Waves Part 3A: Standing Waves

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Baldwin Poole
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1 Waves Part 3A: Standing Waves Last modified: 24/01/2018

2 Contents Links Contents Superposition Standing Waves Definition Nodes Anti-Nodes Standing Waves Summary Standing Waves on a String Standing Waves in an Open-Ended Pipe Standing Waves in a Pipe with One Closed End

3 Superposition Contents If two (or more) waves are travelling in the same medium, in the same direction, then the two displacements y 1 and y 2 will add together: y 1 + y 2 = F (x vt) + G(x vt) = H(x vt) Recall that a travelling wave has the form F (x vt) and the speed v will be the same for all waves in the same medium. The sum, or superposition of two (or more) waves travelling in the same direction is also a wave, in htat same direction. The situation is a little different for waves travelling in different directions.

4 Standing Waves Contents Standing (or Stationary) waves are formed by the superposition of waves travelling in opposite directions, in particular by waves that are otherwise identical. This will most often occur when a wave meets its own reflection. In this case, if y 1 = F (x vt) and y 2 = F (x + vt), then the superposition, y 1 + y 2 = F (x vt) + F (x + vt) will not be a function of x vt and so is not a travelling wave. As usual, let s consider only harmonic waves, with y 1 = A sin(kx ωt) and y 2 = A sin(kx + ωt). Then the superposition will be: y 1 + y 2 = A sin(kx ωt) + A sin(kx + ωt) = 2A sin(kx) cos(ωt) [ Using the identity: sin A + sin B = 2 sin ( A+B 2 ) ( cos A B )] 2

5 Contents The superposition of the waves has the following form: y(x, t) = 2A sin(kx) cos(ωt) As already mentioned, this is clearly not a travelling wave. There is however, still some vibration occuring. To understand what is going on, let s plot this equation over half a period T : t=0 t = 1 12 T t = 1 6 T t = 1 4 T t = 1 3 T t = 5 12 T t = 1 2 T Unlike a travelling wave, where every position vibrates with the same amplitude, in a standing wave the amplitude of the vibration varies with position x, and at some particular values, this amplitude is zero.

6 Nodes Contents Examining the equation, we see that the minimum amplitude occurs when sin(kx) = 0. Remembering that k = 2π λ where λ is the wavelength of the original wave: sin(kx) = 0 kx = nπ n = 0, 1, 2... x = nπ k = n πλ 2 π = n 2 λ At these values of x the amplitude of the vibration will always be zero. These points are called the nodes of the vibration. The distance between two adjacent nodes is 1 2 λ.

7 Anti-Nodes The maximum vibration amplitude occurs when sin(kx) = ±1. Contents sin(kx) = ±1 kx = (n + 1 )π n = 0, 1, x = (n )π k = (n ) πλ 2 π = (2n + 1) λ 4 At these values of x the vibration will have a varying displacement (because of the cos(ωt) factor), but the largest amplitude. These points are called the anti-nodes of the vibration, and are located half-way between the nodes. Note that as in a travelling wave, the individual points of the medium are vibrating around their equilibrium position. BUT the maxima and minima of the vibration are at fixed points. The wave is not travelling, it is stationary.

8 Standing Waves Summary Contents Standing waves consist of vibrations in a medium which have fixed points of minimum (i.e. zero) vibration (the nodes) and maximum vibration (the anti-nodes). In what follows, we will indicate a standing wave by showing the range of maximum vibration: λ 2 anti-nodes λ 2 nodes The distance between two adjacent nodes, or two adjacent anti-nodes is λ the same: 2, where λ is the wavelength of the original harmonic wave creating the standing wave.

9 Contents A standing wave is represented by the function: y(x, t) = sin(100x) cos(600t) Find the amplitude, frequency, wavelength and speed of the component travelling waves. Also determine the distance between nodes in this standing wave, Comparing the general and given equations gives: y = 2A sin(kx) cos(ωt) y = sin(100x) cos(600t) So: 2A = A = m = 2 mm ω = 600 f = ω 2π = 600 = 96 Hz 2π k = 100 λ = 2π k = 2π = m = 6.3 cm 100

10 Contents To find the speed v: v = f λ = = 6.05 m/s And the distance between nodes: 1 2 λ = 1 (0.063) = m = 3.2 cm 2

11 Standing Waves on a String Contents Many musical instruments (e.g. guitar, violin, piano etc) use a length of string (or wire) fixed at both ends. Sound waves are generated by standing waves on these strings. The fact that the ends of the strings are fixed means that there must be nodes located at the end-points. This restriction means that only a limited number of standing waves are possible on the string, and thus a limited number of sound frequencies (i.e. musical notes) can be produced. The possible standing waves are called the modes of vibration of the string. Note that this situation will apply to other objects also. All objects will have a limited set of possible vibrational modes. In general these modes will be much more complicated (occuring in 2 or 3 dimensions) than those we are about to see for the string.

12 Contents The simplest mode of vibration for a string is that with the least number of nodes - only those at the two ends of the string. 1 st Mode L = λ 1 2 = v 2f 1 L f 1 = v 2L The next simplest has three nodes: 2 nd Mode L L = 2 λ 2 2 = v f ( 2 v f 2 = 2 = 2f 1 2L)

13 Contents 3 rd Mode L = 3 λ 3 2 = 3v 2f ( 3 v f 3 = 3 = 3f 1 2L) 4 th Mode L = 4 λ 4 2 = 2v f ( 4 v f 4 = 4 = 4f 1 2L) The pattern should be clear. For any integer n we have for the n-th mode: L = n λ n 2 = nv ( v ) f n = n = nf 1 2f n 2L

14 These vibrational frequencies are the musical notes that the string can create. The lowest frequency, f 1 is the fundamental frequency. Contents In cases like the string where the frequencies of higher modes are integer multiples of the fundamental frequency, then these allowed frequencies are called harmonics. For a stretched string, the frequency of the n-th mode (aka the n-th harmonic) is f n, given by; ( v f n = nf 1 = n = 2L) n T 2L µ T where we remember that for a wave on a string v = µ with T being the tension, and µ the linear mass density, of the string.

15 Contents A mass M = 5 kg is suspended by a string with linear mass density µ = 0.40 g/m over a frictionless pulley as shown below right. Calculate the fundamental frequency that the string will vibrate with. Find also the frequencies of the second and third vibrational frequencies. L = 20 cm M The block M is in equilibrium, so T = Mg = 50 N For the fundamental mode: f 1 = v 2L = 1 2L T µ = 1 50 = 884 Hz 2(0.2) For the second mode: f 2 = 2f 1 = = 1770 Hz And the third mode: f 3 = 3f 1 = = 2650 Hz

16 Standing Waves in an Open-Ended Pipe Contents Other musical instuments (e.g. flutes, organs) produce sound via standing waves in an open-ended pipe, formed by the superposition of sound waves. Again there are physical restrictions on these allowed modes - in this case, each open end of the pipe must be an anti-node. The simplest mode of vibration for the pipe is that with the least number of nodes - only one at the middle of the pipe.. 1 st Mode L L = λ 1 2 = v 2f 1 f 1 = v 2L The speed v here is the speed of sound in the air inside the pipe.

17 Contents 2 nd Mode L L = 2 λ 2 2 = v f ( 2 v f 2 = 2 = 2f 1 2L) 3 rd Mode L = 3 λ 3 2 = 3v 2f ( 3 v f 3 = 3 = 3f 1 2L) 4 th Mode L = 4 λ 4 2 = 2v f ( 4 v f 4 = 4 = 4f 1 2L)

18 Contents These equations should look familiar! The expression for the frequencies of the vibrational modes (harmonics) of the open pipe is the same as for the stretched string: ( v f n = nf 1 = n 2L) In practice, this formula is not quite accurate. Because of the inertia of air at the ends of the pipe, the above formula must be slightly modified by what is called the end correction, involving the radius r of the pipe. ( v ) f n = n 2L where L = L + 1.2r This is an empirical formula, i.e. derived from experiment rather than mathematically.

19 Pipe with One Closed End Contents Another musical possibility is a pipe with one end open and the other closed. As we should expect, the possible vibrational modes are restricted by the conditions that the closed end is a node, and the open an anti-node. The simplest mode of vibration includes only the required node and anti-node: 1 st Mode L L = 1 2 λ 1 2 = v 4f 1 f 1 = v 4L

20 Contents 2 nd Mode L L = 3 2 λ 2 2 = 3v 4f ( 2 v f 2 = 3 = 3f 1 4L) 3 rd Mode L = 5 2 λ 3 2 = 5v 4f ( 3 v f 3 = 5 = 5f 1 4L) 4 th Mode L = 7 2 λ 4 2 = 7v 4f ( 4 v f 4 = 7 = 7f 1 4L)

21 Contents The pattern this time is a little trickier: ( v f n = (2n 1) = (2n 1)f 1 n = 1, 2, L) The frequencies of the higher modes are integer multiples of the fundamental fequency, so are harmonics, but this time only the odd harmonics are present, which makes the naming potentially a little confusing: the second mode is the third harmonic the third mode is the fifth harmonic and so on... For the same reasons as with the open ended pipe an empirical end correction factor is required in this formula: ( v ) f n = (2n 1) 4L where the corrected length is L = L + 0.6r.

Waves Part 3: Superposition

Waves Part 3: Superposition Last modified: 06/06/2017 Superposition Standing Waves Definition Standing Waves Summary Standing Waves on a String Standing Waves in a Pipe Standing Waves in a Pipe with One