A brief note on Superposition and Standing Waves . Read it before a test. Try each question at the end before you open its answer.
Adding waves Superposition: where waves meet, the displacements add, y = y 1 + y 2 y = y_1 + y_2 y = y 1 + y 2 . Two equal waves with a phase difference ϕ \phi ϕ : A = 2 a cos ( ϕ / 2 ) A = 2a\cos(\phi/2) A = 2 a cos ( ϕ /2 ) . Unequal waves: A 2 = a 1 2 + a 2 2 + 2 a 1 a 2 cos ϕ A^2 = a_1^2 + a_2^2 + 2a_1a_2\cos\phi A 2 = a 1 2 + a 2 2 + 2 a 1 a 2 cos ϕ . In phase ( ϕ = 0 \phi = 0 ϕ = 0 ): they help each other. Out of phase ( ϕ = π \phi = \pi ϕ = π ): they cancel. Path difference to phase difference: ϕ = 2 π λ Δ x \phi = \dfrac{2\pi}{\lambda}\Delta x ϕ = λ 2 π Δ x . Intensity I ∝ A 2 I \propto A^2 I ∝ A 2 . Reflection and standing waves Fixed end: the wave comes back inverted. Free end: it comes back upright. Two equal waves going opposite ways add to y = 2 a sin k x cos ω t y = 2a\sin kx\cos\omega t y = 2 a sin k x cos ω t . Nodes never move. Next nodes are λ / 2 \lambda/2 λ /2 apart, and a node is λ / 4 \lambda/4 λ /4 from the nearest antinode. Strings and pipes String tied at both ends: f n = n v 2 L f_n = \dfrac{nv}{2L} f n = 2 L n v , every harmonic, with v = T / μ v = \sqrt{T/\mu} v = T / μ . Pipe open at both ends: f n = n v 2 L f_n = \dfrac{nv}{2L} f n = 2 L n v , every harmonic. Pipe closed at one end: f = ( 2 n − 1 ) v 4 L f = \dfrac{(2n - 1)v}{4L} f = 4 L ( 2 n − 1 ) v , odd harmonics only. A closed end is a displacement node; an open end is an antinode. A worked example A string 0.65 m 0.65\ \text{m} 0.65 m long has a tension of 90 N 90\ \text{N} 90 N and 2.5 g 2.5\ \text{g} 2.5 g of mass in each metre. Find its fundamental.
Speed. v = 90 N 0.0025 kg/m = 189.7 m/s v = \sqrt{\dfrac{90\ \text{N}}{0.0025\ \text{kg/m}}} = 189.7\ \text{m/s} v = 0.0025 kg/m 90 N = 189.7 m/s . Wavelength. In the fundamental, λ = 2 L = 1.3 m \lambda = 2L = 1.3\ \text{m} λ = 2 L = 1.3 m . Frequency. f 1 = 189.7 m/s 1.3 m = 146 Hz f_1 = \dfrac{189.7\ \text{m/s}}{1.3\ \text{m}} = 146\ \text{Hz} f 1 = 1.3 m 189.7 m/s = 146 Hz . Practice with new numbers
A program to try The program works out the example above, and the first three notes of a closed pipe.
Run it. Compare the string's first number with the worked example.
1. Change L_pipe = 0.34 to L_pipe = 0.17. Do the pipe's notes double?
Mistakes to avoid Adding amplitudes when the waves are not in phase. Use 2 a cos ( ϕ / 2 ) 2a\cos(\phi/2) 2 a cos ( ϕ /2 ) . Node spacing of λ \lambda λ . It is λ / 2 \lambda/2 λ /2 . Even harmonics from a closed pipe. It has odd ones only. f ∝ T f \propto T f ∝ T . It is f ∝ T f \propto \sqrt{T} f ∝ T . Check yourself Two equal waves meet with a phase difference of 180 ∘ 180^\circ 18 0 ∘ . What is the amplitude of the sum? Stuck? Show a hint Show the answerNext nodes of a standing wave are 40 cm 40\ \text{cm} 40 cm apart. What is the wavelength? Stuck? Show a hint Show the answerA string has a fundamental of 200 Hz 200\ \text{Hz} 200 Hz . What is its third harmonic? Stuck? Show a hint Show the answerWhich has the lower fundamental: an open pipe or a closed pipe of the same length? Stuck? Show a hint Show the answer