
Oscillations, SHM & Resonance
Deep Physics notes on periodic motion, angular frequency, simple harmonic motion, mass–spring and pendulum systems, oscillation energies, damped and forced motion, resonance and UPSC examination questions.
Periodic motion, oscillation and equilibrium
A motion is periodic when a physical state repeats after a definite interval called the period T. An oscillation is motion back and forth about a reference position, often a stable equilibrium. All ideal simple harmonic motions are periodic oscillations, but not every periodic motion is oscillatory and not every oscillation is simple harmonic. Uniform circular motion, for example, repeats but does not move to and fro along one line.
A pendulum swinging through small angles, a mass attached to a light spring, and electric oscillations in an ideal LC circuit are examples of oscillatory systems. The equilibrium position is where net restoring force vanishes. Stable equilibrium means a slight displacement produces a tendency to return, which is essential for simple mechanical oscillation.
- Frequency f=1/T measures cycles per second (hertz).
- The period is the duration of a complete repeated cycle, not the time to travel from one extreme to the other.
Amplitude, phase, displacement and frequency
The maximum magnitude of displacement from the mean position is the amplitude A. For one-dimensional simple harmonic motion, position may be written x=A cos(ωt+φ), where ω is angular frequency, t time and φ the initial phase. The phase controls the object’s instantaneous position and direction within its cycle; two motions with equal amplitude and period can still be out of phase.
Angular frequency is ω=2π/T=2πf, measured in rad/s. Starting at an extreme at t=0 permits x=A cosωt, while starting at equilibrium toward the positive side permits x=A sinωt. These are the same class of motion with different initial conditions. Never confuse angular frequency ω with ordinary frequency f, nor with the angular displacement of a rotating wheel.
- Amplitude A is a length for a mechanical oscillator.
- Phase is dimensionless when measured in radians.
Defining condition of simple harmonic motion
Simple harmonic motion (SHM) is oscillation in which acceleration is proportional to displacement from equilibrium and directed oppositely: a=−ω²x. Equivalent restoring force is F=−kx for mass m, with ω²=k/m. The negative sign is physical: the acceleration points toward equilibrium, not away from it.
If force is not directly proportional to displacement, motion may still oscillate but will not generally be exactly simple harmonic. The SHO differential equation is d²x/dt²+ω²x=0, with sinusoidal solutions. For a real pendulum at large angular amplitudes, the sine of the angle is not well approximated by the angle, and the motion is not exactly SHM.
- At equilibrium x=0, restoring acceleration is zero but speed is maximal.
- At extreme displacement ±A, speed vanishes momentarily but restoring acceleration is maximal.
Velocity and acceleration throughout an SHM cycle
Differentiating x=A cos(ωt+φ) gives v=−Aω sin(ωt+φ) and a=−Aω² cos(ωt+φ). Consequently vmax=Aω and amax=Aω². At a general position x, the speed magnitude follows v²=ω²(A²−x²), showing how kinetic energy and speed change with displacement.
Maximum speed occurs at the mean position, where displacement and acceleration vanish. At both turning points, v=0 but |a|=Aω². A common exam mistake treats zero velocity as zero net force; this is false for an oscillator at its extremes. Acceleration is exactly opposite displacement in the ideal SHM model.
- Phase difference between displacement and acceleration is π radians.
- The amplitude controls maximum speed linearly and maximum acceleration linearly for fixed ω.
Mass–spring oscillator and spring constant
A mass m attached to an ideal Hookean spring of stiffness k and moving frictionlessly along the spring axis satisfies m a=−kx. Therefore ω=√(k/m), T=2π√(m/k) and f=(1/2π)√(k/m). Doubling mass increases the period by √2, while quadrupling the stiffness halves the period.
For an ideal linear spring, the period is independent of amplitude. This does not mean amplitude has no effect on maximum velocity or total energy; vmax=A√(k/m) and E=½kA². If a spring is cut, its effective spring constant generally changes, so avoid assuming the period remains the same when the physical spring geometry changes.
- Spring k has unit N/m; it is not identical to Young’s modulus.
- Mass–spring period assumes small Hookean deformation and negligible damping.
Simple pendulum: small-angle approximation
A simple pendulum has a point-like bob suspended by a light inextensible string of length L under gravity. Its exact angular dynamics include tangential restoring torque proportional to sinθ. For sufficiently small angular amplitude, sinθ≈θ (radians), yielding angular SHM with period T≈2π√(L/g). In this approximation the bob mass does not affect the period.
Changing length alters period as √L; a fourfold length produces twice the period. Decreasing local gravitational acceleration increases period. This is why a pendulum clock calibrated at one place may run differently if transported to a location with different effective g. For large amplitudes, period increases above the small-angle approximation, so amplitude-independence is not exact for the real pendulum.
- The small-angle period is independent of mass.
- The relevant L is effective pivot-to-centre-of-mass length for an ideal bob approximation.
Energy of simple harmonic motion
For an ideal spring oscillator, elastic potential energy is U=½kx² and kinetic energy K=½mv². The total mechanical energy E=K+U=½kA²=½mω²A² is constant if no damping or external driving occurs. At equilibrium, potential energy relative to that equilibrium is minimum and K=E; at a turning point K=0 and U=E.
At x=A/2, potential energy is U/E=(x/A)²=1/4, so kinetic energy is 3/4 of total. Doubling amplitude at fixed k multiplies total energy by four. The periodic exchange of energy between forms underlies oscillations from tuning forks to elastic machine components.
- At x=±A/√2, U=K=E/2.
- Use energy conservation for speed-at-displacement questions.
Connection between uniform circular motion and SHM
The projection of a particle in uniform circular motion onto a diameter follows sinusoidal SHM if the particle moves at constant angular speed. Its circular radius becomes the SHM amplitude A and its angular speed becomes SHM angular frequency ω. This provides a geometrical explanation for sine and cosine solutions.
The actual particle’s motion on the circle is two-dimensional, but the projection oscillates along a line. The velocity and acceleration components also have sinusoidal relationships. This analogy is useful for interpreting phase and why successive turning points are separated by half a period.
- Projection motion has period equal to the original circle’s revolution period.
- SHM is one-dimensional under projection, even when represented by a circular diagram.
Damped oscillations and energy dissipation
Real oscillators lose mechanical energy through friction, air resistance and internal deformation. In a common viscously damped model, m x¨+b x˙+kx=0; the amplitude decays over time when damping is positive. Under weak damping, the system continues oscillating but with a gradually decreasing envelope, approximately exponential.
If damping is sufficiently strong, the system may return toward equilibrium without oscillating. Critical damping is the boundary between oscillatory and overdamped return in the standard linear model; it is desirable in some measuring instruments and door closers to avoid prolonged overshoot. Damping removes energy, so the motion is not an ideal constant-amplitude SHM.
- Amplitude decay indicates energy leaving mechanical oscillation.
- Damping can modify oscillation frequency and suppress resonance peaks.
Forced oscillation and resonance
A forced oscillator receives an external periodic driving force. Its long-term response oscillates at the driving frequency, with an amplitude determined by driving strength, damping and how closely the frequency matches its natural response. Resonance is a large response near the frequency of maximum amplitude under suitable driving conditions.
In the simplest lightly damped model, maximum response occurs close to the natural angular frequency. More damping produces a smaller, broader resonance peak. Resonance can be useful in musical instruments and tuning systems, but it can also damage structures or machinery if periodic forcing produces excessive vibration. The precise peak frequency depends on damping and the response quantity being measured.
- Resonance does not automatically mean infinite amplitude in a real damped system.
- A strong response can occur even under modest driving when damping is very low.
Phase relations, coupled oscillators and beats
Two oscillations of identical frequency may have a phase difference. If in phase, their displacements reinforce; if opposite in phase they can cancel when amplitudes match. For two nearby frequencies superposed at one point, the resulting signal displays beats: a slowly varying amplitude envelope with beat frequency |f₁−f₂| under the standard sinusoidal superposition model.
Beats are often discussed within waves and sound but the underlying mathematics is the superposition of oscillatory motions. Coupled pendulums exchange energy between modes; real complicated systems may have multiple normal frequencies. For basic UPSC questions, calculate phase intervals, periods and beat frequency using the simple two-component model before advancing to complex coupled equations.
- Two nearby sinusoidal frequencies create slow loudness/amplitude modulation.
- Resonance and beats are distinct phenomena: resonance is driven-response enhancement; beats arise from superposition.
Exam applications: clocks, seismometers and engineering vibrations
Pendulum clocks rely on approximately periodic motion; tuning forks exhibit mechanical vibration, and seismometers detect ground motion by tracking relative motion between a suspended element and its frame. Designers seek to control damping and resonance in buildings and rotating machinery. A stable oscillator can still suffer amplitude loss unless maintained by an energy source.
In questions, identify the restoring force, effective inertia, equilibrium location and damping. Derive the ideal natural frequency from force versus displacement, then compare the physical situation to its assumptions. An oscillation can be periodic without being sinusoidal; SHM adds a strict linear restoring law. Distinguish frequency from angular frequency and amplitude from equilibrium displacement.
- The common pendulum formula needs small amplitude and negligible losses.
- Maximum acceleration and maximum velocity occur at different positions.
Essential formulas and their conditions
Read every equation together with its stated assumptions. Symbols denote physical quantities rather than arbitrary numerical shortcuts.
f=1/Tω=2πf=2π/Tx=A cos(ωt+φ)v=−Aω sin(ωt+φ)a=−ω²xv²=ω²(A²−x²)vmax=Aωamax=Aω²T=2π√(m/k)T≈2π√(L/g)E=½kA²fbeat=|f₁−f₂|Solved numericals and exam applications
Each solution uses stated approximations and is intended for method practice, not mere memorisation.
A 1 kg block is attached to k=100 N/m. T=2π√(1/100)=0.2π ≈ 0.628 s.
Mass quadruples at fixed k: T’ / T = √4 = 2.
Pendulum length rises from 1 m to 4 m at fixed g: the period becomes twice its original value.
A=0.10 m and f=2 Hz. ω=4π rad/s; vmax=Aω=0.4π ≈ 1.26 m/s.
If A=0.05 m and ω=10 rad/s, amax=ω²A=100×0.05=5 m/s².
At displacement A/2, U/E=1/4, so K/E=3/4.
Two near-frequency oscillations at 256 Hz and 260 Hz produce 4 beats/s.
Previous-year UPSC examination concepts
The following entries paraphrase topics recorded by a third-party NDA General Ability Test past-paper index. These are not verbatim original questions or official answer keys; session labels and topic associations should be verified against UPSC’s published papers. This section is not claimed to reproduce EPFO/APFC Recruitment Test PYQs.
UPSC NDA (II) 2022 · indexed Simple Pendulum question
Question focus: how a pendulum’s period changes when local g changes at a higher altitude.
CrackTarget explanation: Use T∝1/√g for fixed length in the small-angle limit. Since effective gravity generally declines with altitude, the pendulum period increases.
Browse indexed UPSC NDA Physics topicsUPSC NDA (I) 2026 · indexed Oscillations question
Question focus: compare the large-amplitude period of a pendulum with its small-angle approximation.
CrackTarget explanation: The exact pendulum does not keep period strictly independent of amplitude; the period increases at appreciable amplitudes, so the small-angle formula underestimates it.
Browse indexed UPSC NDA Physics topicsUPSC NDA (I) 2019 · indexed Simple Pendulum concept
Question focus: relation of a simple pendulum’s frequency or period to length and gravity.
CrackTarget explanation: Write f=(1/2π)√(g/L) and compare the scaling, holding all quantities except the named change constant.
Browse indexed UPSC NDA Physics topicsUPSC-style MCQs: 20 original questions
Attempt first; reveal the correct option and explanation when ready. All questions below are newly authored practice, not verbatim UPSC PYQs.
- amplitude
- period
- acceleration
- mass
Reveal answer and explanation
Answer B. The period is the time for a complete repeat.
- 5 s
- 1 s
- 0.2 s
- 2π s
Reveal answer and explanation
Answer C. T=1/f=0.2 s.
- π rad/s
- 2π rad/s
- 4π rad/s
- 8π rad/s
Reveal answer and explanation
Answer C. ω=2πf=4π rad/s.
- a=+ω²x
- a=−ω²x
- a=constant positive
- a=0 always
Reveal answer and explanation
Answer B. Restoring acceleration is opposite displacement.
- maximum
- zero
- infinite
- equal to angular frequency
Reveal answer and explanation
Answer B. At x=±A, v=0.
- zero
- maximum
- always negative
- undefined
Reveal answer and explanation
Answer B. All ideal spring energy is kinetic at mean position.
- 1/2
- 1/4
- 3/4
- 1
Reveal answer and explanation
Answer B. U/E=x²/A²=1/4.
- 2 times
- 4 times
- half
- unchanged
Reveal answer and explanation
Answer B. E=½kA².
- √(m/k)
- m/k²
- A²
- g always
Reveal answer and explanation
Answer A. T=2π√(m/k).
- 4T
- 2T
- T/2
- T/4
Reveal answer and explanation
Answer C. T varies inversely with √k.
- √m
- m
- 1/m
- independent
Reveal answer and explanation
Answer D. T≈2π√(L/g), with no bob mass.
- halves
- doubles
- quadruples
- is unchanged
Reveal answer and explanation
Answer B. T∝√L.
- halves
- doubles
- quadruples
- unchanged
Reveal answer and explanation
Answer B. T∝1/√g.
- zero
- maximum magnitude
- identical to velocity
- always upward only
Reveal answer and explanation
Answer B. The restoring acceleration magnitude is proportional to |x|.
- uniform speed
- simple harmonic motion
- random walk
- straight-line uniform acceleration
Reveal answer and explanation
Answer B. A cosine projection follows SHM.
- grow indefinitely
- decrease
- remain exactly constant
- switch sign once
Reveal answer and explanation
Answer B. Dissipative mechanisms remove mechanical energy.
- external drive near natural frequency
- zero restoring force
- no periodic forcing
- always zero damping
Reveal answer and explanation
Answer A. Driven response grows near the resonance region.
- 2 Hz
- 4 Hz
- 252 Hz
- 504 Hz
Reveal answer and explanation
Answer B. Beat frequency is |254−250|=4 Hz.
- maximum
- zero
- always g
- −ωA
Reveal answer and explanation
Answer B. a=−ω²x=0 at mean position.
- exact at all angles
- valid only approximately for small angles
- impossible in any motion
- determined by bob weight
Reveal answer and explanation
Answer B. Large-angle pendulum periods rise with amplitude; the familiar formula assumes small angles.
Exam traps and structured revision
- The ‘amplitude independence’ of a pendulum’s period is only a small-angle approximation.
- Zero speed at a turning point does not mean zero acceleration.
- A mass–spring ideal period depends on m and k, not amplitude.
- At the mean position of SHM, speed is maximum and acceleration zero.
- Damping and external forcing change the ideal undamped picture.
- Convert f to ω with the factor 2π.
