Simple Harmonic Motion β Definition & Equations
Definition
Simple Harmonic Motion (SHM) is a periodic oscillation where the restoring force (and hence acceleration) is always directly proportional to displacement from equilibrium, and always directed towards the equilibrium (mean) position: $F = -kx$, giving $a = -\omega^2x$.
Key Characteristics
Velocity is maximum at the mean position ($x=0$) and zero at the extreme positions ($x=\pm A$). Acceleration is zero at the mean position and maximum at the extremes β always directed back towards the centre.
Solved Example
A particle executes SHM with amplitude 5 cm and angular frequency 2 rad/s. Calculate its velocity when its displacement is 3 cm from the mean position.
$v = \omega\sqrt{A^2-x^2} = 2\sqrt{5^2-3^2} = 2\sqrt{25-9} = 2\sqrt{16} = 2(4) = 8\ \text{cm/s}$
Q. Why is the acceleration in SHM always directed opposite to the displacement from the mean position?
Ans: The defining equation of SHM is a = βΟΒ²x, where the negative sign shows acceleration is always opposite in direction to displacement β this is precisely what makes the motion "restoring": whenever the particle moves away from the mean (equilibrium) position in either direction, the resulting acceleration always acts to pull it back towards that centre, which is the fundamental reason the motion oscillates back and forth rather than moving away indefinitely.
Q. At what point in its oscillation does a particle in SHM have maximum speed, and why?
Ans: At the mean (equilibrium) position, x = 0. Since v = Οβ(AΒ²βxΒ²), velocity is maximum when x = 0 (giving v_max = ΟA) β physically, this is because the restoring force (and hence deceleration) is zero exactly at the mean position, so the particle has been continuously accelerated towards this point from the extreme position without yet experiencing any opposing force to slow it down.
Time Period & Frequency in Terms of Angular Frequency
Angular frequency relates to time period and ordinary frequency by $\omega = \dfrac{2\pi}{T} = 2\pi f$. Also, since $a = -\omega^2x$, the time period of any SHM can be found directly from $T = 2\pi\sqrt{\dfrac{x}{a}}$ (using the magnitude of displacement and corresponding acceleration at any point, ignoring signs) β a general, versatile method applicable to any SHM system, not just pendulums or springs specifically.
Solved Example
A particle in SHM has a displacement of 0.02 m and a corresponding acceleration of magnitude 0.8 m/sΒ² at that point. Calculate the time period of its motion.
$T = 2\pi\sqrt{\dfrac{x}{a}} = 2\pi\sqrt{\dfrac{0.02}{0.8}} = 2\pi\sqrt{0.025} = 2\pi(0.158) \approx 0.99\ \text{s}$
Q. Why can the general relation T = 2Οβ(x/a) be applied to find the time period of any SHM system, not just a specific case like a pendulum or spring?
Ans: This relation is derived directly from the defining equation of SHM itself, a = βΟΒ²x (so ΟΒ² = a/x, and T = 2Ο/Ο = 2Οβ(x/a)) β since every SHM system, regardless of its specific physical setup, must obey this same fundamental defining relationship between acceleration and displacement, the formula for time period in terms of x and a applies universally to any system executing simple harmonic motion.
Energy in SHM
Kinetic & Potential Energy
In SHM, energy continuously converts between kinetic and potential forms, while the total mechanical energy remains constant (assuming no damping):
$$KE = \tfrac{1}{2}m\omega^2(A^2-x^2) \qquad PE = \tfrac{1}{2}m\omega^2x^2 \qquad E_{total} = \tfrac{1}{2}m\omega^2A^2$$
Key Insight
KE is maximum at the mean position (where all energy is kinetic) and zero at the extremes (where all energy is potential) β exactly the opposite pattern to PE. The sum, total energy, stays constant throughout the motion.
Solved Example
A particle of mass 0.5 kg executes SHM with amplitude 4 cm and angular frequency 5 rad/s. Calculate its total mechanical energy.
$E_{total} = \tfrac{1}{2}m\omega^2A^2 = \tfrac{1}{2}(0.5)(5)^2(0.04)^2 = \tfrac{1}{2}(0.5)(25)(0.0016) = 0.01\ \text{J}$
Q. At what displacement from the mean position does a particle in SHM have equal kinetic and potential energy?
Ans: Setting KE = PE: Β½mΟΒ²(AΒ²βxΒ²) = Β½mΟΒ²xΒ² β AΒ²βxΒ² = xΒ² β AΒ² = 2xΒ² β x = A/β2 β 0.707A. This occurs at a displacement of A/β2 from the mean position, on either side.
Q. Why does the total energy in SHM remain constant despite kinetic and potential energy continuously changing throughout the motion?
Ans: Because the restoring force in SHM is conservative (analogous to a spring force), no energy is lost to friction or other dissipative effects (assuming an idealised, undamped system) β energy simply and continuously converts back and forth between kinetic and potential forms as the particle moves, but the sum of these two forms at any instant remains exactly constant, consistent with the general principle of conservation of mechanical energy for a conservative force.
Simple Pendulum
Time Period
For small angular displacements (so that $\sin\theta \approx \theta$, keeping the motion approximately simple harmonic), a simple pendulum's time period is:
$$T = 2\pi\sqrt{\dfrac{l}{g}}$$
where $l$ = length of the pendulum, $g$ = acceleration due to gravity. Notably, the time period does not depend on the mass of the bob, or (for small angles) on the amplitude of swing.
Solved Example
Calculate the time period of a simple pendulum of length 1 m. (g = 9.8 m/sΒ²)
$T = 2\pi\sqrt{\dfrac{l}{g}} = 2\pi\sqrt{\dfrac{1}{9.8}} = 2\pi\sqrt{0.102} = 2\pi(0.319) \approx 2.01\ \text{s}$
Q. Why is the small-angle approximation (sinΞΈ β ΞΈ) necessary for a pendulum's motion to be treated as simple harmonic?
Ans: The actual restoring force on a pendulum bob is proportional to sinΞΈ (not ΞΈ itself), which would make its motion not exactly SHM in the strict sense (since true SHM requires a restoring force exactly proportional to displacement/angle). Only for small angles does sinΞΈ β ΞΈ hold as a very good approximation, allowing the pendulum's motion to be treated as approximately simple harmonic and giving the standard T = 2Οβ(l/g) formula β for larger swing angles, this approximation breaks down and the actual period differs from this simple formula.
Q. If the length of a simple pendulum is quadrupled, how does its time period change?
Ans: Since T β βl, quadrupling the length doubles the time period (β4 = 2).
Solved Example
A pendulum has a time period of 2 s on Earth (g = 9.8 m/sΒ²). Calculate its time period on the Moon, where g_moon β 1.63 m/sΒ².
Since $T \propto 1/\sqrt{g}$: $\dfrac{T_{moon}}{T_{earth}} = \sqrt{\dfrac{g_{earth}}{g_{moon}}} = \sqrt{\dfrac{9.8}{1.63}} \approx \sqrt{6.01} \approx 2.45$
$T_{moon} = 2 \times 2.45 \approx 4.9\ \text{s}$ β the same pendulum swings noticeably more slowly on the Moon, due to its weaker gravity.
Q. Why does a pendulum clock run slower on the Moon than on Earth, without any change to the pendulum's length?
Ans: Since T = 2Οβ(l/g), and the Moon's gravitational acceleration is significantly weaker than Earth's, the time period T increases (the pendulum takes longer to complete each swing) for the same length l β a pendulum clock, which is calibrated assuming Earth's gravity, would tick more slowly on the Moon, causing it to run behind (lose time) compared to a correctly functioning clock.
Spring-Mass Oscillations
Time Period
For a mass $m$ attached to an ideal spring of force constant $k$, undergoing SHM:
$$T = 2\pi\sqrt{\dfrac{m}{k}}$$
Series & Parallel Spring Combinations
For springs in series, the effective spring constant is smaller than either individual spring: $\dfrac{1}{k_{eff}} = \dfrac{1}{k_1}+\dfrac{1}{k_2}$. For springs in parallel, the effective spring constant is larger: $k_{eff} = k_1+k_2$.
Solved Example
A 2 kg mass is attached to a spring of force constant 200 N/m. Calculate the time period of its oscillation.
$T = 2\pi\sqrt{\dfrac{m}{k}} = 2\pi\sqrt{\dfrac{2}{200}} = 2\pi\sqrt{0.01} = 2\pi(0.1) \approx 0.628\ \text{s}$
Q. Why does connecting two identical springs in parallel (rather than series) with the same mass result in a shorter time period of oscillation?
Ans: Springs in parallel combine to give a larger effective spring constant (k_eff = kβ+kβ, stiffer overall combination), since both springs now share the load and each contributes its own full restoring force simultaneously. Since T = 2Οβ(m/k), a larger effective k (stiffer spring system) results in a shorter time period β the mass is restored to equilibrium more quickly by the combined, stronger restoring force.
Q. Two identical springs, each with force constant 100 N/m, are connected in series and attached to a 1 kg mass. Calculate the time period of oscillation.
Ans: For series springs: 1/k_eff = 1/100 + 1/100 = 2/100 β k_eff = 50 N/m. T = 2Οβ(m/k_eff) = 2Οβ(1/50) β 2Ο(0.141) β 0.889 s.
Damped Oscillations
Real oscillating systems experience damping β a resistive force (like friction or air resistance) that gradually removes energy from the system, causing the amplitude of oscillation to decrease exponentially over time, eventually bringing the system to rest, unlike the idealised, undamped SHM (which would oscillate forever with constant amplitude).
Solved Example
Explain why a swinging pendulum, if left undisturbed, eventually comes to rest rather than continuing to swing forever, contrary to the idealised SHM prediction.
In practice, a pendulum experiences small resistive forces β air resistance acting on the bob, and friction at the pivot point β which continuously remove a small amount of mechanical energy from the system with each swing, converting it into heat. This is a form of damping: the oscillation amplitude gradually decreases over successive swings until essentially all the initial energy has been dissipated, and the pendulum comes to rest β the idealised, undamped SHM model (assuming perpetual, constant-amplitude oscillation) only applies as an approximation valid over short timescales, ignoring these small but ever-present real-world resistive effects.
Q. What happens to the amplitude of a damped oscillation over time, and what is ultimately responsible for this change?
Ans: The amplitude decreases exponentially over time, eventually approaching zero. This is caused by a resistive (damping) force, such as friction or air resistance, which continuously does negative work on the oscillating system, progressively removing its mechanical energy and converting it into heat, rather than allowing the oscillation to continue indefinitely at constant amplitude.
Wave Motion β Transverse & Longitudinal Waves
Wave Motion β General Idea
A wave is a disturbance that transfers energy from one point to another through a medium (or through vacuum, for electromagnetic waves) without any net transport of matter itself β individual particles of the medium oscillate about their equilibrium positions but don't travel along with the wave.
Transverse Waves
Particles of the medium oscillate perpendicular to the direction of wave propagation β e.g., waves on a string, electromagnetic waves. Characterised by crests (highest points) and troughs (lowest points).
Longitudinal Waves
Particles of the medium oscillate parallel to the direction of wave propagation β e.g., sound waves in air. Characterised by regions of compression (particles bunched closer together) and rarefaction (particles spread farther apart).
Solved Example
Classify sound waves in air and waves on a plucked guitar string as transverse or longitudinal, and justify.
Sound waves in air are longitudinal β air molecules vibrate back and forth along the same direction the sound wave travels, creating alternating compressions and rarefactions.
Waves on a guitar string are transverse β the string's particles move up and down, perpendicular to the direction the wave travels along the string's length.
Q. Why can sound waves not travel through a vacuum, unlike electromagnetic waves?
Ans: Sound is a mechanical (longitudinal) wave that relies on a physical medium (like air, water, or a solid) for its particles to actually vibrate and transmit the disturbance from one point to the next through direct particle-to-particle interaction. In a vacuum, there are no particles present to vibrate or transmit this disturbance, so sound cannot propagate at all β unlike electromagnetic waves, which consist of oscillating electric and magnetic fields that don't require any physical medium and can travel through empty space.
Q. Why can transverse waves not propagate through the bulk of a liquid or gas, unlike through a solid or along a string surface?
Ans: Transverse wave propagation through a bulk medium requires that medium to support shear stress (resistance to a sideways, perpendicular displacement) between adjacent layers, so that one oscillating layer can effectively "pull" the neighbouring layer sideways with it. Liquids and gases cannot sustain shear stress internally (they flow freely under any tangential force), so they cannot support genuine transverse wave propagation through their bulk β only longitudinal waves (which involve compression, not shear) can travel through liquids and gases.
Speed of a Wave & the Wave Equation
Speed of a Wave on a Stretched String
For a transverse wave on a stretched string: $v = \sqrt{T/\mu}$, where $T$ = tension in the string, $\mu$ = mass per unit length (linear mass density).
Speed of Sound in a Medium
Sound speed depends on the medium's elastic properties and density: $v = \sqrt{E/\rho}$ (where $E$ is the relevant elastic modulus) β this is why sound travels fastest in solids (highest elastic modulus), slower in liquids, and slowest in gases.
Solved Example
A wave has a frequency of 50 Hz and a wavelength of 2 m. Calculate its speed.
$v = f\lambda = (50)(2) = 100\ \text{m/s}$
Q. Why does increasing the tension in a guitar string increase the speed of waves along it, and hence raise the pitch (frequency) of the note produced?
Ans: Wave speed on a string is given by v = β(T/ΞΌ) β increasing tension T directly increases wave speed. Since the string's length (and hence wavelength, largely fixed by the string's fixed endpoints) doesn't change, and v = fΞ», a higher wave speed with the same wavelength requires a correspondingly higher frequency β this is exactly why tightening a guitar string raises the pitch of the note it produces when plucked.
Q. Why does sound generally travel faster in solids than in gases?
Ans: Sound speed depends on v = β(E/Ο), where E is the relevant elastic modulus of the medium. Solids have a vastly higher elastic modulus (much stronger intermolecular bonding, resisting compression/deformation far more than gases) β even though solids are also denser than gases, the elastic modulus increases by a proportionally much larger factor than density does, resulting in an overall higher sound speed in solids compared to gases.
Superposition & Standing Waves
Principle of Superposition
When two or more waves overlap at the same point in a medium, the resultant displacement is simply the vector sum of the individual displacements each wave would have produced independently.
Standing (Stationary) Waves
When two identical waves travel in opposite directions along the same medium (e.g., an incident wave and its reflection), they superpose to form a standing wave β a pattern that appears to oscillate in place, with fixed points of zero displacement (nodes) and points of maximum displacement (antinodes), rather than travelling progressively through the medium.
Key Feature β Nodes and Antinodes
Adjacent nodes (or adjacent antinodes) are separated by exactly $\lambda/2$. A node and the nearest antinode are separated by $\lambda/4$. Unlike a travelling wave, every point on a standing wave (except at the nodes) oscillates with the same frequency, but with different, fixed amplitudes depending on position.
Solved Example
A standing wave is formed on a string of length 2 m, fixed at both ends, vibrating in its second harmonic (2 loops/antinodes). Calculate the wavelength of the standing wave pattern.
For a string fixed at both ends vibrating with $n$ loops, the length equals $n \times (\lambda/2)$. Here $n=2$: $2 = 2 \times (\lambda/2) \Rightarrow \lambda = 2\ \text{m}$
Q. Why do the nodes of a standing wave remain permanently at rest (zero displacement) at all times, unlike any point on a simple travelling wave?
Ans: At a node, the two component waves (travelling in opposite directions) always arrive exactly out of phase with each other at every instant, so their individual displacements are always equal in magnitude but opposite in sign β by the principle of superposition, these two contributions continuously and exactly cancel out at that specific point, keeping it at zero displacement permanently, regardless of the fact that the individual component waves themselves are still actively passing through that location.
Q. What is the distance between two adjacent nodes in a standing wave of wavelength 40 cm?
Ans: Ξ»/2 = 40/2 = 20 cm.
Organ Pipes
Closed Organ Pipe (Closed at One End)
A closed end must be a node (air can't move there), and the open end must be an antinode. Only odd harmonics are possible: fundamental frequency $f_0 = v/4L$, with overtones at $3f_0, 5f_0, 7f_0, ...$
Open Organ Pipe (Open at Both Ends)
Both ends must be antinodes. Both odd and even harmonics are possible: fundamental frequency $f_0 = v/2L$, with overtones at $2f_0, 3f_0, 4f_0, ...$ β twice the fundamental frequency of an equally long closed pipe, and with all integer harmonics present, not just odd ones.
Solved Example
Calculate the fundamental frequency of a closed organ pipe of length 0.5 m, given the speed of sound is 340 m/s.
$f_0 = \dfrac{v}{4L} = \dfrac{340}{4(0.5)} = \dfrac{340}{2} = 170\ \text{Hz}$
Q. Why can a closed organ pipe produce only odd harmonics, while an open organ pipe of the same length produces both odd and even harmonics?
Ans: A closed pipe requires a node at the closed end and an antinode at the open end β this specific boundary condition can only be satisfied by odd multiples of the fundamental wavelength pattern, mathematically excluding even harmonics entirely. An open pipe requires antinodes at both ends, a symmetric boundary condition that can be satisfied by both odd and even multiples of the fundamental pattern, allowing the full harmonic series to be produced.
Q. An open organ pipe and a closed organ pipe have the same length. Compare their fundamental frequencies.
Ans: The open pipe's fundamental frequency (v/2L) is exactly twice that of the closed pipe's fundamental frequency (v/4L), for the same length L and same speed of sound v.
Beats
Definition
Beats are periodic variations in loudness (amplitude) heard when two sound waves of slightly different frequencies are superposed β the two waves alternately reinforce (constructive interference, louder) and cancel (destructive interference, quieter) each other as their phase relationship continuously drifts in and out of alignment.
Solved Example
Two tuning forks of frequencies 256 Hz and 260 Hz are sounded together. Calculate the beat frequency heard.
$f_{beat} = |f_1-f_2| = |256-260| = 4\ \text{Hz}$ β 4 beats (loud-soft cycles) will be heard per second.
Practical Application β Tuning Instruments
Musicians use beats to tune instruments: as one string/note is adjusted closer to matching a reference frequency, the beat frequency (heard as a periodic "wah-wah-wah" throbbing) slows down, eventually disappearing entirely when the two frequencies exactly match β providing a precise, audible way to achieve perfect tuning without needing any electronic equipment.
Q. Why do beats disappear (no audible throbbing) when two sound sources have exactly the same frequency?
Ans: Beats arise from the two waves periodically drifting in and out of phase with each other due to their frequency difference β when the frequencies are exactly equal, this phase relationship (whatever it happens to be) never changes over time, so the resulting interference (whether constructive, destructive, or somewhere in between) stays fixed and doesn't vary periodically, meaning no beat throbbing is heard at all.
Q. A tuning fork of unknown frequency, when sounded with a standard 512 Hz fork, produces 6 beats per second. What are the two possible frequencies of the unknown fork?
Ans: Since f_beat = |fββfβ| = 6, the unknown frequency is either 512+6 = 518 Hz or 512β6 = 506 Hz (additional information, like the effect of slightly loading one fork, would be needed to determine which of these two possibilities is correct).
Resonance
Resonance occurs when a system is driven by an external periodic force whose frequency matches the system's own natural frequency of oscillation β under this condition, energy is transferred into the system with maximum efficiency, causing the amplitude of oscillation to build up dramatically, far beyond what a similar force at any other (non-matching) frequency would produce.
Solved Example
Explain why soldiers are traditionally instructed to break step (stop marching in unison) when crossing a bridge.
A bridge, like any physical structure, has its own natural frequency of vibration. If a large group of soldiers marches in perfect step, their rhythmic footsteps could apply a periodic force at a frequency that happens to match the bridge's natural frequency, driving it into resonance β this could cause the bridge's oscillation amplitude to build up dangerously large, potentially leading to structural damage or even collapse. Breaking step ensures the applied force from the soldiers' footsteps becomes essentially random/non-periodic (from the bridge's perspective), preventing this resonance condition from being established.
Q. Why does resonance cause such a dramatically larger amplitude buildup compared to driving a system at a non-matching frequency?
Ans: At the matching (resonant) frequency, each cycle of the driving force does positive work on the oscillating system in a consistently reinforcing, in-phase manner, continuously adding more energy with every cycle without any of it working against the motion. At a non-matching frequency, the driving force and the system's own natural oscillation gradually drift in and out of phase with each other, so the force sometimes does positive work and sometimes negative work on the system β this partial cancellation prevents the same dramatic, continuous energy buildup seen at exact resonance.
Doppler Effect
Definition
The Doppler effect is the apparent change in frequency of a wave observed when there is relative motion between the source of the wave and the observer β frequency appears higher when source and observer approach each other, and lower when they move apart.
Solved Example
Explain, using the Doppler effect, why an ambulance siren sounds higher-pitched as it approaches you and lower-pitched as it moves away.
As the ambulance approaches, each successive sound wave crest is emitted from a position slightly closer to you than the previous one, effectively compressing the wavelength (and increasing the frequency) of the sound reaching your ears β resulting in a higher perceived pitch. As the ambulance moves away, each successive crest is emitted from farther away, effectively stretching the wavelength (decreasing the frequency) β resulting in a lower perceived pitch. This shift from higher to lower pitch as the ambulance passes is the classic, everyday demonstration of the Doppler effect.
Q. Does the Doppler effect require the source's speed to change (accelerate) to produce a change in observed frequency?
Ans: No β even at a constant source speed, as long as there is relative motion between source and observer (source approaching or receding), the observed frequency will be shifted from the source's true, actual frequency. The specific value of this shift depends on the (possibly constant) relative velocity, not on any acceleration β a steadily approaching source produces a constant, but genuinely higher, observed frequency throughout its approach.
Q. Why does the Doppler formula distinguish between the observer's motion and the source's motion (using different signs/terms), rather than treating both identically based only on relative velocity?
Ans: While both source motion and observer motion do produce a frequency shift, they affect the sound wave through physically different mechanisms β a moving source changes the actual wavelength emitted into the medium, while a moving observer changes the rate at which they personally encounter wave crests without altering the wavelength in the medium itself. Because these are physically distinct effects (not always numerically identical in their overall formula behaviour, especially at higher/relativistic speeds), the standard Doppler formula for sound treats them as separate terms rather than assuming they're interchangeable in every respect.
Q. A stationary source emits sound of frequency 300 Hz. If an observer moves towards the source at a speed such that observed frequency becomes 330 Hz, does this represent an increase or decrease in perceived pitch, and why?
Ans: An increase in perceived pitch. As the observer moves towards the source, they encounter successive wave crests more frequently than they would if stationary β this higher rate of encountering crests is perceived as a higher frequency (330 Hz compared to the true 300 Hz), which corresponds to a higher, sharper pitch being heard.
β‘ Formula Cheat-Sheet
SHM velocityv = Οβ(AΒ²βxΒ²)
Total SHM energyE = Β½mΟΒ²AΒ²
Simple pendulum periodT = 2Οβ(l/g)
Spring-mass periodT = 2Οβ(m/k)
Wave speedv = fΞ»
Wave speed on stringv = β(T/ΞΌ)
Closed pipe fundamentalfβ = v/4L (odd harmonics)
Open pipe fundamentalfβ = v/2L (all harmonics)
Beat frequency|fβ β fβ|