Simple Pendulum & Compound Pendulum Simple Pendulum & Compound Pendulum A pendulum is the gentlest gateway into oscillations: a mass suspended so it can swing back and forth under gravity. When displaced a little, gravity pulls it back toward its lowest point; as it passes through, inertia carries it to the other side. This tug-of-war of restoring pull and inertia repeats, creating periodic motion. For small swings, the motion is so regular that the period depends mainly on the length and local gravity, not on the mass or the starting side. That is why pendulums have timed clocks for centuries. But an ideal “simple pendulum” (a point mass on a massless, inextensible string) exists only in our equations. Real bobs have size, strings have stiffness, and air drags. The closer we stay to small angles and slender strings, the better reality matches the simple formula. When the bob itself has size or when the whole rigid body swings about a pivot (like a rod), we get a compound or physical pendulum. If, instead of gravity, the restoring agent is a twisted wire producing a torque proportional to angle, we meet the torsional pendulum—another clean example of angular SHM with many lab uses (e.g., measuring moment of inertia or rigidity modulus). The second’s pendulum (period 2 s) links timekeeping and measurement of g. These ideas interlock: we start with the small-angle equation, then build up to compound and torsional cases, constantly checking limits and noting where the simple rules fail. Along the way, we learn to read formulas as maps: what parameters control the period, how to minimize it, and how conjugate points in a compound pendulum share the same period. These insights are highly testable: NEET often probes sign conventions, small-angle approximations, correct measurement of length (pivot to center of mass), and the dependence of T on L and g—plus practical tweaks like thermal expansion or slight differences leading to beats between two pendulums. remember Think of a playground swing: pull slightly to one side and release—gravity restores, inertia carries, and the swing repeats. Smaller pushes give more regular timing. An ideal point mass (bob) suspended by a massless, inextensible string of length L, oscillating in a vertical plane under gravity. Simple pendulum Any rigid body that can oscillate about a horizontal axis under gravity. Its mass is distributed; its dynamics depend on its moment of inertia and the distance of its center of mass from the pivot. Compound (physical) pendulum Torsional pendulum A body suspended by a wire that twists. The restoring torque is proportional to angular displacement, producing angular SHM. A pendulum with period 2 s (1 s each way), historically used for timekeeping and estimating g. Second's pendulum For angles in radians with |θ| ≲ 10°, sinθ ≈ θ and the motion becomes strictly SHM. Small-angle approximation Moment of inertia (I) Rotational analogue of mass for a given axis; measures resistance to angular acceleration. Radius of gyration (k) Defined by I = m k 2 for an axis; it parameterizes how mass is effectively distributed at distance k from the axis. Conjugate points Two pivot positions on a rigid body, at distances d1 and d2 from the center of mass, that give the same period. They satisfy d1 d2 = k 2 . Exact restoring torque about the pivot; linearized for small angles by sin . Restoring torque (simple pendulum) Applies to the rotational motion of a simple pendulum, describing the torque that pulls it back towards the equilibrium position. Equation and period of a simple pendulum for small angles Point mass bob; massless, inextensible string Small angular displacement: sin No air resistance; pivot friction negligible; motion in a vertical plane '' + g L , = 0, T = 2 L g Rotational dynamics about the pivot. For a point mass at distance L from pivot. Substitute I and . Exact nonlinear pendulum equation. Linearized SHM form. Angular frequency and time period. tip Validity window: small angles (typically |θ| ≲ 10°), light string, point-like bob, negligible air drag and pivot friction. Outside this, the period increases slightly with amplitude and damping. easy Use T = 2 L/g . T = 1.5 s g = 9.8 m/s 2 Length L A simple pendulum has time period T = 1.5 s near sea level where g = 9.8 m/s². Find its length. NCERT Exemplar-style Rearrange for L. Compute numerically. The linearized pendulum obeys d²x/dt² + ω² x = 0 with x ↔ θ and ω² = g/L. d 2 x dt 2 + 2 x = 0 Newton’s second law with Hooke-type force. Equation of motion. Standardized form. Define angular frequency. General SHM differential equation Linear restoring force: F = -k x (or restoring torque proportional to angle) No driving force, no damping With initial phase ϕ, displacement follows x = A cos(ω t + ϕ). For a pendulum’s small-angle motion, replace x with θ. Angle vs linear displacement: Near equilibrium, the arc length s of a pendulum bob relates to angle by s = Lθ (θ in radians). If the swing is small, the bob’s motion along the arc closely resembles SHM with angular variable θ obeying θ'' + (g/L)θ = 0. The linearized mapping is: position x ↔ angle θ, effective spring constant k ↔ m(g/L), and ω = √(g/L). The phase constant ϕ sets the starting angle: θ(t) = Θ cos(ω t + ϕ). Velocity and acceleration follow by differentiation: ω sets how fast the angle cycles; doubling L halves ω by √2 and increases T by √2. For exam work, always ensure θ is in radians when using trigonometric approximations. Approximate percentage change in T Use logarithmic differentiation: ( T/T 1 2 , L/L - 1 2 , g/g ). L/L = +2 % = +0.02 g/g = -0.5 % = -0.005 T = 2 L/g medium Add contributions from L and g. Convert to percentage. By how much (in %) does the period of a simple pendulum change if its length increases by 2% and it is moved to a place where g decreases by 0.5%? NEET-style approximation neet-alert Angles must be in radians for sinθ ≈ θ. Also, pendulum length is measured from the pivot to the center of mass of the bob, not to the bottom or the string knot. Compound pendulum period I is about the pivot; d is distance from pivot to COM; I = m k 2 . This formula provides the period of oscillation for a rigid body pivoted at a point, assuming small angular displacements and negligible dam Rigid body oscillates about a fixed horizontal axis through pivot P Small angles: sin Negligible air resistance and pivot friction Period of a compound (physical) pendulum; equivalent length and conjugate points Torque about P; d is distance from P to COM. Linearized equation. Time period. Parallel axis theorem; define radius of gyration k. Standard form. Equivalent simple-pendulum length. T is minimum when pivot is one radius of gyration from COM. Pairs of pivots with the same period. T = 2 k 2 + d 2 g d , L eq = d + k 2 d , d 1 d 2 = k 2 ( same T) Compound pendulum — key takeaways Equivalent length: L eq = d + k 2 /d. Minimum period when d = k. Conjugate pivot distances satisfy d1 d2 = k 2 and share the same T. If T is known at one d, you can find k and predict T at any other d. For a slender rod about one end, k = L rod/√3 about its COM axis; use parallel axis carefully. Physical meaning of terms: In a compound pendulum, increasing d (moving pivot farther from the COM) both raises the torque arm (stronger restoring torque) and the moment of inertia about the pivot (harder to start rotating). The competition produces a U-shaped T(d) curve with a minimum at d = k, where the torque gain and inertia gain balance. The equivalent length L eq behaves like the length of a simple pendulum swinging with the same period; its decomposition d + k 2 /d highlights conjugate pairs: swap d with k 2 /d and L eq stays unchanged. In labs, by finding two pivot points with the same period (a Kater’s pendulum approach), the distance between these points directly equals L eq—an easy route to measure g with fewer modeling assumptions about the mass distribution. T(d=0.25 m), d min, and T min Use T = 2 k 2 + d 2 g d . m = 2.0 kg k = 0.20 m d = 0.25 m g = 9.8 m/s 2 hard Compute numerator and denominator. Numerical evaluation. Minimum at d = k. Evaluate T at d = k. A rigid body has mass m = 2.0 kg and radius of gyration k = 0.20 m about its COM. It is pivoted at distance d = 0.25 m from the COM. Find its period near g = 9.8 m/s². Also find the pivot position for minimum period and the corresponding minimum T. Physical pendulum application Time period T Distance from COM d U-shaped curve T(d) = 2π√((k² + d²)/(g d)) with a minimum at d = k. control control dependent T(d) custom Time period versus pivot distance for a compound pendulum; note the symmetric conjugate pairs about d = k. Minimum at d = k T min Conjugate point d1 d1 k 2 /d1 Conjugate partner k²/d1 2D PLOT Compound-pendulum period vs distance from CoM T = 2 pi sqrt((k 2 + d 2)/(g d)) Radius of gyration Equivalent length (L eq) The length of a simple pendulum that would have the same period as a given compound pendulum: L eq = d + k 2 $/d. For equal periods at distances d1 and d2 from the COM: d1 d2 = k 2 and L eq = d1 + d2. The line joining the two knife edges of a Kater’s pendulum equals L eq.$ remember Torsional pendulum C is torsion constant (restoring torque per radian): ( = -C , ). Torsional pendulum period Restoring torque proportional to angle: ( = -C , ) Small oscillations; no damping T = 2 I C Rotational SHM equation. Standard SHM form with ( 2 = C/I ). Time period from ( ). Uses of a torsional pendulum: Because T depends on I/C, you can determine I if C is known (or vice versa). For example, mount a disk on a thin wire, measure T, and compute C = 4π² I/T². If the disk’s I is known (I = 1/2 mR² about its central axis), this gives the torsion constant of the wire; repeating with different wire lengths allows estimation of the wire’s rigidity modulus. Torsional oscillations are also the basis of moving-coil galvanometers where a magnetic torque balances torsional restoring torque, and the pointer deflection is proportional to current. N·m/rad m = 0.50 kg R = 0.10 m T = 1.20 s I disk = 1 2 m R 2 Use C = 4 2 I/ T 2 with I = 1 2 m R 2 . Torsion constant C medium Lab-style calculation A circular disk (m = 0.50 kg, R = 0.10 m) is suspended by a wire with torsion constant C. Its period is measured as 1.20 s. Find C. Moment of inertia of the disk. Compute numerically. Second’s pendulum: Set T = 2 s in the simple pendulum formula to get length L = g T 2 /(4π 2). Near sea level where g ≈ 9.8 m/s², L ≈ 0.993 m. Such a pendulum was historically used for timekeeping. Any change in length or in g (due to latitude or altitude) shifts the rate: if L increases (e.g., due to thermal expansion), T increases and the clock loses time. If L is shortened, the clock gains. Because T ∝ √L, a small fractional change in length produces half as large a fractional change in period—handy for quick estimates. Second’s pendulum length Numerically, L 0.993 m for g = 9.8 m/s 2 . At a location where g = 9.78 m/s², what should be the length of a second’s pendulum? If the pendulum is mistakenly made 1.000 m long, does the clock gain or lose time per day? Second’s pendulum application Compute the correct length. Fractional length error. Fractional period increase. New period is longer; clock runs slow. Daily loss estimate. medium g = 9.78 m/s 2 Desired T = 2.0 s Mistaken L = 1.000 m Correct L = g T 2 /(4 2). Fractional change in T is ( 1 2 , L/L ). Correct L and time gain/loss per day Finite amplitude: For larger swings, T increases slightly; first correction term is ≈ (1/16)θ 0 2 for peak angle θ 0 in radians. Air drag: Damping makes amplitude decay; the period shift is usually small for light damping. Buoyancy and string mass: Effective weight reduces slightly; massive strings add distributed inertia; lab manuals include small correction factors. Pivot friction: Slightly increases effective period and dissipates energy. Real-world corrections (qualitative) The period of a pendulum is completely independent of amplitude. Only for small angles (in radians) is T effectively independent of amplitude. For larger angles, T increases; the small-angle approximation breaks down. Pendulum length is the string length. Length means pivot to the center of mass of the bob. A spherical bob’s center lies at its geometric center, not at the hook. Two pendulums with close frequencies produce beats: f b = |f 1 − f 2|. Useful for fine-tuning lengths. f b = |f 1 - f 2| Two harmonic signals. Sum-to-product identity. Energy is ( A 2 ), giving two maxima per envelope period. Beat frequency equals the absolute frequency difference. Beat frequency for two close frequencies Equal amplitudes; same type of oscillation Frequencies close so that amplitude modulation is slow Using beats to tune pendulums: If two pendulums have slightly different lengths (and thus slightly different frequencies), their swings drift in and out of step. You can count the number of times they line up in the same phase per unit time to get the beat frequency f b . Since for a simple pendulum f ≈ (1/2 )√(g/L), increasing L lowers f. If the beats are 6 per minute relative to a standard, you know how much to lengthen or shorten the string to match frequencies. This is an excellent practical trick for calibrating clocks or laboratory pendulums without sophisticated timing tools. y = 2A sin(kx) cos(ωt). Included for cross-links: resonance and normal modes underlie many oscillators; a pendulum is a single-degree-of-freedom oscillator, not a standing wave. Standing waves vs pendulum motion: A pendulum is a lumped oscillator governed by a single coordinate (θ). A standing wave on a string involves many coupled oscillators; its shape varies with x and t as y(x, t) = 2A sin(kx) cos(ωt). Despite being different systems, both share the SHM core: every point in a standing wave executes SHM with the same ω but different amplitudes. This shared mathematical backbone explains why ideas like resonance and Q factor recur across oscillations. f obs = f s (v + v o )/(v − v s $). Not directly for pendulums, but useful when oscillatory sources move; included for unit coherence. Doppler and oscillations: The Doppler effect concerns frequency shifts when a source or observer moves relative to the medium. Although a swinging pendulum does not by itself create a traveling wave, the idea of perceived frequency change is relevant to many wave-based oscillation contexts (like sound-emitting oscillators). For mechanics of pendulums, Doppler is usually not needed directly; know it as a separate, wave-focused result in this unit. neet-alert Do not mix linear and angular variables: use s = Lθ only for small angles, and keep θ in radians. Using degrees inside sin(θ) ≈ θ is a classic error. Comparison snapshot Simple pendulum Gravity via arc component θ'' + (g/L) θ = 0 T = 2π√(L/g) Point mass, massless string, small θ Compound pendulum Gravity; rigid body rotation θ'' + (mgd/I) θ = 0 T = 2π√(I/(mgd)) Rigid body, small θ Torsional pendulum Twist of wire: τ = −Cθ θ'' + (C/I) θ = 0 T = 2π√(I/C) Linear torsion, small θ Oscillator Restoring agent Equation (small-angle) Time period Key assumptions Frequency relation For a simple pendulum under small-angle approximation; inversely proportional to ( L ). The measured period is used to determine the effective length of a compound pendulum or the torsion constant of a wire. Measuring g with a pendulum: For a simple pendulum, g = 4π² L/T². If timing multiple oscillations, reduce random error by timing N swings and using T = (total time)/N. With a Kater’s pendulum (two knife edges), adjust until periods about both pivots match; then the distance between knife edges equals L eq, and g = 4π² L eq/T² without knowing k or d individually. This differential method suppresses systematic errors from mass distribution, making it a standard metrology technique. Relates a physical pendulum to a simple pendulum of the same period. Equivalent length (compound pendulum) This formula determines the equivalent length used to calculate the period of oscillation for a compound pendulum, assuming small angle osci m/s² Distance between knife edges = 1.000 m = L eq T = 2.005 s Use g = 4 2 L eq / T 2 . hard A Kater’s pendulum has knife edges separated by 1.000 m. The periods about both edges are adjusted to be equal: T = 2.005 s. Estimate g. Kater’s pendulum principle Compute numerically. Small angle, small sin: for tiny θ (in radians), sinθ ≈ θ. Think “SAS” to remember when pendulum formulas stay simple. Memory hook Amplitude dependence (beyond small-angle): The exact period of a simple pendulum grows with amplitude. A series expansion gives T ≈ 2π√(L/g)[1 + (1/16)θ 0² + (11/3072)θ 0⁴ + …], with θ 0 in radians. Even at 10° (≈ 0.1745 rad), the fractional increase is about (1/16)(0.1745)² ≈ 0.0019, i.e., ~0.19%. This is why the small-angle formula works so well in everyday labs. For careful timekeeping (e.g., clock escapements), designs try to keep the drive small and periodic to avoid large-amplitude drifts. Variation of g and its effect: Gravity varies slightly with latitude and altitude; g is smaller at the equator and at higher altitudes. Since T ∝ 1/√g, moving a pendulum clock from pole to equator makes it lose time. Typical fractional changes are on the order of 10⁻³, so a second’s pendulum could drift a minute or more per day if not adjusted. Precision devices either recalibrate length or use compensation schemes (like temperature-compensated pendula) to stabilize T. Thermal expansion and clock rate: If the pendulum rod lengthens with temperature by ΔL = α L ΔT, then ΔT period/T ≈ 1/2 α ΔT. Steel has α ≈ 12×10⁻⁶/°C, so a 10 °C rise gives ΔT/T ≈ 6×10⁻⁵ (about 5 s per day). Precision clocks use low-expansion alloys (Invar) or gridiron compensation (different metals with opposing expansions) so the effective length stays nearly constant and the clock keeps accurate time across seasons. a = −ω²x. For pendulums under small-angle approximation, replace x by θ and ω² by g/L to get angular acceleration α = −(g/L) θ. Characteristic equation: r 2 + ω 2 = 0. General real solution. Combine constants into amplitude and phase. x = A ( t + ) Undamped, unforced SHM Initial conditions encoded in amplitude A and phase ϕ Displacement solution of SHM: x = A cos(ω t + ϕ) Oppositely directed waves. Use sin(u)+sin(v) identity. y = 2A (kx) ( t) Two counter-propagating waves with equal A, ω, and k Linear medium so superposition applies Standing wave superposition: y = 2A sin(kx) cos(ωt) Classical Doppler for sound: f obs = f s (v + v o)/(v − v s) Velocities collinear with wave propagation v s, v o ≪ v (no shock waves); medium at rest f obs = f s , v + v o v - v s Effective wavelength from moving source. Relative speed of wavefronts toward observer. Observed frequency. Boundary checks: L → 0 makes T → 0; L → ∞ makes T → ∞. For a compound pendulum, d → 0 makes T blow up (no restoring torque about COM), and d → ∞ also makes T large (huge inertia). Minimum sits at d = k. tip Energy view: At an extreme, the pendulum’s kinetic energy is zero and potential energy is maximized relative to the lowest point. As it passes through equilibrium, potential energy is minimum and kinetic energy peaks. For small angles, the potential energy near equilibrium is approximately quadratic in θ (U ≈ 1/2 m g L θ²), matching the energy form of a spring (1/2 k x²) with effective k eff = m g/L in angular terms. This energy-shuttling picture justifies why the motion is sinusoidal under small-angle conditions. Free-body and torque clarity: The forces on a simple pendulum bob are just tension along the string and weight mg downward. The tangential component of weight is m g sinθ directed toward equilibrium. Tension does no work in the tangential direction, so only gravity changes the bob’s kinetic energy. About the pivot, the torque is −m g L sinθ, leading directly to the equation of motion. In the compound case, the same gravity force acts at the COM; all that changes is the moment arm d and the body’s moment of inertia about the pivot. Equal L nominally Beat rate = 4 min -1 = 4/60 s -1 f = 1/T = (1/2 ) g/L For small differences, (f b = |f 1 - f 2| ), and ( f/f 1 2 ( g/g - L/L) ). Fractional mismatch in frequency and hence in g or L hard Beats as calibration insight Two simple pendulums have equal lengths but slightly different bobs. One loses 4 beats per minute relative to the other. Estimate their fractional difference in effective g or L assuming small differences. Nominal frequency. Beats per second. Illustrative estimate near second’s pendulum. Interpreting mismatch. A 26.6% net mismatch is huge—real clocks avoid such large errors. Smaller beat rates are typical. neet-alert Do not cancel mass m in the wrong place for a compound pendulum. T = 2π√(I/(mgd)) uses I about the pivot. If you insert I cm instead of I pivot, you must include the md² term via the parallel axis theorem. Worked strategy summary: 1) Identify the oscillator type (simple, compound, torsional). 2) Write the small-angle equation (θ'' + ω²θ = 0) with the correct ω²: g/L, mgd/I, or C/I. 3) Plug into T = 2π/ω and simplify. 4) Mind definitions: L (pivot to COM), d (pivot to COM), I (about the pivot), k (I = m k²). 5) For percentage-change questions, linearize with logarithmic differentiation. 6) For conjugate-point questions, use d1 d2 = k² and L eq = d1 + d2. This recipe answers 90% of NEET pendulum numericals cleanly. Pendulum glossary recap ideal pendulum Point mass on massless string under gravity; T = 2π√(L/g). Simple pendulum physical pendulum Rigid body about a pivot; T = 2π√(I/(mgd)). Compound pendulum twisting oscillator Twist-restored angular SHM; T = 2π√(I/C). Torsional pendulum Second’s pendulum T = 2 s; L ≈ 0.993 m near g = 9.8 m/s². Radius of gyration (k) I = m k² for a given axis. Pivot distances d1, d2 with the same period, satisfying d1 d2 = k². Conjugate points