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 Closed End Interference Definition Young s Double Slit Experiment What Sort of Wave is Light? Michelson-Morley Experiment Electromagnetic Spectrum Gravity Waves
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 the waves is also a wave. This applies to any waves at all, but we will be only looking further at two particularly interesting cases, where the waves being added have almost exactly the same properties: Standing Waves, where the waves only differ in the direction of travel. Interference, where the difference is in the phase of the waves.
Standing Waves
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 most often occurs 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 : sin A + sin B = 2 sin ( A+B 2 ) ( cos A B )] 2
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 has 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.
Minima: The minimum of the equation 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 vibration will always be zero, they are called the nodes of the vibration. The distance between two adjacent nodes is 1 2 λ.
Maxima: The maximum value occurs when sin(kx) = ±1. sin(kx) = ±1 kx = (n + 1 )π n = 0, 1, 2... 2 x = (n + 1 2 )π k = (n + 1 2 ) 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.
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.
A standing wave is represented by the function: y(x, t) = 0.004 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 = 0.004 sin(100x) cos(600t) So: 2A = 0.004 A = 0.002 m = 2 mm ω = 600 f = ω 2π = 600 = 96 Hz 2π k = 100 λ = 2π k = 2π = 0.063 m = 6.3 cm 100
To find the speed v: v = f λ = 96 0.063 = 6.05 m/s And the distance between nodes: 1 2 λ = 1 (0.063) = 0.032 m = 3.2 cm 2
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 puts restrictions on which standing waves are possible on the string. 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 than those we are about to see for the string.
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)
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 now 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
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 this where the frequencies of other 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.
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) 0.4 10 3 For the second mode: f 2 = 2f 1 = 2 884 = 1770 Hz And the third mode: f 3 = 3f 1 = 3 884 = 2650 Hz
Standing Waves in a Pipe Contents Other musical instuments 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.
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)
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 theoretically.
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 condition 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
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)
The pattern this time is a little trickier: ( v f n = (2n 1) = (2n 1)f 1 n = 1, 2, 3... 4L) 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 there is an empirical end correction factor required in this formula: ( v ) f n = (2n 1) 4L where the corrected length is L = L + 0.6r.
Interference of Waves
Interference Contents The term interference usually refers to the superposition of waves with different phases, but otherwise identical. Consider two waves described by the equations: y 1 = A sin(kx ωt) and y 2 = A sin(kx ωt + φ) The following plots show the superposition y = y 1 + y 2 calculated for several example values of φ: φ = 0 φ = π 3 + = + = φ = 2π 3 + = φ = π + =
In each case, the superposition has the same wavelength and frequency, but with different amplitudes. The maximum amplitude occurs when φ = 0 (the waves are in phase). This is known as constructive interference. The minimum amplitude occurs when φ = π (the waves are out of phase). This is known as destructive interference. Intermediate values of φ give partially destructive interference. Since adding a phase of 2nπ (where n is an integer) doesn t change the function, more generally we have: φ = 2nπ constructive interference φ = (2n + 1)π destructive interference
The phase difference between two waves is often due to a difference in the distance travelled by the two waves. As an example consider the situation shown at right. A point source S is producing sound waves, which are then detected by a microphone M. Two sound waves reach M - one directly and one after reflecting off a wall at W. S W M M measures the superposition of these waves: y M = A sin(kx ωt) + A sin(k(x + ) ωt) ( ) k = 2A cos sin(kx ωt + k 2 2 ) A sin(kx ωt + k 2 ) = SW + WM SM is the difference in path lengths of the two waves.
The sound measured at M has the same frequency as emitted at S, but the amplitude A depends on the path difference. We should remember from previous lectures that the loudness or intensity of sound is proportional to amplitude squared, so the intensity measured at M is (I 0 is the maximum intensity): ) ) I M = I 0 cos 2 ( k 2 = I 0 cos 2 ( π λ Looking at this formula, it is clear that the maximum intensity (i.e. constructive interference) will occur when cos ( ) π λ = ±1. max volume is when = nλ (n = 0, ±1, ±2...) Also, the minimum intensity (i.e. destructive interference) occurs when cos ( ) π λ = 0 min volume is when = (n + 1 )λ (n = 0, ±1, ±2...) 2
Young s Double Slit Experiment Contents Interference plays an important role in a very historic experiment performed by Thomas Young in 1801. Since the time of Newton (late 1600 s) there had been much controversy about whether light is a wave or a beam of particles. (Newton himself believed the latter - he named the individual particles of light corpuscles ) Young was able to (at least temporarily) settle the question with his famous Double Slit experiment. The apparatus is shown at right. A point light source is placed in front of a screen containing two thin, closely spaced slits. Another screen is placed behind the first screen. screen If Newton is correct that light consists of beams of particles coming from the source, then we would expect to only see light making it through to two points on the second screen as shown.
If light is a wave, the behaviour is very different. The properties of the slits are important - a wave transmitted through a thin slit will appear the same as produced by the slit being a point source - i..e circular wavefronts. The light source must be placed equidistant from the two slits. This ensures that the wavefronts transmitted by the two slits are in phase. Light from the two slits reaches all points on the screen. But the phase difference (and hence the interference) between the two sources will vary across the screen. When the wavefronts arrive together there will be a bright line seen (constructive interference), and when out of phase there will be darkness (destructive interference). dark bright
So, with these two very different predictions, Young performed the experiment and observed a series of alternating bright and dark lines (often called fringes ) as expected in the wave model of light. Light must therefore be a wave! The brightness of the light seen on the screen is given by the intensity of the wave. The calculation is exactly the same as the previous sound example: ( ) π I = I 0 cos 2 λ Note that the centre of the screen, between the slits, always has a bright fringe. Other maxima, as before, are found where = nλ, and minima when = (n + 1 2 )λ. = 3λ = 5 λ 2 = 2λ = 3 λ 2 = λ = 1 λ 2 = 0 = 1 λ 2 = λ = 3 λ 2 = 2λ = 5 λ 2 = 3λ
To determine the position on the screen for a particular value of, we need to do a little geometry, as shown below. Let d be the separation of the slits, L be the distance between the slits and the screen and consider the point P a distance y from the centre of the screen. L θ y d θ When L d, then the purple and orange lines become close to parallel and the green and blue triangles will be similar. From the green triangle: y = L tan θ L sin θ (true when θ is small) and from the blue triangle: sin θ = d
Combining these two formulas gives the connection between y and : y = L sin θ = L d As shown earlier, maxima occur when = nλ, so bright fringes are seen on the screens at the positions: y max = nλl d (n = 0, ±1, ±2...) and minima have = (n + 1 2 )λ, so dark fringes are located at: y min = (n + 1 2 )λl d (n = 0, ±1, ±2...) In both cases the fringes are evenly spaced, with separation λl d, and the dark fringes lying midway between the bright. Remember these formulas are approximations, valid when L d
Two thin slits, 1.00 mm apart are illuminated by a laser, causing an interference pattern on a screen 2.00 m behind the slits. The distance between two adjacent bright fringes on the screen is measured to be 1.26 mm. What is the wavelength of the laser? From the previous page, we know the separation of two adjacent slits is: δy = λl d λ = δy d L = (1.26 10 3 ) (1.00 10 3 ) 2.00 = 6.30 10 7 m = 630 nm Note: as done here, it is very common to express the wavelength of light in nanometres (1 nm = 1 10 9 m).
What Sort of Wave is Light? Contents After Young established that light was indeed a wave, the obvious next question is: A wave in what?. Other waves require a medium to travel in - air for sound, water for an ocean wave etc. What is the medium for light waves? This mysterious substance was known as the luminiferous ether (or just ether). Since we can see light from distant galaxies, the ether must be everywhere around us. Measuring and understanding the properties of the ether was very important. A key observation is that since the Earth is spinning on its axis, and also moving around the Sun, the Earth must be moving relative to the ether. A famous experiment trying to measure the effects of this relative motion was the 1887 Michelson-Morley experiment.
Michelson-Morley Experiment Contents The Michelson interferometer is another apparatus that relies on interference of light waves. The basic idea is shown below right. mirror Light shines on a beam splitter (a half-mirror that reflects 50% of light while the rest passes directly through). The two split beams then get reflected back by two regular mirrors before recombining at a light detector. The two beams have a path difference of: d 1 beam splitter d 2 mirror = 2(d 1 d 2 ) detector
The detector will detect fringes based on the value of - bright when = nλ and dark when = (n + 1 2 )λ. Because of the motion of the apparatus relative to the ether, we expect to be constantly changing, but in the actual experiment, no change is observed. This result raises serious doubt about the existence of the ether. Also in 1887, Heinrich Hertz produced and detected electromagnetic waves consisting of vibrating electric and magnetic fields, and requiring no other medium. We now understood visible light to be one of these electromagnetic waves, which also include radio waves, microwaves, X rays etc.
Electromagnetic Spectrum Contents Electromagnetic waves consist of vibrating electic and magnetic fields. In a vacuum they travel at a constant speed: c = 2.99792 10 8 m/s Different ranges of wavelengths are given different names as shown below (Note that these ranges are only approximately defined, and will vary a little in usage): frequency (Hz) 10 22 10 21 10 20 10 19 10 18 10 17 10 16 10 15 10 14 10 13 10 12 10 11 10 10 10 9 10 8 10 7 10 6 Gamma Rays X Rays Ultra- Violet Infrared Microwaves Radio 10 14 10 13 10 12 10 11 10 10 10 9 10 8 10 7 10 6 10 5 10 4 10 3 10 2 10 1 10 0 10 1 10 2 10 3 wavelength (m) 400 450 500 550 600 650 wavelength (nm) Some familiar examples: Dental X-ray 10 19 Hz Microwave oven 2.45 GHz AM radio 1 MHz FM radio/tv 100 MHz WiFi 2.4 GHz Mobile phones 900, 1800 MHz
Gravity Waves Contents In 1916, Albert Einstein, as part of his General Theory of Relativity predicted the existence of gravitational waves. These are oscillations in space itself. In 2015, the LIGO (Laser Interferometer Gravitational-Wave Observatory) experiment in the USA detected these waves for the first time. The design of this experiment is essentially the same as the Michelson-Morley experiment, with a passing gravity wave changing one of the path lengths d 1 or d 2, and thus causing a change in the interference pattern seen in the detector. In the original Michelson-Morley experiment these distances d 1 and d 2 were less than a metre, and the whole experiment fit comfortably on a benchtop. LIGO consists of two interferometers, each of which has d i = 4 kilometres! The detector is able to measure a change in path length of only 10 18 m - about one thousandth the size of a proton.