Oscillations and Harmonic Motion
Simple harmonic motion, pendulum, damped and forced oscillator, resonance. Fundamental chapter for waves and optics.
Complete Theory
5A system undergoes simple harmonic motion when the restoring force is proportional to the displacement and directed toward the equilibrium position (Hooke's law, ). This is the most important model in oscillatory physics because it describes the behavior of every stable system near an equilibrium point.
The equation of motion can be written in canonical form as: This is a linear homogeneous second-order ODE with constant coefficients. The characteristic equation has complex conjugate roots , which lead to sinusoidal solutions. The general solution is a linear combination of sine and cosine, or equivalently: where is the amplitude (maximum displacement) and the initial phase (angular position at on the phasor diagram).
Initial conditions: the constants and are determined from the initial position and velocity: The phase is chosen in the interval based on the quadrant determined by the signs of and .
Fundamental properties:
The equation of motion can be written in canonical form as: This is a linear homogeneous second-order ODE with constant coefficients. The characteristic equation has complex conjugate roots , which lead to sinusoidal solutions. The general solution is a linear combination of sine and cosine, or equivalently: where is the amplitude (maximum displacement) and the initial phase (angular position at on the phasor diagram).
Initial conditions: the constants and are determined from the initial position and velocity: The phase is chosen in the interval based on the quadrant determined by the signs of and .
Fundamental properties:
- Isochronism: the period does not depend on amplitude. Large and small oscillations share the same period — a property discovered by Galileo while observing a swinging lamp in the Pisa cathedral.
- Natural angular frequency: increases with spring stiffness and decreases with mass. A stiffer spring → faster oscillations; a larger mass → slower oscillations.
- Frequency: in hertz (Hz), the number of complete oscillations per second.
- Mass attached to an ideal spring (the prototypical model).
- Motion of a diatomic molecule near equilibrium (Morse potential approximated by a parabola).
- LC circuit (inductor-capacitor): the equation is formally identical, with (charge) ↔ , ↔ , ↔ .
- Simple pendulum for small angles (see §3).
Differentiating with respect to time yields the instantaneous velocity and acceleration:
Mechanical energy: the elastic force is conservative (depends only on position), so the total mechanical energy is conserved. The kinetic energy and potential energy are: Adding them, and recalling : . The total energy is therefore proportional to the square of the amplitude.
Cyclic energy exchange:
Velocity-position relation: from energy conservation we obtain a useful time-independent relation: The sign depends on the direction of motion. This formula shows that velocity is maximum at the center (, ) and zero at the endpoints ().
Energy diagram: the plot is a parabola. The total energy is a horizontal line. The intersection points () define the turning points , where . The particle oscillates between these two points like a ball rolling in a parabolic bowl, continuously converting kinetic energy into potential energy and back.
Mechanical energy: the elastic force is conservative (depends only on position), so the total mechanical energy is conserved. The kinetic energy and potential energy are: Adding them, and recalling : . The total energy is therefore proportional to the square of the amplitude.
Cyclic energy exchange:
- When (endpoints): , , — all energy is potential.
- When (equilibrium): , , — all energy is kinetic.
- At any intermediate position: .
Velocity-position relation: from energy conservation we obtain a useful time-independent relation: The sign depends on the direction of motion. This formula shows that velocity is maximum at the center (, ) and zero at the endpoints ().
Energy diagram: the plot is a parabola. The total energy is a horizontal line. The intersection points () define the turning points , where . The particle oscillates between these two points like a ball rolling in a parabolic bowl, continuously converting kinetic energy into potential energy and back.
The simple pendulum consists of a point mass suspended from an inextensible massless string of length . The restoring force is the tangential component of the weight: . Writing Newton's second law along the tangent to the circular path:
This is a nonlinear equation (due to ). To solve it analytically, we use the small-angle approximation: for (i.e., ), the Taylor series expansion gives . Keeping only the first term:
which is formally identical to the SHM equation () with:
Fundamental properties:
Historical applications:
Fundamental properties:
- Mass independence: the period of a simple pendulum does not depend on mass . Galileo deduced this experimentally by observing that pendulums of different masses but equal length oscillate with the same period (for equal amplitude). The mass cancels out in the equation of motion because it appears in both the restoring force () and the inertia ().
- Approximate isochronism: for small oscillations, is independent of amplitude. For larger amplitudes, the actual period is longer than the approximation (corrected by ).
- Dependence on : by measuring for a known , one can determine precisely. This is the principle behind pendulum gravimeters used in geophysics.
Historical applications:
- Galileo (1583): discovers pendulum isochronism by observing a lamp in the Pisa cathedral.
- Huygens (1656): invents the pendulum clock, improving time-measurement precision by two orders of magnitude.
- The meter was originally defined as the length of a pendulum with (the definition has since changed).
In the presence of a viscous damping force (proportional and opposite to velocity, like the force exerted by a fluid at low speed), the equation of motion becomes:
Dividing by and introducing standard parameters:
is the damping coefficient (in ). The characteristic equation has roots , whose discriminant determines the damping regime.
The three damping regimes:
Physical examples:
The three damping regimes:
- Underdamped (): the discriminant is negative, giving complex conjugate roots with . The solution is a sinusoidal oscillation with exponentially decaying amplitude: The exponential envelope describes energy dissipation. The time constant is the time for the amplitude to drop by a factor .
- Critically damped (): the roots are real and equal . The solution is , which returns to equilibrium in the shortest possible time without oscillating. This is the ideal regime for car shock absorbers, door dampers, and spring-based measuring instruments.
- Overdamped (): real and distinct roots, both negative. The solution is , with a slower return to equilibrium compared to the critical case.
Physical examples:
- Car shock absorbers: tuned near critical damping for a rapid return without oscillations after a bump.
- Door closers: viscous fluid mechanism preventing the door from slamming and oscillating.
- Galvanometer: the needle is critically damped to reach the measurement value quickly without oscillating.
- Seismometers: often overdamped to avoid spurious oscillations.
When a damped oscillator is driven by a periodic external force , the equation of motion becomes:
The general solution is the sum of two contributions:
Resonance: the amplitude reaches its maximum when the denominator is minimized, i.e. at: At resonance, the amplitude is . As , — the oscillator "absorbs" energy from the driving force with maximum efficiency.
Phase at resonance: when , we have (velocity is in phase with the driving force, so the power is maximal). For , (motion in phase with the driving force). For , (motion opposite in phase).
Quality factor and bandwidth: The bandwidth (difference between the frequencies at which ) is related to by: A high means a narrow, tall resonance peak; a low means a broad, low peak.
Examples and applications:
- Transient: the homogeneous solution (damped oscillation) that decays after a time .
- Steady state (particular solution): an oscillation at the driving frequency , with amplitude and phase .
Resonance: the amplitude reaches its maximum when the denominator is minimized, i.e. at: At resonance, the amplitude is . As , — the oscillator "absorbs" energy from the driving force with maximum efficiency.
Phase at resonance: when , we have (velocity is in phase with the driving force, so the power is maximal). For , (motion in phase with the driving force). For , (motion opposite in phase).
Quality factor and bandwidth: The bandwidth (difference between the frequencies at which ) is related to by: A high means a narrow, tall resonance peak; a low means a broad, low peak.
Examples and applications:
- Tacoma Narrows Bridge (1940): the bridge entered resonance with periodic wind gusts, producing increasingly large torsional oscillations until collapse. Classic example of the dangers of resonance in civil engineering.
- RLC circuit: resonance in an circuit is analogous: , . Resonant circuits are the basis of radio tuning.
- Musical instruments: the soundboard of a violin or guitar selectively amplifies certain frequencies (acoustic resonance).
- Crystal glass: an opera singer can shatter it if the voice matches the glass's resonance frequency with sufficient amplitude.
- Magnetic resonance imaging (MRI): atomic nuclei in a magnetic field absorb radio waves at the Larmor frequency (), the principle behind nuclear magnetic resonance and medical imaging.
Worked Examples
2Example 1Mass-spring oscillator: period, energy, and maximum velocity
Given
,
(initial amplitude),
Find
Natural frequency , period , frequency
Equation of motion
Total energy and maximum speed
Speed at
Step-by-step solution
1Problem setup: a mass of 0.5 kg is attached to a spring with spring constant . The mass is displaced 8 cm from equilibrium and released from rest (). We want to fully characterize the resulting harmonic motion.
2Why ? Newton's second law with Hooke's force gives , or . Comparing with the standard SHM equation identifies . The square root arises because has units and must be an angular frequency in . Substituting: .
3Why and ? The cosine repeats when increases by , hence and . Frequency counts oscillations per second: .
4Why ? The general solution has two constants to fix. With we have , so (or ). Since we choose , and . The position is .
5Why ? The elastic force is conservative, so mechanical energy is conserved. At the mass is at rest (), so .
6Why ? At equilibrium () potential energy is zero and kinetic energy is maximal: . Solving: .
7Why is at not ? Potential energy scales with , not . At : . So , and .
8Check: , , ✓.
✓ Final result: , , , ,
Example 2Simple pendulum — Earth vs Moon
Given
,
Find
Period on Earth
Period on Moon
How much does a pendulum clock lose on the Moon in 1 hour?
Step-by-step solution
1Problem setup: a simple pendulum of length is taken from Earth to the Moon. We compare its period in the two environments and determine how much a pendulum clock (which counts oscillations to mark time) would run slow on the Moon compared to an accurate Earth clock.
2Period on Earth using the simple pendulum formula (valid for small angles, ): . Independent of mass and amplitude (isochronism). — almost exactly 2 s for .
3On the Moon gravity is ~6× smaller (), so the pendulum swings more slowly: . Ratio .
4Oscillations in 1 real hour: on Earth: oscillations (showing exactly 60 minutes). On Moon: oscillations.
5Time lost: the lunar clock "thinks" each oscillation lasts 2.007 s (as it would on Earth), so it displays instead of 60 min. It loses per real hour.
✓ Final result: , ; pendulum clock on Moon loses 35.6 min per hour
Exercises with Solutions
3Exercise 1SHM with initial conditionsHard
Problem to solve
A mass-spring oscillator (, ) is set in motion with and . (a) Determine the amplitude and initial phase of the motion. (b) Calculate the position at time . (c) Find the first time () when the mass passes through .
Given data
m = 0.3\,\mathrm{kg}k = 120\,\mathrm{N/m} (starts at equilibrium) (initial velocity)
Step-by-step solution
1Setup: we have a mass-spring oscillator with special initial conditions: the mass starts at the equilibrium position () with a positive initial velocity . We use the general formulas linking and to the initial conditions.
2Compute : .
3Amplitude : . The total energy .
4Initial phase : . Since , we need to take the limit. If and , we are in the quadrant where and , so . The equation of motion is .
5Position at : . The mass is about 11.4 cm from equilibrium, in the positive direction.
6First time : solve . Dividing: . The sine function equals for (and also , but is the first after ). Hence .
✓ Final answer: , , first at
Exercise 2Damping — underdamped regimeHard
Problem to solve
A damped mass-spring oscillator has , , damping constant . (a) Determine the damping regime by comparing and . (b) Calculate the damped angular frequency . (c) Find the time required for the amplitude to drop to half its initial value. (d) Compute the quality factor and interpret it.
Given data
m = 0.5\,\mathrm{kg}k = 50\,\mathrm{N/m}b = 1.0\,\mathrm{N\cdot s/m}
Step-by-step solution
1Setup: we have a mass-spring oscillator with viscous damping. The three quantities determining the dynamics are , , . From these we compute , , and compare them to establish the regime.
2Natural angular frequency: .
3Damping coefficient: .
4Comparison vs : , so we are in the underdamped regime. Oscillations decay exponentially but the system still oscillates.
5Damped angular frequency: . Notice because damping is weak (). The damped period is only slightly longer than the undamped period .
6Amplitude halving time: the amplitude decays as . We want such that : . Therefore . After about 0.7 s the amplitude has halved; after it is reduced to (12.5% of the initial value).
7Quality factor: . A is relatively low, indicating significant damping: the system loses about radians of energy per cycle (about 20% of the energy per oscillation). For comparison, a tuning fork has , a quartz crystal .
✓ Final answer: Underdamped, , ,
Exercise 3Physical pendulum — rotating diskVery Hard
Problem to solve
A solid homogeneous disk of mass and radius is suspended from a horizontal pivot at a point on its rim. The disk oscillates as a physical pendulum. (a) Calculate the moment of inertia of the disk about the pivot. (b) Determine the period of small oscillations. (c) Find the length of a simple pendulum that would have the same period.
Given data
M = 2\,\mathrm{kg}R = 0.3\,\mathrm{m} (pivot-to-CM distance)
Step-by-step solution
1Setup: a homogeneous rigid disk oscillates about a pivot at its rim. This is a physical pendulum, not a simple one, because the mass is not concentrated at a single point. We must use , where is the moment of inertia about the pivot and the pivot-CM distance.
2Moment of inertia about the CM: for a solid disk, .
3Parallel axis theorem (Steiner): to shift the inertia from CM to the pivot (distance ), use .
4Oscillation period: .
5Equivalent simple pendulum: a simple pendulum of length has period . Equating:
6Check: ✓. The equivalent simple pendulum is longer than the radius (0.3 m) but shorter than the diameter (0.6 m). Intuitively, the distributed mass of the disk makes the system "slower" than a point mass at the same distance from the pivot, so a longer simple pendulum is needed to match the period.
7Note: if the disk were hung at the CM, and it would not oscillate (neutral equilibrium). If hung very close to the rim, the period would be shorter. The minimum period is obtained for from the CM (center of percussion).
✓ Final answer: , ,
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Integrative Problems
Problems combining all chapters — exam levelProblem 1Tower, Ballistic Pendulum, and Keplerian OrbitEXTREME
A cannon is placed on top of a tower tall and fires a projectile of horizontally at .
The projectile strikes and embeds in a wooden block hanging from a rope of length (ballistic pendulum), at ground level.
The Earth-Moon system is then used as a reference for Kepler's third law.
The projectile strikes and embeds in a wooden block hanging from a rope of length (ballistic pendulum), at ground level.
The Earth-Moon system is then used as a reference for Kepler's third law.
📌 Problem data
(a)Uniformly Accelerated Motion(b)Inelastic Collision(c)Potential Energy + Pendulum(d)Moment of Inertia — Rigid Body(e)Gravitation — Kepler's Third Law
Problem 2Spring, Rolling Disk, Inclined Plane Collision, and ConservationEXTREME
A spring (, compressed ) launches a solid disk (, ) up an inclined plane (, , ) that rolls without slipping.
At the top the disk is launched horizontally and strikes a pendulum (, ) — perfectly inelastic collision. What is asked (solved below, a→e): (a) the disk's speed at the top of the plane; (b) the range and impact speed of the horizontal launch; (c) the speed after the inelastic collision with the pendulum and the energy lost; (d) the pendulum's maximum angle, the maximum tension, and whether it completes the loop; (e) the full energy balance (from spring to maximum angle).
At the top the disk is launched horizontally and strikes a pendulum (, ) — perfectly inelastic collision. What is asked (solved below, a→e): (a) the disk's speed at the top of the plane; (b) the range and impact speed of the horizontal launch; (c) the speed after the inelastic collision with the pendulum and the energy lost; (d) the pendulum's maximum angle, the maximum tension, and whether it completes the loop; (e) the full energy balance (from spring to maximum angle).
📌 Problem data
(a)Energy + Rigid Body (rolling)(b)Kinematics — Projectile(c)Inelastic Collision + CM(d)Pendulum Dynamics + Forces(e)Conservation Laws — Complete Energy Balance