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Chapter 10 — Oscillations and Waves

Class 11 · Physics

Overview

This unit introduces oscillations and waves, central concepts in physics that describe periodic motion and the transfer of energy without net transport of matter. You will learn what makes a system oscillate, how to model simple harmonic motion (SHM), and how energy changes between kinetic and potential forms during oscillation. The unit covers free, damped and forced oscillations, resonance and practical examples like mass-spring systems and pendulums. Building on oscillations, it develops the idea of wave motion: transverse and longitudinal waves, wave speed, wavelength and frequency, and the mathematical wave equation for one-dimensional waves. Important phenomena such as superposition, interference, standing waves and resonance in pipes and strings are explained with real-world applications including musical instruments, engineering vibrations and communication systems. The unit emphasises problem solving: deriving time periods, calculating energies, analysing damping and resonance, and sketching waveforms. Understanding oscillations and waves provides a foundation for later topics in optics, acoustics and modern physics, and is useful for practical skills such as measuring frequencies, designing systems to avoid harmful resonance, and interpreting wave-based signals.

Learning Objectives

  • Describe and classify different types of oscillations and waves.
  • Apply equations of simple harmonic motion to calculate displacement, velocity and acceleration.
  • Derive expressions for time period and frequency for mass-spring systems and simple pendulums.
  • Explain energy transfer in oscillatory motion and calculate kinetic and potential energy in SHM.
  • Analyse damped and forced oscillations and explain the concept of resonance and quality factor.
  • Formulate and solve the one-dimensional wave equation and relate wave speed to frequency and wavelength.
  • Use the principle of superposition to explain interference and standing waves in strings and pipes.
  • Solve problems involving wave phenomena such as beats, nodes and antinodes in standing waves.

Topics in this chapter

18 topics · tap a topic title to jump straight to it.

🏃1

Nature of Oscillatory Motion

What is oscillatory motion?
Oscillatory motion is any motion that repeats itself at regular intervals about an equilibrium position. When a system is taken out of equilibrium by a disturbance, restoring forces tend to bring it back; inertia causes overshoot and the cycle repeats. Common examples include swings, vibrating strings, tuning forks and electrical LC circuits. Observing these systems shows repeated patterns of displacement, velocity and acceleration.

Important descriptors
To describe oscillations we use a few standard quantities. Amplitude is the maximum displacement from equilibrium and gives the scale of motion. Period is the time for one complete cycle and frequency is the number of cycles per second. Phase specifies the state of the oscillation at a reference time. Angular frequency relates to frequency by ω = 2πf and is useful when motion is sinusoidal.

Restoring force and stability
For small departures from equilibrium many systems have a restoring force proportional to displacement: F ≈ -kx. This linear relationship produces simple harmonic motion (SHM) with sinusoidal time behaviour. If restoring force is weaker or stronger than linear, motion is non-linear and may show amplitude-dependent frequency or complex behaviour. Stability of equilibrium determines whether small disturbances decay (stable), grow (unstable) or lead to cycles.

Types of oscillations
Free oscillations occur with initial energy and no external driving; they proceed at natural frequencies determined by system properties. Damped oscillations lose amplitude over time due to friction, viscosity or resistance. Forced oscillations happen when an external periodic force acts; the steady-state oscillation typically follows the driving frequency and can show resonance when driving matches a natural mode. Coupled oscillations occur when two or more oscillators interact, leading to energy exchange and collective normal modes with distinct frequencies.

Mathematical and physical importance
Mathematically, oscillations introduce second-order differential equations and sinusoidal solutions, which reappear in wave theory and signal analysis. Physically, oscillations relate directly to waves: a wave carries a repeating disturbance through space while each point of the medium oscillates in time. Studying oscillations prepares you to understand wave propagation, interference, standing waves and resonance in larger systems. Practical skills include measuring period, estimating damping, and designing systems to control unwanted vibrations.

📌 Examples
  • A child on a swing completing one to-and-fro motion in 2 seconds has period T = 2 s and frequency f = 0.5 Hz.
  • A mass displaced on a spring and released shows oscillations; measuring the time for several oscillations gives the period.
  • A tuning fork struck once vibrates and produces sound that decays over time—an example of damped oscillation.
🧮 Formulas
  1. Frequency f = 1/T
  2. Angular frequency ω = 2πf = 2π/T
📊 Visual ideas
A plot of displacement x versus time t for simple harmonic motion, showing sinusoidal curve with labelled amplitude, period and phase.
A sketch showing amplitude decreasing over time for a damped oscillation.
🏃2

Simple Harmonic Motion (SHM) — Definition and Characteristics

Formal idea of SHM
Simple Harmonic Motion (SHM) is a type of periodic motion in which the restoring force is proportional to the displacement and directed towards the equilibrium position. This proportionality leads to a linear differential equation whose solutions are sinusoidal functions. SHM is a good approximation for many systems when displacements are small and forces vary linearly with displacement.

Physical examples
Examples include a mass attached to a spring (horizontal or vertical) for small displacements, small-angle oscillations of a simple pendulum, and small vibrations of strings and membranes. In each case the system oscillates back and forth, passing through equilibrium where speed is maximum and stopping momentarily at turning points where potential energy is maximum.

Deriving SHM from Newton's laws
Consider a mass m on a spring with spring constant k. Hooke's law gives restoring force F = -kx. Newton's second law m d²x/dt² = F yields m d²x/dt² + kx = 0, or d²x/dt² + (k/m) x = 0. This standard form shows that acceleration is proportional to negative displacement, the mathematical hallmark of SHM. The natural angular frequency is ω = √(k/m), which sets the timescale of oscillation.

Solution form and physical meaning
The general solution is x(t) = A cos(ωt + φ), where A is amplitude and φ is phase constant determined by initial conditions. This form shows motion is sinusoidal with fixed amplitude A in the ideal, undamped case. Velocity v(t) and acceleration a(t) follow by differentiation: v = -Aω sin(ωt + φ), a = -Aω² cos(ωt + φ) = -ω² x. The acceleration being proportional to -x means acceleration and displacement are exactly out of phase: when displacement is maximum, acceleration points toward equilibrium.

Key properties and observations
Frequency and period are independent of amplitude for ideal linear SHM, a fact that simplifies many calculations and distinguishes linear SHM from nonlinear oscillations. Maximum speed is Aω and maximum acceleration is Aω². Energy in SHM oscillates between kinetic and potential forms but total mechanical energy stays constant if no non-conservative forces act. SHM forms the basis for analysing small vibrations, waves and oscillatory circuits because many complex motions can be decomposed into SHM components.

📌 Examples
  • A block on a spring with k = 100 N/m and mass m = 0.5 kg has ω = √(k/m) = √(200) ≈ 14.14 rad/s and period T = 2π/ω ≈ 0.444 s.
  • If x(t) = 0.05 cos(20t + π/6) m, then amplitude A = 0.05 m, angular frequency ω = 20 s⁻¹, period T = π/10 s and initial phase φ = π/6.
🧮 Formulas
  1. Equation of motion: d²x/dt² + ω² x = 0
  2. Solution: x(t) = A cos(ωt + φ)
  3. ω = √(k/m), T = 2π/ω, f = 1/T
📊 Visual ideas
Displacement x versus time t showing x(t) = A cos(ωt + φ) with labeled A and T.
Velocity v versus time t and acceleration a versus time t plotted alongside displacement to show phase relations.
🧴3

Mathematical Solution and Phase Relationships in SHM

Solving the SHM differential equation
The core differential equation for SHM is d²x/dt² + ω² x = 0. This linear, constant-coefficient, second-order ordinary differential equation has characteristic equation r² + ω² = 0 with roots r = ± i ω. Using standard solution methods we write the general real solution as x(t) = C1 cos(ωt) + C2 sin(ωt). Constants C1 and C2 are fixed by initial displacement x(0) and initial velocity v(0).

Amplitude-phase form and relation to constants
It is often convenient to combine C1 and C2 into amplitude-phase form x(t) = A cos(ωt + φ). The amplitude A and phase φ relate to C1 and C2 by A = √(C1² + C2²) and φ = arctan(-C2/C1) with attention to the correct quadrant. Using initial conditions x(0) = x0 and v(0) = v0 one finds C1 = x0 and C2 = v0/ω, so A = √(x0² + (v0/ω)²) and φ = arctan(-v0/(ω x0)) after choosing the right branch for arctan. This explicit procedure is practical for real problems where initial displacement and velocity are given.

Velocity and acceleration — phase differences
Different kinematic quantities are phase-shifted relative to displacement. From x(t) = A cos(ωt + φ), velocity v(t) = -Aω sin(ωt + φ) which leads displacement by π/2 in phase (or lags by π/2 depending on sign convention). Acceleration a(t) = -Aω² cos(ωt + φ) = -ω² x(t) is in antiphase with displacement (a and x are 180° out of phase). These phase relations explain why when x is maximum (turning point) velocity is zero and acceleration is maximal directed towards equilibrium.

Complex notation and practical use
Complex exponentials simplify algebra when adding oscillations or solving linear forced-response problems. Write solutions as x(t) = Re[Z e^{iωt}], where Z is complex amplitude. The physical displacement is the real part; algebra with exponentials allows easy multiplication and division by i. Despite this convenience, remember to take the real part at the end. Phase diagrams or phasors illustrate amplitude and phase: represent A cos(ωt + φ) as a rotating vector of length A at angle φ, projecting on the real axis to give the instantaneous displacement.

Summary and application
Mastering the mathematical solution and phase relationships of SHM enables you to handle initial-value problems, predict timings of extrema and zeros, relate kinetic and potential energy variations, and prepare for superposition of modes in coupled systems and for wave analysis where similar sinusoidal forms appear in space and time.

📌 Examples
  • Given x(0) = 0.02 m and v(0) = 0, for ω = 10 s⁻¹, C1 = 0.02, C2 = 0, so x(t) = 0.02 cos(10t).
  • For x0 = 0, v0 = 0.1 m/s and ω = 5 s⁻¹, A = √(0² + (0.1/5)²) = 0.02 m and φ = arctan(-v0/(ωx0)) = -π/2, thus x(t) = 0.02 cos(5t - π/2) = 0.02 sin(5t).
🧮 Formulas
  1. x(t) = C1 cos(ωt) + C2 sin(ωt) = A cos(ωt + φ)
  2. A = √(x0² + (v0/ω)²), φ = arctan(-v0/(ωx0))
  3. v(t) = -Aω sin(ωt + φ), a(t) = -Aω² cos(ωt + φ)
📊 Visual ideas
Three curves on same axes: x(t), v(t) and a(t) showing relative phase shifts (x and a opposite, v leading x by 90°).
Vector (phasor) diagram showing A and components C1, C2 used to build amplitude and phase.
🏃4

Energy in Simple Harmonic Motion

Forms of energy in SHM
In ideal simple harmonic motion (no friction or damping), mechanical energy is conserved and continuously exchanges between kinetic and potential forms. In a mass-spring system kinetic energy resides in the mass's motion and potential energy in the spring's deformation. The total energy remains constant and depends only on amplitude, not on time.

Expressions for energies
For a spring with constant k, potential energy at displacement x is U = 1/2 k x². Kinetic energy of the mass m moving with speed v is K = 1/2 m v². Using the SHM expressions x(t) = A cos(ωt + φ) and v(t) = -Aω sin(ωt + φ), we get time-dependent forms U(t) = 1/2 k A² cos²(ωt + φ) and K(t) = 1/2 m A² ω² sin²(ωt + φ). Because ω² = k/m these combine to a constant total E = 1/2 k A² = 1/2 m A² ω².

Physical interpretation of the exchange
At extreme positions x = ±A, potential energy is maximal and kinetic energy is zero; the mass momentarily stops and changes direction. At equilibrium x = 0, potential energy is zero and kinetic energy is maximal. During motion energy flows from spring to mass and back, but the sum remains constant. This exchange happens twice during each full displacement cycle, because energies depend on square of sinusoidal functions and therefore vary with frequency 2ω.

Using energy to solve problems
Energy methods often simplify calculations. To find speed at a given displacement avoid solving differential equations: equate total energy E = 1/2 k A² = 1/2 k x² + 1/2 m v² and solve for v. Similarly, amplitude after energy loss can be found from remaining total energy. For small damping use exponential decay E(t) ≈ E0 e^{-2γt} to estimate amplitude reduction over time.

Mass-spring corrections and examples
Real springs have mass and internal friction. For a heavy spring, part of the spring's mass contributes to kinetic energy; an effective mass correction adds a fraction of spring mass to m. Also, non-linear springs give potential not strictly proportional to x², changing energy expressions and motion. Despite these complexities, the core concept — energy oscillating between kinetic and potential forms while total energy remains constant for an ideal system — provides a powerful tool for analysis and design.

📌 Examples
  • A mass-spring with k = 200 N/m and A = 0.05 m has total energy E = 1/2 k A² = 0.5 × 200 × 0.0025 = 0.25 J.
  • If ω = 20 s⁻¹ and A = 0.01 m, v_max = Aω = 0.2 m/s and K_max = 1/2 m v_max² (use given m) to find kinetic energy at equilibrium.
🧮 Formulas
  1. U = 1/2 k x²
  2. K = 1/2 m v²
  3. Total energy E = 1/2 k A² = 1/2 m A² ω²
  4. v_max = Aω, a_max = Aω²
📊 Visual ideas
Plot of U(t) and K(t) versus time showing complementary sin² and cos² behaviour and constant E.
Energy distribution vs displacement: U increasing with x² and K decreasing correspondingly.
🔬5

Damped Oscillations

Why damping occurs
Ideal oscillators keep oscillating forever, but real systems lose energy to the environment. Energy may dissipate due to friction between surfaces, viscous drag in fluids, internal friction within materials, or electrical resistance in circuits. This energy loss reduces amplitude over time and is called damping. Modelling damping helps predict how long oscillations last and how structures respond to external disturbances.

Equation with damping
For a mass-spring with a damping force proportional to velocity, F_d = -b v, the equation of motion becomes m d²x/dt² + b dx/dt + k x = 0. Dividing by m gives d²x/dt² + (b/m) dx/dt + (k/m) x = 0. Introducing γ = b/(2m) and ω0 = √(k/m), the equation is d²x/dt² + 2γ dx/dt + ω0² x = 0. The character of the solution depends on γ relative to ω0.

Underdamped motion
If γ < ω0 the system is underdamped and oscillates while amplitude decays exponentially: x(t) = A e^{-γt} cos(ω' t + φ) where ω' = √(ω0² - γ²). The exponent e^{-γt} sets the envelope of decreasing amplitude. Underdamped systems still complete many oscillations before energy is mostly lost; this is typical for lightly damped mechanical and electrical resonators.

Critical and overdamping
Critical damping occurs at γ = ω0 and yields the fastest return to equilibrium without overshoot; solutions are non-oscillatory and involve terms like t e^{-γt}. Useful in applications like door closers or instrument design where rapid settling is desired. Overdamping (γ > ω0) also produces non-oscillatory decay but slower return than critical damping. Choosing damping properly balances response speed and overshoot risks.

Energy decay and quality factor
Damping causes mechanical energy to decrease with time. For underdamped motion energy decays roughly as E(t) = E0 e^{-2γt}. The quality factor Q = ω0/(2γ) measures how many oscillations occur before energy falls significantly; high Q means slow energy loss and sharp resonance, low Q means broad response and rapid damping. Engineers use Q to design filters, oscillators and mechanical systems.

Measuring damping
Experimentally one can measure amplitude of successive peaks and compute logarithmic decrement δ = ln(A_n/A_{n+1}). For weak damping, δ ≈ 2π/Q. Measuring the decay time constant or peak amplitudes gives γ and hence b. Practical control of damping includes viscous dampers, friction pads, tuned mass dampers in buildings, and electrical resistors in circuits.

📌 Examples
  • A damped oscillator with m = 1 kg, k = 100 N/m and b = 1 kg/s has γ = 0.5 s⁻¹ and ω = 10 s⁻¹ giving ω' ≈ √(100 - 0.25) ≈ 9.987 s⁻¹ (underdamped).
  • If amplitude reduces from 0.1 m to 0.05 m in 10 cycles, logarithmic decrement δ = (1/10) ln(0.1/0.05) = (1/10) ln 2 ≈ 0.0693.
🧮 Formulas
  1. Equation: m d²x/dt² + b dx/dt + k x = 0
  2. Damping constant γ = b/(2m), damped frequency ω' = √(ω² - γ²)
  3. \[Energy decay E(t) = E0 e^{-2γt}\]
    \[Q = ω/2γ\]
📊 Visual ideas
Displacement x versus time t showing exponentially decaying amplitude envelope for underdamped motion.
Comparison sketches of underdamped, critically damped and overdamped responses beginning from the same initial displacement.
💪6

Forced Oscillations and Resonance

Driven systems and steady-state
In forced oscillations an external periodic force drives the system. The standard form is m d²x/dt² + b dx/dt + k x = F0 cos(ω t), with driving angular frequency ω. The complete solution has a transient part that decays (if damping exists) and a steady-state part that oscillates at the driving frequency. After transients vanish, the system vibrates at ω, not at its natural frequency, but with amplitude and phase determined by the drive.

Amplitude response function
The steady-state amplitude A(ω) is given by A(ω) = F0 / √[(k - m ω²)² + (b ω)²]. This expression shows amplitude depends strongly on driving frequency: near the natural frequency ω0 = √(k/m) the term (k - m ω²) approaches zero and amplitude can become large if damping b is small. The shape of A(ω) versus ω is called the resonance curve or frequency response curve.

Resonant frequency and damping effect
For light damping the maximum amplitude occurs near ω ≈ ω0, often slightly lower for non-negligible damping. Damping limits maximum amplitude and broadens the resonance peak. The quality factor Q = ω0/(2γ) = m ω0 / b quantifies sharpness of resonance: larger Q means a taller, narrower peak and slower energy loss. In practice, engineering systems balance Q to achieve desired selectivity versus stability.

Phase relations and power transfer
The steady-state motion lags the driving force by a phase φ where tan φ = (b ω)/(k - m ω²). At low frequencies φ ≈ 0 (in-phase), at resonance φ = π/2 (quadrature) and at high frequencies φ ≈ π (out-of-phase). The average power absorbed from the driver depends on both amplitude and phase; power transfer is maximal near resonance when the driver performs work each cycle effectively overcoming damping losses.

Applications and control
Resonance is exploited in radio tuners, musical instruments, and sensors to amplify certain frequencies. However uncontrolled resonance can be destructive: bridges, engines and buildings must avoid resonant forcing. Common control methods include increasing damping, detuning natural frequencies, adding absorbers (tuned mass dampers), or designing forcing to avoid resonance frequencies. Understanding forced oscillations lets engineers predict responses and design safe, efficient systems.

📌 Examples
  • Driven mass-spring with m = 0.5 kg, k = 50 N/m, b = 0.2 kg/s and driving F0 = 1 N: natural ω0 = 10 s⁻¹; compute amplitude at ω = 10 s⁻¹ using formula.
  • A radio tuner adjusts circuit parameters to make the resonance frequency match broadcast frequency, selecting desired station.
🧮 Formulas
  1. Amplitude A(ω) = F0 / √[ (k - mω²)² + (bω)² ]
  2. Phase tan φ = bω / (k - mω²)
  3. Resonant frequency approximately ω0 = √(k/m), Q = ω0 / Δω
📊 Visual ideas
Amplitude A versus driving frequency ω showing resonance peak; indicate ω0 and bandwidth Δω.
Phase φ versus ω showing transition from 0 to π with φ = π/2 at resonance.
⚖️7

Mass-Spring System: Vertical and Horizontal Configurations

Mass-spring as model system
The mass-spring system is a prototypical oscillator used to illustrate SHM, damping and forced response. Consider a mass m attached to a spring of constant k. If the spring obeys Hooke's law and motion is constrained to one dimension, the restoring force is linear and the system is analytically tractable. Orientation (vertical or horizontal) changes static equilibrium but not the dynamics of small oscillations.

Horizontal arrangement
On a horizontal frictionless surface the equilibrium position is where the spring is relaxed. Displacing the mass by x produces a restoring force -kx and equation m d²x/dt² = -kx yields SHM with ω = √(k/m) and period T = 2π√(m/k). This simple picture is used for many theoretical examples and laboratory measurements of k or m.

Vertical arrangement and gravity
For vertical spring the weight mg stretches the spring until static equilibrium position x_eq satisfies k x_eq = mg. If the mass is displaced by small amount y relative to this equilibrium and released, the equation for y is m d²y/dt² = -k y, identical in form to the horizontal case. Thus the small-oscillation period is still T = 2π√(m/k) and independent of gravity. This often surprises beginners but follows from linearisation about the shifted equilibrium.

Multiple springs and effective constants
Practical systems may use springs in series or parallel. For springs in series k_eff = (k1 k2)/(k1 + k2) because extensions add under same force. For parallel springs k_eff = k1 + k2 because forces add under same extension. Using these rules designers tune effective stiffness and hence natural frequency. Also the mass of the spring itself can contribute to the inertia; for light springs an effective mass correction is applied by adding a fraction of the spring mass to m.

Amplitude and extremes
Given amplitude A, the maximum speed and acceleration of mass follow from v_max = A ω and a_max = A ω² using ω = √(k/m). Energy viewpoint gives total mechanical energy E = 1/2 k A². Real systems include damping and nonlinearity; large amplitudes may lead to deviations from SHM because Hooke's law no longer strictly holds.

Experimental notes
Measurements of period with varying mass or spring constant validate T = 2π√(m/k). Plotting T² versus m gives a straight line whose slope relates to k. Mass-spring models also serve as analogues for electrical LC circuits and other oscillatory systems, linking mechanics to broader physics topics.

📌 Examples
  • A 0.2 kg mass attached to a spring of k = 80 N/m has period T = 2π√(m/k) = 2π√(0.2/80) ≈ 0.99 s.
  • Two springs of 100 N/m and 200 N/m in series give k_eff = (100×200)/(100+200) = 20000/300 ≈ 66.67 N/m.
🧮 Formulas
  1. ω = √(k/m), T = 2π√(m/k)
  2. k_series = (k1 k2)/(k1 + k2), k_parallel = k1 + k2
  3. v_max = Aω, a_max = Aω²
📊 Visual ideas
Vertical mass-spring sketch showing equilibrium position x_eq and small displacement y about equilibrium.
Plot of displacement x versus time showing sinusoidal motion for the mass-spring system.
🕐8

Simple Pendulum and Its Time Period

Definition and small-angle behaviour
A simple pendulum consists of a point mass (bob) suspended by a light, inextensible string of length L from a fixed pivot. When displaced by a small angle θ and released, the bob oscillates under the influence of gravity. For small angles (θ measured in radians, θ ≲ 10°), the restoring tangential component of weight is approximately -mgθ, producing motion that closely follows SHM.

Derivation of the equation of motion
The tangential component of force for angle θ is -mg sin θ. For small θ, sin θ ≈ θ, so tangential acceleration is d²s/dt² = L d²θ/dt² and Newton's law gives mL d²θ/dt² = -mg θ. Simplify to d²θ/dt² + (g/L) θ = 0. This has the SHM form with angular frequency ω = √(g/L). Thus the period of small oscillations is T = 2π√(L/g), independent of mass and (to first order) independent of amplitude.

Physical interpretation and consequences
The period depends only on pendulum length and local gravity. This allows measurement of g by timing oscillations for known L, or vice versa. The independence from mass arises because gravitational force and inertial resistance both scale with mass and cancel out in the equation of motion. The independence from amplitude is only approximate; for larger amplitudes period increases slightly because sin θ ≠ θ.

Limits of the approximation
For larger angles the small-angle approximation fails and the exact period involves an elliptic integral. Practically, for θ up to about 10°, error is under 1% and the simple formula is accurate enough for many classroom experiments. Keep amplitude small to avoid systematic errors when determining g experimentally.

Simple pendulum vs physical pendulum
A simple pendulum assumes a point mass at distance L from pivot. If the mass is extended, rotational inertia matters and the system becomes a physical pendulum with period T = 2π√(I/(mgd)), where I is moment of inertia about pivot and d distance from pivot to centre of mass. For a point mass I = mL² and d = L this expression reduces to the simple pendulum result.

Applications and experiments
Pendulums are used in clocks, seismometers and experiments to measure g. To improve accuracy measure time for many oscillations and divide, reducing fractional timing errors. Temperature and air resistance can slightly change effective length and damping; in precision work these are accounted for.

📌 Examples
  • A pendulum of length 1 m has period T = 2π√(1/9.8) ≈ 2.01 s.
  • If T is measured as 2 s for a pendulum in a lab, using g = 9.8 m/s² gives L = g(T/2π)² ≈ 0.994 m.
🧮 Formulas
  1. For simple pendulum: T = 2π√(L/g), ω = √(g/L)
  2. Physical pendulum: T = 2π√(I/(mgd))
📊 Visual ideas
Pendulum sketch showing length L, displacement angle θ and small-angle approximation where arc ≈ Lθ.
Plot of period T versus length L showing square-root dependence.
⚗️9

Physical Pendulum and Compound Systems

Extending to rigid bodies
Real oscillating bodies are often extended objects rather than point masses. A physical pendulum is any rigid body oscillating about a horizontal axis not through its centre of mass. The restoring torque for small angular displacement θ is τ ≈ -mgd θ where d is the distance from pivot to the centre of mass. Using rotational dynamics, I d²θ/dt² = -mgd θ, where I is moment of inertia about the pivot. This leads to SHM with angular frequency ω = √(mgd/I) and period T = 2π√(I/(mgd)).

Calculating I and d for common shapes
Compute I for given geometry about the pivot using standard formulae or parallel-axis theorem. For a uniform rod of length l pivoted a distance a from its centre, I = (1/12) m l² + m a² and d = a. Substituting gives T in terms of l, m and a. For a disk pivoted at rim, I = 1/2 m R² and d = R, so T = 2π√(R/(2g)). These formulae allow design and prediction of behaviour for beams, levers and instruments.

Equivalent simple pendulum and centre of oscillation
Any physical pendulum has an equivalent simple pendulum length L_eq such that T = 2π√(L_eq/g), with L_eq = I/(m d). This concept permits using familiar simple pendulum relations to interpret physical pendulum behaviour. Measuring period about different pivots can identify the centre of oscillation; the distance between pivot and centre of oscillation is reversible: switching pivot and oscillation centre yields the same period. This principle underlies the reversible pendulum used for accurate g measurements.

Design and sensitivity
Physical pendulums are used in sensors, clocks and experimental apparatus. Sensitivity to small changes in mass distribution or pivot location is exploited in balances and suspension-based instruments. For precise timing, damping should be minimized and the small-angle condition preserved. Large amplitude or non-rigid mounting can introduce nonlinearities and energy losses.

Practical examples and measurements
Use measured T and known geometry to calculate I or g. For a uniform rod pivoted at one end, I = (1/3) m l² and d = l/2 gives T = 2π√(2l/(3g)). For design, choosing pivot location and mass distribution adjusts period to desired values. Understanding physical pendulums extends the simple pendulum concept to realistic experimental and engineering systems.

📌 Examples
  • A uniform rod of length 1 m pivoted about one end: I = (1/3) m l², d = l/2; T = 2π√(I/(mgd)) = 2π√((1/3)m l²/(m g l/2)) = 2π√(2l/(3g)).
  • For a disk of radius R pivoted at rim, I = 1/2 m R², d = R; T = 2π√((1/2 m R²)/(m g R)) = 2π√(R/(2g)).
🧮 Formulas
  1. Physical pendulum: T = 2π√(I/(mgd)), ω = √(mgd/I)
  2. Equivalent length L_eq = I/(md)
📊 Visual ideas
Diagram of a physical pendulum showing pivot, centre of mass, distance d and small angular displacement θ.
Graph of T versus pivot position a for a uniform rod showing minimum at centre of oscillation.
🔬10

Coupled Oscillations and Normal Modes (Qualitative)

What coupling means
Coupled oscillators are systems in which individual oscillators interact so that motion of one affects others. Coupling may be mechanical (springs between masses), electromagnetic (coupled circuits), or through any medium that allows energy transfer. Coupled systems have collective behaviour that cannot be understood by studying each oscillator in isolation.

Normal modes concept
Normal modes are particular collective motions in which all parts oscillate sinusoidally at a common frequency with fixed relative amplitudes. In a normal mode the system behaves as if it were a single oscillator with its own characteristic frequency (eigenfrequency). For N coupled degrees of freedom there are N normal modes. Any initial motion can be expressed as a superposition of these modes, each evolving independently in linear systems.

Simple two-mass example
Consider two identical masses each attached to fixed supports by outer springs k and coupled to each other by a central spring k_c. There are two normal modes: (1) in-phase mode where both masses move together—coupling spring is unstretched and frequency equals that of a single mass on one spring; (2) out-of-phase mode where masses move oppositely—coupling spring stretches more and frequency is higher. Calculating these frequencies involves solving simultaneous equations and finding eigenvalues.

Energy exchange and beats
If initial conditions do not correspond to a single normal mode, the motion is a superposition and energy oscillates between components. For nearly equal frequencies this produces beats—amplitude modulation at the beat frequency which equals the difference between mode frequencies. Beats are observable in coupled tuning forks and other nearly-tuned oscillators and provide a way to measure small frequency differences.

Mathematical outline and generalisation
To find normal modes write equations of motion for each coordinate, assume solutions proportional to e^{iωt} and reduce to an algebraic eigenvalue problem. Solving gives allowed frequencies and amplitude ratios (mode shapes). This procedure generalises to chains of masses, continuous systems (strings, rods) and even molecules where normal modes correspond to vibrational spectra used in spectroscopy.

Practical significance
Understanding coupled oscillations is essential in designing bridges, buildings and machines where collective resonances can cause structural failure. It also underlies the operation of musical instruments, the analysis of molecular vibrations and the design of filters and coupled resonators in electronics. Studying normal modes simplifies complex dynamics into independent oscillatory components.

📌 Examples
  • Two identical masses m coupled by spring kc and each attached to wall by k; modes are in-phase with frequency √(k/m) and out-of-phase with frequency √((k+2kc)/m).
  • Beats heard when two tuning forks of slightly different frequencies are struck together; loudness varies at beat frequency Δf.
🧮 Formulas
  1. For symmetric two-mass system: ω1 = √(k/m), ω2 = √((k + 2kc)/m)
  2. Beat frequency f_beat = |f1 - f2|
📊 Visual ideas
Sketch of two-mass system showing mode shapes: (a) both masses moving same direction (in phase), (b) masses moving opposite directions (out of phase).
Amplitude vs time plot showing beats: envelope varying at beat frequency with faster oscillations inside.
🏃11

Wave Motion: Definitions and Basic Quantities

Definition and essential idea
A wave is a disturbance that travels through space and time, carrying energy and sometimes information, while individual particles of the medium undergo local oscillations about equilibrium. Waves can be mechanical, requiring a medium (sound, waves on strings), or non-mechanical, like electromagnetic waves which can travel in vacuum. Understanding basic wave quantities allows you to describe and relate time and space behaviour of waves.

Key measurable quantities
Wavelength λ is the spatial period: distance between successive points with same phase (crests, troughs or compression centres). Period T is the temporal period at a fixed point and frequency f = 1/T is how many oscillations occur per second. Wave speed v is the rate at which a particular phase travels through space. For sinusoidal waves these are related by v = f λ. Angular frequency ω = 2π f and wave number k = 2π/λ are convenient when writing sinusoidal expressions.

Mathematical representation
A one-dimensional harmonic wave travelling in the positive x-direction can be written y(x,t) = A cos(kx - ωt + φ). The phase ψ = kx - ωt + φ determines the state of oscillation; points of constant ψ move with velocity v = ω/k. The choice of sign in the phase argument sets the direction: kx - ωt moves to increasing x, kx + ωt moves to decreasing x. The amplitude A sets displacement magnitude; φ sets initial phase.

Dependence of wave speed on medium
Wave speed depends on restoring forces and inertia of the medium. For transverse waves on a stretched string v = √(T/μ) where T is tension and μ mass per unit length. For sound in a gas v ≈ √(γ P / ρ) or more practically v ≈ 331 + 0.6 θ m/s with temperature θ in °C. In solids waves can be longitudinal or transverse with speeds determined by elastic moduli and density. In dispersive media v may depend on frequency, causing wave packets to spread.

Physical meaning and experiments
When a wave passes, energy flows through the medium although particles return to near their starting positions in many cases. Measuring λ and f lets you determine v; measuring v and known medium properties allows inference of other quantities like tension or density. Basic wave understanding is foundational for studying superposition, interference, standing waves, resonance and applications such as musical instruments and communication systems.

📌 Examples
  • A wave with λ = 0.5 m and f = 10 Hz has speed v = f λ = 5 m/s.
  • A sinusoidal transverse wave on a string y = 0.02 cos(4πx - 40πt) has k = 4π rad/m, ω = 40π rad/s, so v = ω/k = 10 m/s and f = ω/(2π) = 20 Hz.
🧮 Formulas
  1. v = f λ, ω = 2πf, k = 2π/λ, y(x,t) = A cos(kx - ωt + φ)
  2. String wave speed v = √(T/μ) (T = tension, μ = mass per unit length)
📊 Visual ideas
A snapshot y versus x showing a sinusoidal wave with labelled wavelength λ and amplitude A.
Space-time diagram showing movement of crest from x1 at time t1 to x2 at t2 to illustrate wave speed.
🌊12

Transverse and Longitudinal Waves

Classification by particle motion
Waves are often classified by the direction of particle oscillation relative to wave propagation. In transverse waves particles oscillate perpendicular to the direction the wave travels; in longitudinal waves they oscillate parallel to it. Understanding the distinction helps identify appropriate models and boundary conditions for different physical situations.

Transverse waves — examples and properties
On a taut string or membrane, disturbances move along the medium while each element moves up and down. Mathematical form y(x,t) = A cos(kx - ωt) describes transverse displacement. Polarisation is a property of transverse waves: because oscillation direction lies in a plane perpendicular to travel, one can restrict or analyse vibration directions. Energy transport in transverse waves involves both kinetic energy of motion and potential energy stored by stretching or bending of the medium.

Longitudinal waves — examples and properties
In longitudinal waves, particles oscillate along the same axis as wave travel. Sound waves in gases and liquids are longitudinal: regions of compression (higher pressure) and rarefaction (lower pressure) move through the medium. A longitudinal displacement function s(x,t) = s_m cos(kx - ωt) describes particle motion. For sound, pressure variations often lead displacement by a phase of about π/2. Speed depends on medium's compressibility and density: v = √(B/ρ) where B is bulk modulus; for ideal gases v ≈ √(γ P/ρ).

Representations and measurement
Transverse waves are conveniently sketched as crests and troughs in the plane perpendicular to motion; longitudinal waves are shown as alternating compressed and expanded regions along the direction of travel. Wavelength in longitudinal waves equals distance between successive compressions. Experiments with slinky springs make the difference easy to see: pushing the slinky along its length produces longitudinal pulses, while moving one end sideways produces transverse pulses.

Special considerations
Some media support both types of waves: solids can carry both longitudinal (compressional) and transverse (shear) waves with different speeds. In fluids, shear modes are not supported. Polarisation is relevant for transverse waves in optics and mechanics. Recognising wave type determines boundary conditions when constructing standing waves and predicting node/antinode positions in experiments and instruments.

📌 Examples
  • A slinky stretched and pushed longitudinally shows compressions traveling along its length — an example of longitudinal wave.
  • A rope shaken up and down produces transverse waves; if shaken sideways at fixed frequency, crests and troughs travel along rope.
🧮 Formulas
  1. Longitudinal wave speed in gas: v = √(γ P / ρ) (approx.), or v = √(B/ρ) generally.
  2. Transverse wave on string: v = √(T/μ).
📊 Visual ideas
Sketch of transverse wave showing crests and troughs and labelled wavelength λ and amplitude A.
Longitudinal wave diagram showing compressions and rarefactions along x-axis with labelled λ.
🌊13

Wave Equation and Solutions for One-Dimensional Waves

Derivation idea from a string element
Consider a small segment of a stretched string of length dx under tension T. Taking vertical displacement y(x,t) and using force balance with small-angle approximations (slope ≈ ∂y/∂x) leads to T ∂²y/∂x² dx = μ dx ∂²y/∂t², where μ is linear mass density. Cancelling dx gives the one-dimensional wave equation ∂²y/∂x² = (1/v²) ∂²y/∂t² with wave speed v = √(T/μ). This fundamental partial differential equation describes propagation of transverse disturbances on the string.

General solution form
The wave equation is linear and admits general solutions y(x,t) = f(x - vt) + g(x + vt), where f and g are arbitrary twice-differentiable functions representing right-travelling and left-travelling waveforms. This shows that any disturbance splits into components moving in both directions without change of shape if v is constant and medium linear. For harmonic waves choose f and g as sinusoids to get y(x,t) = A cos(kx - ωt + φ) or sums of such terms.

Harmonic waves and dispersion
For a harmonic plane wave y = A cos(kx - ωt) to satisfy the wave equation requires ω/k = v, giving dispersionless relation ω = v k. In more complex media dispersion may occur where wave speed depends on frequency, causing wave packets to spread. For many mechanical waves in simple media dispersion is absent and all frequency components travel at same speed.

Superposition and standing waves
By linearity, the sum of any two solutions is a solution. Two equal-amplitude harmonic waves travelling opposite directions produce a standing wave y(x,t) = 2A sin(kx) cos(ωt) or similar form depending on phase. Standing waves have fixed nodes where amplitude is always zero and antinodes where oscillation is maximal. Boundary conditions such as fixed ends quantise allowed k and therefore allowed wavelengths and frequencies in bounded media.

Applying boundary conditions
For a string fixed at both ends of length L, y(0,t) = y(L,t) = 0 gives allowed wavelengths λ_n = 2L/n and frequencies f_n = n v/(2L). These discrete modes form a basis: any initial deformation can be expressed as a sum of normal modes. This approach extends to pipes, rods and membranes with appropriate conditions, and forms the basis of musical tones and resonance in engineering.

📌 Examples
  • A string fixed at both ends of length L supports standing waves with frequencies f_n = n v/(2L). For L = 1 m and v = 100 m/s, the fundamental f1 = 50 Hz, second harmonic f2 = 100 Hz, etc.
  • Right-moving harmonic wave y = 0.01 cos(πx - 200πt) has k = π rad/m, ω = 200π rad/s, so v = ω/k = 200 m/s and f = ω/(2π) = 100 Hz.
🧮 Formulas
  1. Wave equation: ∂²y/∂x² = (1/v²) ∂²y/∂t²
  2. General harmonic solution: y(x,t) = A cos(kx - ωt + φ), with v = ω/k
  3. Standing wave frequencies for fixed ends: f_n = n v/(2L), λ_n = 2L/n
📊 Visual ideas
Sketch of standing wave on a string fixed at both ends showing nodes and antinodes for n = 1 and n = 2.
Plot of a travelling sinusoidal wave y(x,t) at two different times showing displacement shifted by vt.
🔬14

Superposition, Interference and Beats

Principle of superposition
When two or more waves coexist in a linear medium, the resultant displacement at any point and time is the algebraic sum of the individual displacements. This is the superposition principle and it leads directly to interference phenomena. It holds when the underlying wave equation is linear and amplitudes are small enough that nonlinear effects are negligible.

Interference of coherent waves
If two sinusoidal waves of same frequency and amplitudes A1 and A2 meet, their phase difference Δφ determines the resultant amplitude R. Using trigonometric identities, R = √(A1² + A2² + 2A1A2 cos Δφ). For equal amplitudes A, R = 2A cos(Δφ/2). If Δφ = 0 waves add constructively, if Δφ = π they cancel destructively. Spatial or temporal phase differences produce interference patterns such as bright and dark fringes in optics or loud and soft spots in acoustics.

Beats from slightly different frequencies
When two waves of nearly equal frequency f1 and f2 superpose, the result shows amplitude modulation at the beat frequency f_beat = |f1 - f2|. Using the identity cos α + cos β = 2 cos((α-β)/2) cos((α+β)/2) the sum becomes a rapidly oscillating carrier at average frequency with a slowly varying envelope at difference frequency. Beats are audible in sound and used practically to tune instruments by listening for the beat rate between a reference tone and the instrument's tone.

Conditions for stable interference
For sustained, visible or audible interference the sources must be coherent — they must maintain a constant phase relation. Lasers and carefully controlled oscillators provide coherence; ordinary lamps or random-phase sources do not produce stable interference patterns. Equal or comparable amplitudes improve visibility of interference fringes or beat envelopes.

Applications and experiments
Interference underlies many measurement techniques: interferometers measure small distances and refractive index changes by observing fringe shifts. Beats help tune musical instruments and detect small frequency differences in electronics. Noise-cancelling headphones use destructive interference by generating sound waves out of phase with ambient noise. Practically, mastering superposition and interference lets you predict resultant waveforms and design experiments or devices that harness or avoid interference effects.

📌 Examples
  • Two sound sources of 440 Hz and 442 Hz produce beats at frequency 2 Hz; amplitude oscillates twice per second.
  • Two coherent light waves with phase difference π produce destructive interference at a point leading to dark fringe in a double-slit experiment.
🧮 Formulas
  1. Resultant amplitude R = √(A1² + A2² + 2A1A2 cos Δφ), for A1 = A2 = A, R = 2A cos(Δφ/2)
  2. Beat frequency f_beat = |f1 - f2|
📊 Visual ideas
Graph of resultant amplitude vs time for two close frequencies showing slow envelope (beats) modulating fast carrier oscillation.
Schematic of two-source interference on a screen showing constructive and destructive fringes.
🌊15

Standing Waves on Strings and Pipes

How standing waves form
Standing waves arise when two waves of the same frequency and amplitude travel in opposite directions and interfere. This often happens when a travelling wave reflects from a boundary. The result is a spatial pattern that oscillates in time but does not translate: nodes are fixed points with zero amplitude and antinodes are points of maximum amplitude. Standing waves are central to resonance in strings and air columns.

Strings fixed at both ends
For a string fixed at x = 0 and x = L boundary conditions require y(0,t) = y(L,t) = 0. Only sinusoidal modes that satisfy these conditions exist: λ_n = 2L/n and f_n = n v/(2L) where n is a positive integer. The fundamental (n=1) has one antinode at the centre; higher harmonics have additional nodes and antinodes. These discrete frequencies are the harmonics of the string and determine musical notes produced by plucked or bowed instruments.

Pipes open or closed
Pipes support longitudinal standing waves. For a pipe open at both ends displacement antinodes occur at ends so allowed wavelengths are λ_n = 2L/n, same as a string; frequencies f_n = n v/(2L). For a pipe closed at one end displacement node at closed end and antinode at open end produce allowed wavelengths λ_n = 4L/n with n odd, leading to frequencies f_n = n v/(4L) for n = 1,3,5,... Thus closed pipes generate only odd harmonics, giving characteristic timbre of instruments like clarinets. End correction must be considered: antinodes form slightly outside open ends, effectively increasing L by an amount dependent on pipe radius.

Nodes, antinodes and mode shapes
Mode shapes show where nodes and antinodes occur along the medium. For the nth harmonic on a string there are n half-wavelength segments and n+1 nodes including ends. The spacing of nodes relates to harmonic number: adjacent nodes are separated by λ/2. Plucking or exciting the string at different positions favours certain harmonics; placing a pluck at a node suppresses the corresponding harmonic.

Practical implications
Standing wave resonance amplifies particular frequencies in instruments and cavities but can be problematic in buildings or pipes if resonant frequencies match forcing frequencies. Designers use damping, geometry changes or material selection to control resonances. Experimentally, measuring resonant frequencies provides a way to determine wave speed and medium properties.

📌 Examples
  • A guitar string length 0.65 m with wave speed 520 m/s has fundamental f1 = v/(2L) ≈ 400 Hz (approximate A note).
  • An open organ pipe of length 0.85 m with sound speed 340 m/s has fundamental f1 = v/(2L) ≈ 200 Hz.
🧮 Formulas
  1. String or open pipe: f_n = n v/(2L), λ_n = 2L/n
  2. Pipe closed at one end: f_n = n v/(4L) with n = 1,3,5..., λ_n = 4L/n
📊 Visual ideas
Mode shape sketches for first three harmonics of a string fixed at both ends showing node and antinode positions.
Longitudinal standing wave in a closed-open pipe showing node at closed end and antinode at open end.
🔊16

Sound Waves: Characteristics and Speed in Air

Nature of sound and perceptual qualities
Sound is a mechanical longitudinal wave that propagates through elastic media like air, liquids and solids. The basic physical quantities of sound are frequency (perceived as pitch), amplitude (related to loudness), wavelength and speed. The waveform and harmonic content determine timbre, the quality that distinguishes different instruments playing the same note.

Speed of sound in air
The speed of sound in an ideal gas under adiabatic conditions is v = √(γ P / ρ), where γ is the ratio of specific heats, P is pressure and ρ density. For practical purposes at ordinary temperatures, v depends mainly on temperature and is approximated by v ≈ 331 + 0.6 θ m/s where θ is temperature in °C. Humidity and altitude also affect speed slightly by changing air composition and density.

Intensity and decibel scale
Sound intensity I is power transmitted per unit area and is proportional to the square of wave amplitude. Because human hearing responds logarithmically, sound levels are expressed in decibels: β = 10 log10(I/I0), with reference I0 = 10⁻¹² W/m². An increase of 10 dB represents ten times greater intensity and is typically perceived as about twice as loud. Understanding intensity is important for safety, room acoustics and sound engineering.

Doppler effect
The Doppler effect describes the change in observed frequency when source and observer move relative to each other. For speeds small compared to sound speed, f' ≈ f (v ± v_o)/(v ∓ v_s) where signs follow motion towards or away. This effect has practical uses in radar, medical ultrasound, and astronomy, and explains frequency shifts heard from passing vehicles.

Applications and safety
Sound waves are essential in music, communications and sensing. Ultrasound (above human hearing) is used in medical imaging and industrial inspection. Excessive exposure to high sound levels can damage hearing; regulations and engineering controls manage occupational and environmental noise. Understanding wave behaviour helps design concert halls, loudspeakers and instruments with desired acoustic properties.

📌 Examples
  • At 20°C, speed of sound v ≈ 331 + 0.6×20 = 343 m/s.
  • A source at 1000 Hz approaches at 10 m/s; observer stationary: approximate observed frequency f' = f v/(v - v_s) ≈ 1000×343/(343 - 10) ≈ 1030 Hz.
🧮 Formulas
  1. Speed in gas: v = √(γ P / ρ) approximately v ≈ 331 + 0.6 θ m/s
  2. Sound intensity level β (dB) = 10 log10(I/I0), I0 = 10⁻¹² W/m²
  3. Doppler formula: f' = f (v ± v_o)/(v ∓ v_s)
📊 Visual ideas
Pressure variation versus position for a plane longitudinal sound wave showing compressions and rarefactions.
Plot of intensity in dB versus distance showing approximate decrease with distance from a point source (inverse square law for intensity).
🎵17

Resonance in Tubes and Musical Instruments

Resonant air columns basics
Air columns in tubes resonate at discrete frequencies determined by length, boundary conditions and the speed of sound. For an open-open tube antinodes occur at the open ends and nodes inside; allowed wavelengths are λ_n = 2L/n and frequencies f_n = n v/(2L). For an open-closed tube the closed end forces a node and the open end an antinode, producing allowed wavelengths λ_n = 4L/n with n odd and frequencies f_n = n v/(4L) for n = 1,3,5.... End correction increases effective length slightly because antinodes occur a short distance outside the open end; for a circular pipe the correction is roughly 0.6 times pipe radius per open end.

Instrument examples and harmonic content
Different instruments approximate different boundary conditions. Flutes and organ pipes behave like open-open tubes producing both even and odd harmonics. Clarinets approximate open-closed geometry, emphasising odd harmonics and producing a distinct tonal colour. Brass instruments use lip excitation and a flaring bell to select and radiate harmonics; changing the effective length via valves or slides shifts pitch. String instruments combine string modes with resonant soundboards and cavities to produce rich harmonic spectra.

Tuning and practical adjustments
Musicians tune by adjusting effective length or tension to align resonant frequencies with desired notes. Makers shape mouthpieces, tone holes and body cavities to control harmonic content and volume. End correction must be considered for precise tuning and for laboratory measurements of sound speed: measuring resonant lengths for known frequencies lets one calculate v accurately when end corrections are included.

Resonance hazards and engineering controls
Resonant amplification of sound or mechanical vibrations can damage structures or instruments. Examples include amplified vibrations in pipes or cavities and structural resonances excited by periodic forces. Engineers control harmful resonance by adding damping materials, changing geometry, shifting natural frequencies and using absorbers. In concert hall design resonances and standing waves are managed to achieve desirable acoustics rather than destructive peaks.

Measurements and experiments
Laboratory experiments with tuning forks and adjustable-length tubes allow measurement of resonant modes and calculation of sound speed. Fourier analysis of recorded sounds reveals harmonic content and helps compare instruments. Understanding resonance in tubes and instruments connects physical wave theory with musical practice and engineering design.

📌 Examples
  • An open organ pipe of length 0.85 m at 20°C (v ≈ 343 m/s) has fundamental f1 = v/(2L) ≈ 202 Hz.
  • A clarinet (approx. closed-open) of length 0.6 m has fundamental f1 ≈ v/(4L) ≈ 143 Hz (approximate for illustration).
🧮 Formulas
  1. Open-open pipe: f_n = n v/(2L), λ_n = 2L/n
  2. Open-closed pipe: f_n = n v/(4L) for n = 1,3,5..., λ_n = 4L/n
  3. End correction: L_eff = L + 0.6r for open end radius r (approximate)
📊 Visual ideas
Sketch of standing wave patterns in open-open and open-closed pipes showing node/antinode positions for first few harmonics.
Diagram showing end correction with effective antinode slightly outside open end.
🌊18

Wave Optics Connections: Brief Introduction to Harmonics and Fourier View

Periodic motion and harmonic decomposition
Many oscillatory motions and waveforms are not simple sinusoids but can be represented as sums of sinusoidal components called harmonics. Fourier analysis states that any reasonable periodic function can be written as an infinite series of sines and cosines with frequencies that are integer multiples of a fundamental frequency. This viewpoint links time-domain shapes to frequency-domain spectra and is a powerful tool across physics and engineering.

Harmonics and timbre
In musical acoustics the tones produced by an instrument are combinations of a fundamental frequency and overtones (harmonics). A string fixed at both ends supports modes with frequencies f_n = n f1; the specific amplitudes and phases of these harmonics determine the instrument's timbre. Plucking or bowing location and technique change the mix of harmonics, producing varied sounds even at the same pitch.

Fourier series and practical use
For a periodic function of period T the Fourier series expresses it as Σ [a_n cos(2π n t/T) + b_n sin(2π n t/T)]. The coefficients a_n and b_n quantify the contribution of each harmonic. In practice, Fourier transforms extend this idea to non-periodic signals, decomposing a signal into continuous frequency components. Linear systems respond independently to each frequency component, simplifying analysis: treat each harmonic separately and superpose results due to linearity.

Connections to optics and diffraction
In wave optics the diffraction pattern of an aperture is the Fourier transform of the aperture function. Interference patterns can be predicted using harmonic components. Concepts like spatial harmonics, grating orders and spectral decomposition are direct applications of Fourier ideas. Thus the same mathematical tools describe vibrations of strings and diffraction of light, showing unity across wave phenomena.

Experimental illustrations
Simple experiments demonstrate Fourier ideas: record a plucked string and use a frequency analyser (or smartphone app) to view harmonics. Synthesise square or sawtooth waveforms by adding odd or all harmonics with appropriate amplitudes and phases. Observing how filtering removes or emphasises components helps understand equalisers, instrument design and signal processing.

📌 Examples
  • Plucking a guitar string at the midpoint suppresses even harmonics and emphasises odd harmonics, changing tone.
  • A square wave can be approximated by adding odd harmonics of a fundamental sine wave series with appropriate amplitudes.
🧮 Formulas
  1. Harmonic frequencies for string fixed at both ends: f_n = n v/(2L) = n f1
  2. Fourier representation (qualitative): periodic function = Σ [a_n cos(nωt) + b_n sin(nωt)]
📊 Visual ideas
Schematic frequency spectrum showing fundamental and higher harmonics with labelled amplitudes.
Waveform of a square wave approximated by first few odd harmonics plotted against time.

Key Concepts

Oscillation
A repeated to-and-fro motion about an equilibrium position.
Simple Harmonic Motion (SHM)
Periodic motion where restoring force is proportional to displacement and directed towards equilibrium.
Amplitude
Maximum displacement from equilibrium in an oscillation.
Period
Time taken for one complete oscillation.
Frequency
Number of oscillations per unit time, f = 1/T.
Angular frequency
Rate of change of phase, ω = 2πf.
Damping
Process by which oscillation amplitude decreases due to energy loss.
Resonance
Large amplitude response when driving frequency matches system's natural frequency.
Quality factor (Q)
Dimensionless measure of sharpness of resonance; Q = ω/2γ for weak damping.
Wave
A disturbance that transfers energy through a medium or space without net transport of matter.
Wavelength
Distance between successive points in phase on a wave, denoted λ.
Wave speed
Speed at which a disturbance or phase travels, v = f λ.
Transverse wave
Wave in which particles oscillate perpendicular to direction of propagation.
Longitudinal wave
Wave in which particles oscillate parallel to direction of propagation.
Standing wave
Result of superposing two waves of same frequency traveling opposite directions, producing nodes and antinodes.
Node
Point on a standing wave with zero amplitude at all times.
Antinode
Point on a standing wave where amplitude is maximum.
Beat frequency
Frequency of amplitude modulation when two waves of slightly different frequencies superpose, equal to |f1 - f2|.
Wave equation
Partial differential equation ∂²y/∂x² = (1/v²) ∂²y/∂t² governing one-dimensional wave propagation.

Practice Questions

  1. A mass of 0.5 kg is attached to a spring of constant 200 N/m and displaced 0.02 m from equilibrium and released. Find the angular frequency, period and maximum speed. / 0.5 किग्रा द्रव्यमान को 200 N/m के स्थिरांक वाले वसंत से जोड़ा गया है और इसे 0.02 m व्यवहारिक विचलन पर छोड़ दिया जाता है। कोणीय आवृत्ति, अवधी और अधिकतम वेग ज्ञात कीजिए।
    Show answer

    Angular frequency ω = √(k/m) = √(200/0.5) = √400 = 20 s⁻¹. Period T = 2π/ω = 2π/20 = π/10 ≈ 0.314 s. Maximum speed v_max = A ω = 0.02 × 20 = 0.4 m/s. / कोणीय आवृत्ति ω = √(k/m) = √(200/0.5) = 20 s⁻¹। अवधी T = 2π/ω = 2π/20 ≈ 0.314 s। अधिकतम वेग v_max = Aω = 0.02×20 = 0.4 m/s।

  2. Write the equation of motion for a simple pendulum of length L for small angles and derive its time period. / छोटे कोणों के लिए लंबाई L वाले सरल लोलक का गति समीकरण लिखिए और इसका अवधी व्युत्पन्न कीजिए।
    Show answer

    For small angle θ, tangential restoring torque gives equation d²θ/dt² + (g/L) θ = 0. This is SHM with ω = √(g/L). Therefore period T = 2π/ω = 2π√(L/g). / छोटे कोणों के लिए स्पर्शीय प्रत्यास्थ बल से d²θ/dt² + (g/L) θ = 0। यह SHM है जिसका ω = √(g/L)। अतः अवधी T = 2π√(L/g)।

  3. Two waves y1 = 0.03 cos(200π t) m and y2 = 0.03 cos(202π t) m are superposed. Find the beat frequency and write the resultant expression showing beats. / दो तरंगें y1 = 0.03 cos(200π t) m और y2 = 0.03 cos(202π t) m सुपरपोज़ की जाती हैं। बीट आवृत्ति ज्ञात कीजिए और बीट दिखाते हुए संयुक्त अभिव्यक्ति लिखिए।
    Show answer

    Frequencies are f1 = 200π/(2π) = 100 Hz and f2 = 202π/(2π) = 101 Hz. Beat frequency f_beat = |f2 - f1| = 1 Hz. Using trig identity, resultant y = 2×0.03 cos(π t) cos(201π t) = 0.06 cos(π t) cos(201π t), showing envelope cos(π t) with beat frequency 1 Hz. / आवृत्तियाँ f1 = 100 Hz, f2 = 101 Hz। बीट आवृत्ति 1 Hz है। योग y = 0.06 cos(π t) cos(201π t) जो 1 Hz के बीट का आवरण दर्शाता है।

  4. A string of length 0.8 m fixed at both ends has wave speed 320 m/s. Calculate the fundamental frequency and the frequency of the third harmonic. / दोनों सिरों पर स्थिर 0.8 m लंबी तार पर तरंग गति 320 m/s है। मूल आवृत्ति और तीसरे हार्मोनिक की आवृत्ति ज्ञात कीजिए।
    Show answer

    Fundamental f1 = v/(2L) = 320/(2×0.8) = 320/1.6 = 200 Hz. Third harmonic f3 = 3 f1 = 600 Hz. / मूल आवृत्ति f1 = 200 Hz, तीसरा हार्मोनिक f3 = 600 Hz।

  5. Explain qualitatively why resonance can be dangerous for structures and give one engineering method to reduce harmful resonance. / गुणात्मक रूप से समझाइए कि अनुनाद संरचनाओं के लिए क्यों हानिकारक हो सकता है और हानिकारक अनुनाद कम करने का एक अभियांत्रिकी तरीका बताइए।
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    Resonance amplifies motion when external forcing frequency matches a structure’s natural frequency, causing large amplitudes and stresses that can lead to damage or collapse (example: bridge failure). To reduce harmful resonance engineers add damping (dissipates energy), change stiffness or mass to shift natural frequency, or avoid periodic forcing at the resonant frequency. / अनुनाद तब खतरनाक होता है जब बाहरी आवेग की आवृत्ति संरचना की स्वाभाविक आवृत्ति से मेल खा जाती है; इससे आयाम बड़े हो जाते हैं और तनाव बढ़कर क्षति हो सकती है। हानिकारक अनुनाद कम करने के लिए अभियांत्री damping जोड़ते हैं, कठोरता या द्रव्यमान बदलते हैं ताकि स्वाभाविक आवृत्ति हट जाए, या आवेग को नियंत्रित करते हैं।

  6. A damped oscillator has equation m d²x/dt² + b dx/dt + k x = 0 with m = 1 kg, b = 2 kg/s and k = 5 N/m. Determine whether motion is underdamped, overdamped or critically damped and find damped angular frequency if underdamped. / मित कंपन समीकरण m d²x/dt² + b dx/dt + k x = 0 में m = 1 kg, b = 2 kg/s और k = 5 N/m हैं। बताइए कि गमन-प्रकार क्या है और यदि underdamped है तो damped कोणीय आवृत्ति ज्ञात कीजिए।
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    Compute discriminant b² - 4mk = 4 - 20 = -16 < 0, so motion is underdamped. γ = b/(2m) = 1 s⁻¹. Natural ω = √(k/m) = √5 ≈ 2.236 s⁻¹. Damped frequency ω' = √(ω² - γ²) = √(5 - 1) = √4 = 2 s⁻¹. / निर्णायक b² - 4mk = -16 < 0, अतः underdamped है। γ = 1 s⁻¹, ω = √5 ≈ 2.236 s⁻¹, ω' = 2 s⁻¹।

  7. Derive expression for energy conservation in an ideal mass-spring SHM and show total energy in terms of amplitude. / आदर्श द्रव्यमान-वसंत SHM में ऊर्जा संरक्षण का व्यंजक व्युत्पन्न कीजिए और इसे आवृत्ति के परिप्रेक्ष्य में आयाम के रूप में दिखाइए।
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    For mass-spring, K = 1/2 m v² and U = 1/2 k x². Using x = A cos(ωt + φ) and v = -Aω sin(ωt + φ) with ω² = k/m, total E = K + U = 1/2 m A² ω² sin²(...) + 1/2 k A² cos²(...) = 1/2 k A²[sin² + cos²] = 1/2 k A². Thus total energy is constant and equal to 1/2 k A². / मास-स्प्रिंग के लिए K = 1/2 m v², U = 1/2 k x²। x और v के रूप में रखने पर E = 1/2 k A² जो स्थिर है।

  8. A tuning fork of frequency 440 Hz and another of 444 Hz are sounded together. How many beats per second are heard? / 440 Hz और 444 Hz की दो ट्यूनिंग फोर्क एक साथ बजती हैं। प्रति सेकंड कितने बीट सुने जायेंगे?
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    Beat frequency = |444 - 440| = 4 Hz, so 4 beats per second are heard. / बीट आवृत्ति = 4 Hz; प्रति सेकंड 4 बीट सुनाई देंगे।

  9. For a pipe open at both ends of length 1.5 m at 20°C, find the wavelength and frequency of the second harmonic. (Take v = 343 m/s) / 20°C पर दोनों सिरों खुली लंबाई 1.5 m वाले पाइप के दूसरे हार्मोनिक की तरंगदैর্ঘ्य और आवृत्ति ज्ञात कीजिए। (v = 343 m/s लें)
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    For open-open pipe λ_n = 2L/n. For n = 2, λ_2 = 2L/2 = L = 1.5 m. Frequency f2 = v/λ_2 = 343/1.5 ≈ 228.7 Hz. / दूसरे हार्मोनिक के लिए λ = 1.5 m और f ≈ 228.7 Hz।

  10. A transverse wave on a string has equation y = 0.01 cos(8π x - 400π t) where x in metres and t in seconds. Find amplitude, wavelength, frequency and wave speed. / एक तार पर अनुप्रस्थ तरंग y = 0.01 cos(8π x - 400π t) (x मीटर में, t सेकंड में) है। आयाम, तरंगदैর্ঘ्य, आवृत्ति और तरंग गतिकी ज्ञात कीजिए।
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    Amplitude A = 0.01 m. Wave number k = 8π ⇒ λ = 2π/k = 2π/(8π) = 1/4 m = 0.25 m. Angular frequency ω = 400π ⇒ frequency f = ω/(2π) = 200π/π = 200 Hz. Wave speed v = ω/k = (400π)/(8π) = 50 m/s. / आयाम 0.01 m, λ = 0.25 m, f = 200 Hz, v = 50 m/s।

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