Waves · Lesson 5 of 8
Standing Waves on Strings
“Derive stationary patterns, locate nodes and antinodes, and calculate the normal modes and tension of a fixed-end string.”
• Derive a standing-wave equation from equal waves travelling in opposite directions. • Determine position-dependent amplitude, node spacing, antinode spacing and relative phase. • Use fixed-end boundary conditions to obtain allowed wavelengths and harmonic frequencies. • Interpret a numerical standing-wave equation as two travelling waves and infer string tension.
Opposite directions can produce a fixed pattern
A wave on a string reflects back and forth between the ends. For particular frequencies, the continuing incident and reflected contributions form a stable pattern with fixed places that never move. Consider equal waves y₁ = a sin(kx − ωt) and y₂ = a sin(kx + ωt). Their direction of travel is opposite, but the frequency and wave number are the same. Add their displacements at a common x,t and use sin(U − V) + sin(U + V) = 2 sin U cos V.
The x-dependence and t-dependence now multiply as separate factors. There is no kx − ωt phase that must slide along the string. At each position the prefactor is constant in time, so that element oscillates with angular frequency ω, but different positions have different amplitudes. A sequence of photographs shows loops growing, flattening and reversing around the same zero-motion locations. When the whole string is instantaneously straight, the non-node elements can be moving fastest. A zero-displacement snapshot therefore does not make every point a node.
An ideal standing wave has no net time-averaged energy transport along the string because its two travelling constituents carry equal energy in opposite directions. Energy is still present and changes between kinetic and elastic forms. A real driven string can receive energy to replace damping losses. If the opposing waves have unequal amplitudes, the cancellation at the would-be nodes is incomplete and a net travelling component remains.
Nodes and antinodes have fixed coordinates
A node is a position whose displacement stays zero for all time, so its amplitude must vanish: sin(kx) = 0. Sine vanishes at integer multiples of π, giving kx = nπ. Substitute k = 2π/λ to obtain x = nλ/2. Adjacent nodes are separated by λ/2. These coordinates depend on the chosen node at x = 0, but their spacing does not.
At an antinode the displacement is not always maximum; the amplitude is maximum. The element still passes through equilibrium twice per cycle. Adjacent antinodes are λ/2 apart, just like adjacent nodes. Wavelength here refers to the wavelength of either travelling constituent. It equals twice the node spacing, not the distance from a node to its neighbouring antinode.
Frequency is shared, but phase depends on the loop
Within one interval between adjacent nodes, sin(kx) has a fixed sign. All non-node elements there reach their positive extremes together and cross equilibrium in the same direction: they are in phase. Across the next node, the spatial factor changes sign. The next loop therefore moves in the opposite direction at the same instant and is π out of phase. Loops separated by two nodes return to the same phase. The phase of a point that never oscillates is undefined, so nodes must be excluded from a statement about oscillation phase.
| Feature | Uniform harmonic progressive wave | Single ideal standing mode |
|---|---|---|
| Amplitude | Same a at all positions in the uniform ideal model | A(x) varies; zero at nodes |
| Relative phase | Changes continuously with separation, kΔx | Same within a loop; neighbouring loops differ by π |
| Frequency | Same f for all elements | Same f for all non-node elements |
| Pattern | Travels | Nodes and antinodes stay fixed |
| Net average energy flow | Nonzero for one travelling wave | Zero for equal opposite constituents |
Fixed ends select the wavelengths
Let a uniform string have length L, with endpoints x = 0 and x = L fixed. The first condition is already built into sin(kx). The second demands sin(kL) = 0, hence kL = nπ. The n = 0 solution gives a string with no nontrivial vibrating shape, so the allowed modes begin at n = 1. The length must hold an integer number of half-wavelengths: L = nλ/2.
A normal mode is a collective motion with one natural frequency and a definite spatial pattern. The lowest mode is the fundamental or first harmonic: f₁ = v/(2L). It has one loop, endpoints as nodes and an antinode halfway along. The second harmonic has two loops with an additional node at the centre, and frequency 2f₁. In general the nth harmonic has n loops, n antinodes, and n + 1 nodes including the endpoints. The first six harmonics therefore have loop counts 1 through 6, while their frequencies are f₁ through 6f₁.
The first overtone is the next mode above the fundamental. For an ideal string it is the second harmonic, so “first overtone” and “first harmonic” are different labels. A real plucked or bowed string generally contains several normal modes at once. The plucking or bowing position changes the relative strengths of the modes; it does not make a fixed-end ideal string acquire arbitrary natural frequencies. If a particular mode has a node at a driving point, that point couples poorly to that mode. The harmonic mixture helps determine the character of a musical note.
Problem
A string of length 1.5 m and mass 3.0 × 10⁻² kg has y = 0.06 sin[(2π/3)x] cos(120πt), with x,y in metres and t in seconds. Classify the pattern, find the two travelling constituents, and calculate the tension.
- 1.The separate spatial and time factors identify a standing mode. Here k = 2π/3 rad m⁻¹, ω = 120π rad s⁻¹, and the maximum standing amplitude is 0.06 m.
- 2.Since 2a = 0.06 m, each constituent has amplitude a = 0.03 m. The identity gives y = 0.03 sin[(2π/3)x − 120πt] + 0.03 sin[(2π/3)x + 120πt].
- 3.Each constituent has λ = 2π/k = 3.0 m, f = ω/(2π) = 60 Hz, and speed v = ω/k = 180 m s⁻¹. Their propagation directions are opposite.
- 4.The length is λ/2, so this is the fundamental. μ = 0.030/1.5 = 0.020 kg m⁻¹. From F = μv², tension is 0.020(180)² = 648 N.
- 5.The amplitude of a point at x = 0.375 m is 0.06|sin[(2π/3)(0.375)]| = 0.06 sin(π/4) = 0.04243 m. Its instantaneous displacement additionally depends on cos(120πt).
- 6.All interior points in this fundamental are within the same loop and therefore in phase, with unequal amplitudes. For a higher mode containing interior nodes, adjacent loops would instead have opposite phase.
Problem
A wire has mass 3.5 × 10⁻² kg, linear density 4.0 × 10⁻² kg m⁻¹, and fundamental frequency 45 Hz between rigid supports. Find the wave speed and tension.
- 1.First find the vibrating length, not its mass density per volume: L = m/μ = 0.035/0.040 = 0.875 m.
- 2.At the fundamental L = λ/2, so v = 2Lf₁ = 2(0.875)(45) = 78.75 m s⁻¹, approximately 79 m s⁻¹.
- 3.Then F = μv² = 0.040(78.75)² = 248.06 N, approximately 248 N. Retain unrounded speed in this step.
Classifying more than one standing mode
The function 2 cos(3x) sin(10t), with consistent units, is a single standing pattern: its zero-amplitude positions are fixed and its nonzero points oscillate at one frequency. A sum such as cos x sin t + cos(2x) sin(2t) combines two standing modes. In a compatible wave-speed model each term has ω/k = 1, but the sum is not one single-frequency normal mode. Its detailed shape changes as the two time factors evolve. More generally, a snapshot showing a few zero crossings does not distinguish a mode from an arbitrary superposition; examine the time dependence and boundaries.
Quick Quiz
Quiz
In y = 2a sin(kx) cos(ωt), the physical amplitude at x is
Amplitude is the maximum magnitude over time at the chosen position.
Adjacent nodes in one standing mode are separated by
Nodes occur when kx changes by π, giving Δx = π/k = λ/2.
Non-node points in neighbouring loops of a single standing mode are
The spatial coefficient changes sign across a node.
A fixed-end string has fundamental f₁. Its third harmonic is
A string supports every positive integer harmonic in this model.
A standing pattern has maximum antinode amplitude 6 cm. Each of its equal travelling constituents has amplitude
The maximum standing amplitude is 2a.
Practice Problems
- Add a sin(kx − ωt) and a sin(kx + ωt) explicitly. Identify the physical amplitude as a function of position and explain why the resulting pattern does not translate.
- For y = 0.040 sin(πx) cos(50πt), x,y in metres and t in seconds, list nodes and antinodes on 0 ≤ x ≤ 2 m. State the phase relation of points at x = 0.25 m and x = 1.25 m.
- Derive the allowed wavelengths and frequencies of a string of length L fixed at both ends. Explain why n = 0 is excluded and draw its first three harmonics.
- For a 1.5 m string of mass 0.030 kg with y = 0.06 sin[(2π/3)x] cos(120πt), give both travelling constituent equations, their wavelength, frequency, speed, and the string tension.
- For the same 1.5 m string, compare frequency, phase and amplitude across its interior. Find the amplitude at x = 0.375 m and distinguish it from instantaneous displacement.
- A fixed-end wire of mass 0.035 kg and μ = 0.040 kg m⁻¹ has fundamental frequency 45 Hz. Calculate length, wave speed and tension without rounding intermediate values.
- Classify 2 cos(3x) sin(10t), 3 sin(5x − 0.5t) + 4 cos(5x − 0.5t), and cos x sin t + cos(2x) sin(2t), with suitable units. Explain which is a single travelling wave, a single standing mode or a superposition of modes.
- Explain why an ideal standing wave can have energy but no net time-averaged energy transport, and why a string that is straight at one instant need not consist entirely of nodes.
Key Takeaways
• Equal opposite travelling waves make a standing pattern with A(x) = 2a|sin(kx)|. • Node–node and antinode–antinode spacing are λ/2; nearest node–antinode spacing is λ/4. • Non-node points in one loop are in phase; adjacent loops are π out of phase. • Fixed-end conditions require λₙ = 2L/n and fₙ = nv/(2L); a general string vibration can mix modes.