct.
Solution 3:
Phase lag satisfies $\tan\phi = \frac{b \omega}{m(\omega_0^2 – \omega^2)}$.
As $\omega \to 0$, the numerator goes to 0 while the denominator tends to $m \omega_0^2 > 0$.
Therefore, $\tan\phi \to 0 \implies \phi \to 0$. Force and displacement are in phase. Hence, (A) is correct.
Solution 4:
Energy is proportional to the square of amplitude: $E \propto A^2$.
In $10\text{ s}$, amplitude halves ($A \to A/2$), so energy becomes $E \to E/4$ in $10\text{ s}$.
To drop to $1/16 = (1/4)^2$, two such intervals are needed:
$$t = 2 \times 10\text{ s} = \mathbf{20\text{ s}}$$
Hence, (B) is correct.
Solution 5:
Deadbeat or critically damped motion allows the pointer to reach its steady reading without overshoot or oscillation in the minimum possible time. Hence, (C) is correct.
Solution 6:
At resonance, $\phi = \pi/2$ (displacement lags force by $90^\circ$).
Since velocity leads displacement by $\pi/2$, velocity and driving force are in phase ($\Delta\phi = \frac{\pi}{2} – \frac{\pi}{2} = 0$). Hence, (A) is correct.
Solution 7:
(A) True: $\omega_d = \sqrt{\omega_0^2 – \gamma^2} < \omega_0$.
(B) True: $\omega_r = \sqrt{\omega_0^2 – 2\gamma^2} < \omega_0$.
(C) True: At $\omega = \omega_0$, $\tan\phi \to \infty \implies \phi = \pi/2$.
(D) False: Higher $Q$ implies less damping, making the resonance curve narrower and sharper.
Hence, (A, B, C) are correct.
Solution 8:
(A) True: In steady state, transient natural oscillations have died out, leaving only the driver frequency $\omega$.
(B) True: For $\gamma \ge \omega_0$, the characteristic roots are real, resulting in non-oscillatory decay.
(C) True: Average power absorption is $\langle P \rangle = \frac{1}{2} b \omega^2 A^2$, which peaks at $\omega = \omega_0$.
(D) False: The amplitude depends explicitly on $m$ through both the inertia term and $\omega_0$.
Hence, (A, B, C) are correct.
Solution 9:
Time taken for 100 cycles: $t = 100 T_d \approx 100 \times \frac{2\pi}{\omega_0}$.
Given $A(t) = A_0 e^{-\gamma t} = A_0/e \implies \gamma t = 1$.
$$\gamma \left(100 \times \frac{2\pi}{\omega_0}\right) = 1 \implies \frac{\omega_0}{2\gamma} = 100\pi$$
Since Quality Factor $Q = \frac{\omega_0}{2\gamma}$:
$$Q = 100\pi = 100 \times 3.1416 \approx \mathbf{314}$$
Answer is 314.
Solution 10:
At velocity resonance ($\omega = \omega_0$), average power supplied and dissipated is:
$$\langle P \rangle = \frac{F_0^2}{2b}$$
Given $F_0 = 10\text{ N}$ and $b = 1.0\text{ kg/s}$:
$$\langle P \rangle = \frac{10^2}{2(1.0)} = \frac{100}{2} = \mathbf{50\text{ W}}$$
Answer is 50.