Superposition and Stationary Waves
In a linear wave model, overlapping displacements add point by point. Equal sinusoidal waves travelling in opposite directions can form stationary nodes and antinodes. A uniform string fixed at both ends supports harmonics with neighbouring nodes half a wavelength apart.
What a learner can do afterwards
- Applies the superposition principle to two pulses meeting, including the case where they cancel
- Marks nodes and antinodes on the first three harmonics of a string and gives each wavelength
- Finds the frequency of a harmonic from the length of the string and the wave speed
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In a linear wave model, add signed displacements at the same position and time. Positive means one chosen direction; negative means the opposite. Equal opposite displacements cancel, but unequal ones leave a remainder. Two identical sinusoidal waves travelling in the same direction and in phase produce twice the amplitude with the same wavelength.
Two pulses of equal size and opposite shape can give zero total displacement when they overlap exactly. In an ideal linear medium, the pulses then continue unchanged. Cancellation of displacement at one instant does not destroy their energy. Two equal-amplitude sinusoidal waves of the same frequency travelling in opposite directions form a stationary pattern.
A displacement node has zero displacement at all times. An antinode has the greatest oscillation amplitude and lies halfway between neighbouring nodes in this ideal stationary pattern. Antinodes still pass through zero displacement during each cycle. The pattern stays in place even though most string points move.
Neighbouring displacement nodes are half a wavelength apart. If their separation is g, then g = wavelength / 2, so wavelength = 2g. Do not confuse a node-to-node gap with a node-to-antinode gap, which is a quarter wavelength.
For a uniform string fixed at both ends, each end is a displacement node. The first harmonic has one half-wave in length L: nodes at 0 and L, and an antinode at L/2. The second has two half-waves, so its wavelength is L. Its nodes are at 0, L/2 and L, with antinodes at L/4 and 3L/4.
The third harmonic has three half-waves in length L. Its nodes are at 0, L/3, 2L/3 and L; antinodes are at L/6, L/2 and 5L/6. Since L = 3 × wavelength / 2, its wavelength is 2L/3. For example, a 0.90 m string has third-harmonic wavelength 0.60 m.
Third-harmonic nodes, ordered along the string. These cells list positions; they do not draw its oscillating shape.
For harmonic number n on this fixed-end string, L = n × wavelength / 2, so wavelength = 2L/n. Frequency is wave speed divided by wavelength: f_n = nv/(2L). Here v is the component travelling-wave speed, not a speed of the stationary pattern. The second harmonic has f_2 = v/L.
Add signed displacements in a linear model. Fixed-end string harmonics fit whole numbers of half-waves. Their wavelengths are 2L/n and frequencies are nv/(2L).
Original lesson, questions and flashcards by the dataset authors.
Rewritten readings were written for this dataset. Videos credited to Khan Academy.
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