Stationary wavesEdexcel A-Level Physics: Revision notes
Section 1
Formation of a stationary wave
A stationary (standing) wave is formed when two progressive waves of the same frequency and similar amplitude travel in opposite directions and superpose. This usually happens when a wave is reflected and meets the incoming wave, for example on a string fixed at one or both ends, or in a pipe.
Unlike a progressive wave, a stationary wave does not transfer energy along the medium, and its pattern stays in place. All points between adjacent nodes oscillate in phase, and points in adjacent loops are in antiphase.
Section 2
Nodes and antinodes
- A node is a point of zero amplitude. The two waves are always in antiphase there, so they cancel.
- An antinode is a point of maximum amplitude. The waves are in phase there and reinforce.
Adjacent nodes (or adjacent antinodes) are separated by λ/2. A node and the next antinode are separated by λ/4.
The amplitude varies with position: zero at a node, maximum at an antinode, and in between it follows a sine shape. At a point a distance x from a node the amplitude is proportional to sin(2πx/λ).
Nodes are not points where the string is momentarily at rest. They never move at all.
Section 3
Harmonics on a string
A string of length L fixed at both ends has a node at each end, so only certain wavelengths fit: L = nλ/2, giving λ = 2L/n and
f = n v/(2L), n = 1, 2, 3, ...
The fundamental (first harmonic, n = 1) has one loop: λ = 2L and f₁ = v/2L. The second harmonic has two loops, λ = L and f = 2f₁, and so on.
The frequencies that fit are the resonant frequencies of the string.
Section 4
Speed of a transverse wave on a string
The speed of transverse waves on a taut string depends on the tension and on the mass per unit length:
v = √(T/μ)
where T is the tension in N and μ is the mass per unit length in kg m⁻¹. A greater tension raises the speed, and a heavier string lowers it.
For the fundamental, f₁ = (1/2L)√(T/μ). So f ∝ 1/L, f ∝ √T and f ∝ 1/√μ.
If the tension is quadrupled, v and f double, because they depend on √T.
Section 5
Core practical: vibrating strings
A vibration generator drives a string that passes over a pulley. Masses on the end provide the tension (T = mg).
- Effect of tension: keep L and μ fixed, vary T, find the fundamental frequency at resonance (largest amplitude). Plot f² against T: a straight line through the origin, gradient 1/(4L²μ).
- Effect of length: keep T and μ fixed, vary L and plot f against 1/L.
- Effect of μ: use strings of different μ, found by weighing a measured length. Plot f² against 1/μ.
Find resonance by approaching from above and below, and repeat to reduce random error.
Section 6
Worked example
A wire of length 0.60 m has μ = 4.0 × 10⁻⁴ kg m⁻¹ and tension 36 N.
v = √(36 ÷ 4.0 × 10⁻⁴) = 300 m s⁻¹
λ = 2L = 1.2 m, so f₁ = 300 ÷ 1.2 = 250 Hz
If the tension rises to 144 N, v doubles to 600 m s⁻¹ and f₁ = 500 Hz.
That's the notes covered.
Carry on to the next subtopic.
Exam questions on Stationary waves
- A taut wire of length 0.60 m is fixed at both ends and plucked so that it vibrates with a single loop. The wire has a mass per unit length of 4.0 × 10⁻⁴ kg m⁻¹ and is under a tension of 36 N. Treat both fixed ends as nodes.The tension is increased to 144 N with the length unchanged. Calculate the new fundamental frequency.2 marks
- A string is stretched between two fixed points 1.50 m apart. A vibration generator near one end drives the string at 120 Hz, and a stationary wave with exactly three loops is seen between the fixed ends. Treat both ends as nodes.Explain how the stationary wave is formed on the string.2 marks
- A student investigates how the fundamental frequency f of a vibrating string depends on its tension T. A string of vibrating length 0.800 m passes over a pulley and carries hanging masses that provide the tension. For each tension she adjusts the frequency of a vibration generator until the string shows a single loop of maximum amplitude. She plots f² against T and obtains a straight line through the origin with gradient 391 Hz² N⁻¹.Use the gradient to determine the mass per unit length of the string.3 marks
Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).