Damped & Forced Oscillations Resonance Damped and Forced Oscillations: Resonance and Q-Factor Everyday oscillators are never perfectly elastic. A guitar string’s note fades, a pendulum’s swing slows, a car’s suspension settles after a bump. This loss is damping: energy leaves the oscillator due to friction, air drag, or internal resistance. If we keep supplying energy periodically—like pushing a swing at regular intervals—the system is “forced.” The most dramatic phenomenon in forced motion is resonance, where the response amplitude becomes large when the driving frequency matches a special frequency of the system. This peak can be sharp or broad depending on damping. In a world of real oscillators, understanding how amplitude decays, how steady-state motion sets in under a periodic force, how phase shifts, and how “sharpness” is quantified by the Q factor decides whether structures vibrate safely or break, whether musical instruments sound rich or dull, and whether measuring devices can separate close frequencies. This lesson builds intuition first, then brings in the math: the damped equation of motion, the steady-state amplitude and phase for a sinusoidal drive, resonance frequency, power and bandwidth, and Q. Along the way we connect to beats and standing waves so you can see how the same sine math shows up in different clothes. remember Big picture: Damping bleeds energy out; forcing feeds energy in; resonance is the sweet (or dangerous) spot where input best matches the system so energy build-up is maximal. Damping Loss of mechanical energy of an oscillator per cycle due to resistive forces (e.g., air drag, viscosity, internal friction). Damping Constant (b) Parameter in a viscous damping force F d = -b v with SI unit kg/s ; larger b means stronger damping. = b/(2m) , sets the rate of exponential amplitude decay e - t in underdamped motion. Decay Constant (β) Natural Frequency (ω₀) Frequency of the undamped oscillator, 0 = k/m for a mass–spring system. Actual oscillation frequency under light damping: d = 0 2 - 2 . Damped Frequency (ω d) Angular frequency of the external periodic force in a forced oscillator. Driving Frequency (ω) Peak response (amplitude or power) when driving frequency is near the system’s characteristic frequency; for displacement amplitude, the peak is slightly below 0 if damping is small. Resonance Quality Factor (Q) A measure of sharpness of resonance: Q = 0 m 0 b for weak damping. Bandwidth (Δω) Frequency width between the half-power points of a resonance curve; for light damping b/m . Start with a free oscillator. For a mass–spring system of mass m , spring constant k , and viscous damping constant b , the equation is m ,x'' + b ,x' + k ,x = 0 . If damping is light, the solution is a sinusoid multiplied by a decaying exponential: x(t) = A 0 e - t ( d t + ) with = b/(2m) and d = 0 2 - 2 , where 0 = k/m . The exponential envelope sets how fast the peaks shrink; the cosine term sets the oscillation itself. This form gives you three quick insights: (1) The amplitude halves every fixed time interval (a time constant idea). (2) The frequency shifts slightly below 0 due to damping. (3) The total mechanical energy falls faster than amplitude, because energy is proportional to amplitude squared. These simple observations are the anchor for everything forced and resonant that follows. Damped free oscillator (viscous) Second-order linear ODE governing damped free oscillations. This equation represents the ideal, undamped oscillation frequency before considering energy loss due to damping or external forces. Underdamped solution Sinusoid with exponentially decaying envelope for b 2 < 4mk (underdamped). This solution describes the position of a mass attached to a spring undergoing damped harmonic motion, provided the damping coefficient is s Viscous damping: F d = -b v Underdamped case: b 2 < 4 m k Linear restoring force: F s = -k x Envelope of underdamped motion decays exponentially: A(t) = A 0 e - t Try solution x=e rt to reduce to a characteristic quadratic. Complex conjugate roots for underdamped motion. General real solution. The multiplicative factor e - t is the amplitude envelope. A(t) = A 0 e - t , = b 2m Two time scales matter. First, the oscillation period T d = 2 / d . Second, the decay time A = 1/ = 2m/b , over which the amplitude drops by a factor e . Because energy E is proportional to A 2 , its decay time is E = 1/(2 ) = m/b : energy falls twice as fast in the exponent as amplitude. A useful measure is the logarithmic decrement , defined as = ( A n A n+1 ) . For underdamped viscous systems, peaks are separated by T d , so = T d = 2 d . These relations let you estimate b from how quickly successive peaks shrink in a recorded oscillation, a common lab and exam task. Energy decays with time constant E = m/b . Energy decay in underdamped motion Use this relationship to determine how quickly the amplitude of an underdamped oscillator diminishes over time due to viscous damping. tip Valid range: The exponential decay model requires viscous damping (force proportional to velocity). With dry (Coulomb) friction, amplitude typically decays roughly linearly per cycle, not exponentially. Now add a sinusoidal driving force F(t) = F 0 ( t) . The full equation is m x'' + b x' + k x = F 0 t . The steady-state (long-time) motion is also sinusoidal at the same driving frequency, but with a new amplitude X( ) and a phase lag ( ) relative to the driver. Transients die out with the same e - t envelope, leaving only the steady response. The amplitude is largest near resonance. Importantly, resonance is not a single idea: the displacement amplitude, velocity amplitude, and power absorption all have peaks that occur very near 0 but are not exactly identical when damping is noticeable. For light damping (the NEET regime), these differences are tiny, and the language “resonance at 0 ” for power absorption is standard. Equation of motion with sinusoidal driving force. Driven damped oscillator This equation describes the motion of a mass attached to a spring, subjected to damping and an external sinusoidal driving force. Steady-state amplitude and phase Amplitude response and phase lag of displacement relative to driving force. This formula describes the steady-state amplitude and phase shift of a mass attached to a spring and damper system when subjected to an exte rad/s Angular frequency ω Amplitude X(ω) Resonance curve: a single peak near ω ≈ ω0 that becomes sharper and taller as b decreases (higher Q). control dependent Amplitude versus driving frequency for different damping values (qualitative). Xmax ω ≈ √(ω0 2 - b 2 /(2m 2)) Displacement resonance X = F0/(b ω0) At natural frequency (phase = 90°) ω = ω0 custom 2D PLOT Driven oscillator amplitude vs frequency (resonance) X = F0 / sqrt((w0 2 - w 2) 2 + (gamma w) 2) F0 Driving force / mass w0 Natural frequency gamma Damping At the peak of the displacement resonance, the derivative of X( ) with respect to is zero. Solving gives r = 0 2 - b 2 2m 2 , slightly below 0 . Phase evolves smoothly from 0 (low ) to (very high ), crossing /2 exactly at = 0 . The power absorbed from the driver is maximal at 0 , and its half-maximum points define the bandwidth = 2 - 1 . For weak damping, a powerful trio of equalities emerges: Q = 0 = m 0 b and = b m = 2 . These are central to NEET-style numericals because they connect an experimentally measured resonance curve (read off the plot) to microscopic parameters m and b . neet-alert Trap: Do not confuse the damped natural frequency d = 0 2 - 2 (free response) with the resonance frequency of forced displacement r 0 2 - b 2/(2m 2) . For light damping d r 0 , but they are not identical. = b m , Q = 0 = m 0 b ( for weak damping ) Average power absorbed is proportional to 2 b X 2 for a mechanical oscillator. Use r 0 for weak damping. For light damping and near resonance, approximate 0 in slowly varying factors. Condition for half-power points using the amplitude expression. Linearize ( 2 - 0 2) 2 0( - 0) near 0 . Bandwidth–damping–Q relations for weak damping. For light damping, bandwidth Δω = b/m and Q = ω0/Δω = m ω0 / b Weak damping: b is small so ωr ≈ ω0 Half-power points defined where average power is half its maximum Steady-state sinusoidal driving Physically, large Q means the oscillator stores energy well compared with what it dissipates per cycle: the resonance peak is tall and narrow. A small Q means strong damping and a broad, gentle resonance. This is not just vocabulary. In medical ultrasound or radio tuning, a higher Q allows finer frequency selection (narrow bandwidth). In car suspensions, a lower Q makes the ride comfortable by avoiding large oscillations after a road bump. In lab problems, Q immediately tells you how many cycles the energy takes to drop significantly: per radian of oscillation, the fractional energy loss is roughly 1/Q when Q 1 . Combining this with lets you switch between time domain (decay) and frequency domain (resonance curves) descriptions efficiently. Regime Condition Motion Representative solution form Example Underdamped b 2 < 4mk Oscillatory with decaying amplitude x(t) = A 0 e - t ( d t + ) Tuning fork in air Critically damped b 2 = 4mk Fastest non-oscillatory return to equilibrium x(t) = (C 1 + C 2 t) e - 0 t Door closer Overdamped b 2 > 4mk Non-oscillatory, slow return x(t) = C 1 e r 1 t + C 2 e r 2 t (r 1,2 < 0) Galvanometer damping vane Damping regimes Resonance metrics (light damping) Displacement resonance frequency ω r = 0 2 - b 2/(2m 2) Slightly below ω0 Peak displacement amplitude X max F 0/(b , 0 ) At ω ≈ ω0 Bandwidth Δω = b/m = 2β Half-power width Quality factor Q = ω0/Δω = m ω0 / b Sharpness of resonance Phase at ω = ω0 φ = 90 Displacement lags driver by quarter cycle Quantity Expression Comment m (≈ 0.68 cm) A(t) For underdamped viscous damping: A(t) = A 0 e - t with = b/(2m) . m = 0.50 kg b = 0.20 kg/s A 0 = 5.0 cm = 0.050 m t = 10 s easy A 0.50 kg mass on a lightly damped spring has initial amplitude 5.0 cm. The damping constant is b = 0.20 kg/s. What is the amplitude after 10 s? A driven damped oscillator has m = 1.0 kg, k = 100 N/m, b = 2.0 kg/s, and is driven by F0 = 5.0 N at ω = 8.0 rad/s. Find the steady-state amplitude X and phase lag φ. medium X = F 0 (k - m 2) 2 + (b ) 2 , = b k - m 2 . m = 1.0 kg k = 100 N/m b = 2.0 kg/s F 0 = 5.0 N = 8.0 rad/s X(ω), φ(ω) m, degrees Q, b, and ω0 1 = 9.5 rad/s 2 = 10.5 rad/s m = 0.20 kg For weak damping: = 2 - 1 , Q = 0/ , b = m , and 0 1 2 . hard Bandwidth–Q relation, standard NEET style A resonance curve for a mechanical oscillator shows half-power frequencies at ω1 = 9.5 rad/s and ω2 = 10.5 rad/s. If the mass is 0.20 kg, estimate Q and the damping constant b. Also estimate ω0. dimensionless, kg/s, rad/s custom Phase lag of displacement relative to driver across frequencies. ω ≪ ω0 In phase φ ≈ 0 φ = π/2 ω = ω0 Quadrature ω ≫ ω0 Out of phase φ ≈ π control dependent Monotonic phase increase from 0 at low ω to π at very high ω; exactly π/2 at ω = ω0. Phase φ(ω) rad Angular frequency ω rad/s 2D PLOT Phase lag vs driving frequency phi = atan2(gamma w, w0 2 - w 2) phi gamma Damping w0 Natural frequency Phase tells you how the oscillator’s peaks line up with the driver’s peaks. At low frequency, the spring dominates, so the mass follows almost in step ( 0 ). Near 0 , inertia and spring balance, and the motion lags by a quarter cycle ( = /2 ). At very high frequency, inertia dominates: the mass cannot keep up, and displacement is almost opposite to the driving force ( ). This continuous phase rotation is a hallmark of resonant systems and allows you to diagnose where you are on the resonance curve even if you cannot easily read amplitude (e.g., in noisy data). Displacement x(t) Time t control dependent A cosine oscillation multiplied by a decaying exponential envelope e -β t . custom Underdamped free oscillation showing exponential envelope. x = A0 cos φ t = 0 Initial condition Amplitude down by e t = 1/β x ≈ A0/e Bandwidth–ω0–Q relation BW–Q triangle: For light damping, think of a triangle linking BandWidth (Δω), natural frequency (ω0), and Quality (Q): Q = ω0/Δω. If one grows, at least one other must adjust. At resonance, amplitude becomes infinite. Only an undamped ( b=0 ) ideal oscillator can diverge. Any real damping caps the peak: X max F 0/(b , 0) for light damping. Resonance always occurs exactly at the damped natural frequency ωd. Displacement resonance occurs at r = 0 2 - b 2/(2m 2) , not at d = 0 2 - 2 . For light damping they are very close but distinct. A larger Q means larger bandwidth. It is the opposite for weak damping: Q = 0/ . Higher Q gives a narrower (smaller) bandwidth. Half-power point means power drops to 50% of maximum, but amplitude drops only to 1/ 2 of its peak (not to half). Many errors come from mixing up power and amplitude. neet-alert Connection to beats: When two close frequencies superpose, you hear slow variations in loudness—the beats. In a forced oscillator with a slowly drifting driver, the amplitude swells and falls similarly as the driver passes through resonance. The mathematics of sums of sines underlies both: near resonance, the response is largest because the driver keeps adding energy in step with the motion for many cycles. In beats, the local phase alignment between the two waves periodically aligns and misaligns; in resonance, the driver’s constant phase and the system’s lag ( ) arrange so that average power transfer is maximized near 0 . Two close frequencies produce beats at f b = |f 1 - f 2| , giving amplitude modulation that echoes the idea of near-resonant build-up and fade. f b = |f 1 - f 2| Sum-to-product identity. Beat frequency for two close tones: f b = |f 1 - f 2| Same amplitude A Frequencies f1 and f2 close to each other Linear superposition Connection to plain SHM: The displacement equation x = A ( t + ) is the seed from which both damping (by multiplying with e - t ) and forcing (by replacing A and with frequency-dependent steady-state values) grow. Get fluent with this basic sinusoid, and damped/forced results will feel like natural extensions rather than new worlds. Core SHM solution x = A ( t + ) —the building block for damped and driven oscillations. General SHM solution: x(t) = A cos(ω t + φ) Linear restoring force: F = −kx No damping or driving One-dimensional motion x(t) = A ( t + ) Connection to standing waves: A standing wave is the superposition of two identical traveling waves moving in opposite directions. While resonance here comes from boundary conditions (nodes and antinodes), the mathematics of sinusoidal time dependence is the same. The profile does not travel; instead, each point oscillates in time as ( t) with position-dependent amplitude 2A (kx) . Recognizing this pattern makes it easier to remember that energy in a standing wave oscillates locally rather than propagates along the medium. Standing wave pattern y = 2A (kx) ( t) arises from two counter-propagating equal waves. y(x,t) = 2A (kx) ( t) Use u + v = 2 u+v 2 u-v 2 . Standing wave equation: y = 2A sin(kx) cos(ω t) Two waves with equal amplitude A and frequency ω Opposite directions of propagation Linear medium (superposition) Solving forced-oscillation numericals systematically: (1) Extract m, k, b, F 0, . (2) Compute 0 = k/m and, if needed, = b/(2m) . (3) If the question asks for steady-state displacement amplitude/phase, use X( ) and ( ) . (4) If power or bandwidth is asked, locate 1, 2 and use = 2 - 1 and Q = 0/ . (5) For free decay tasks, use A(t) = A 0 e - t , = 2 / d , and E(t) e -2 t . Keep sign conventions consistent and watch units: N·s/m for b , rad/s for . Checklist for resonance-curve questions Read ω0 from the center of the peak: ω0 ≈ √(ω1 ω2). Read bandwidth: Δω = ω2 − ω1 (half-power points). Compute Q = ω0/Δω and damping b = m Δω. At ω = ω0, phase φ = 90° and X ≈ F0/(b ω0) for small b. If asked for energy decay, use τE = m/b = 1/Δω. Record successive peak amplitudes A1 and A2 separated by one damped period Td. Compute logarithmic decrement: δ = ln(A1/A2). Compute β = δ/Td and b = 2mβ. Optionally, estimate ωd from Td = 2π/ωd, then ω0 = √(ω d 2 + β 2). Convert to frequency-domain sharpness: Δω = 2β and Q = ω0/Δω. Procedure to estimate b and Q from time-domain decay remember Two mirrors of the same truth: Time domain ( e -β t decay) and frequency domain (Lorentzian-like resonance with width Δω = 2β) tell the same physics. Choose the lens that makes the problem shortest. Edge cases matter in exams. As b 0 (no damping), the bandwidth 0 , so the resonance peak becomes infinitely sharp and tall—this is the mathematical reason amplitudes can blow up in ideal models. As 0 (quasi-static drive), X F 0/k : the system just follows the force like a static deflection. As , inertia dominates and X F 0/(m 2) , which is tiny. In the free motion, as 0 , oscillation ceases; the system is critically damped. Always ask: Which term (spring, damper, mass) is ruling the limit you are in? Dimensional sanity checks are fast and save marks. In X = F 0/ (k - m 2) 2 + (b ) 2 , the denominator inside the square root has dimensions of force squared because k x and m 2 x are both forces for a given x , and b x is also a force. Thus, X carries the dimension of length as it should. For = b/m , the units are ( kg/s )/( kg ) = 1/ s , matching angular frequency units. Build the habit: a 10-second check can rescue a 4-mark calculation. Practical cues: When adjusting a lab oscillator, if increasing the damper opening (reducing b ) makes the resonance peak thinner and taller, Q is rising. If the peak’s center does not move much, 0 is set by k and m ; to shift it, change mass or spring. If the system rings (oscillates) for many cycles after a pulse, Q is high. If it returns in one smooth swoop with no overshoot, it is critically damped or overdamped—good for instruments that must avoid oscillations, like galvanometers. Concept map summary: Free damped motion—set by m, b, k —gives A(t), d , and energy decay. Forced motion—same m, b, k plus F 0, —gives X( ), ( ) and the resonance curve. Linking both is = b/(2m) : it controls how transients die and how sharp the steady-state resonance is via = 2 . The Q factor bridges measurements from either domain. With these anchors, most NEET tasks reduce to substituting into the right expression and interpreting the physically meaningful limit. Logarithmic decrement Relates peak-to-peak decay to damping. Used exclusively for analyzing the amplitude decay of a system undergoing damped Simple Harmonic Motion (SHM). Exact for displacement-amplitude peak of a driven viscous oscillator. Resonance frequency (displacement peak) This formula determines the resonance frequency for a damped harmonic oscillator when subjected to an external driving force. Be careful with sign conventions. In = b k - m 2 , the denominator changes sign at = 0 . As you cross that point, the arctan jumps; use a continuous phase convention (from 0 to ) to avoid 180° mistakes. When calculating r , remember it exists only for underdamped systems; if b is too large, the concept of a sharp displacement peak loses meaning and the amplitude just rolls over smoothly. In that regime, bandwidth and Q are still defined via half-power points, but the peak is flat and broad. Worked-example strategy for mixed data: If a problem gives a decay plot (time domain) but asks for resonance bandwidth (frequency domain), compute from logarithmic decrement or a fit to A(t) , then report = 2 . If a resonance curve is given and the question asks for the damping constant, read 1, 2 , compute , then b = m . If mass is unknown but k and 0 are known, extract m = k/ 0 2 first. Track units ruthlessly; they are your best error detector. Quality factor and cycles-to-decay: In one radian of oscillation, the fractional energy loss is about 1/Q for large Q . Over one cycle ( 2 radians), the fractional loss is roughly 2 /Q . Thus, the number of cycles required for the energy to fall by a factor e is approximately Q/ . This thumb rule is quick and good enough for estimation questions where exact constants are not demanded. Useful when questions target maximum of velocity or acceleration response. Steady-state velocity and acceleration amplitudes Describes the maximum (amplitude) velocity and acceleration of a damped harmonic oscillator driven by an external force, assuming steady-sta Sometimes, the problem targets maximum velocity or acceleration instead of displacement. For light damping, the velocity response is very close to peaking at 0 because V( ) = X( ) simply multiplies the displacement response by , which changes slowly near the peak. Acceleration peaks a little above 0 for the same reason. However, most NEET items restrict to displacement peak and bandwidth, so unless specifically asked, you can focus on X( ) and the half-power definitions. A final word on modeling: Viscous damping is the standard because it linearizes the math and matches many fluids and internal-friction cases near small speeds. Not all systems are exactly viscous, and real materials can have frequency-dependent loss. For NEET, stick to F d = -b v unless stated otherwise. If the problem mentions “rough surface with constant friction,” do not use exponentials; expect amplitude to reduce by a fixed amount per cycle and adjust work–energy calculations accordingly. Key Terms Recap Viscous damping Resistive force proportional to velocity: F d = -b v . = b/(2m) ; amplitude envelope e - t . Decay constant (β) 0 = k/m for mass–spring. Natural frequency (ω0) d = 0 2 - 2 . Damped frequency (ωd) r = 0 2 - b 2/(2m 2) . Resonance frequency (displacement) Bandwidth (Δω) Half-power width; for weak damping = b/m . Quality factor (Q) Q = 0/ = m 0/b (weak damping). = (A n/A n+1 ) = 2 / d . Logarithmic decrement (δ)