Wave Phenomena Physics 15c. Lecture 9 Wave Reflection Standing Waves

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1 Wave Phenomena Physics 15c Lecture 9 Wave Reflection Standing Waves

2 What We Did Last Time Energy and momentum in LC transmission lines Transfer rates for normal modes: and The energy is carried by the EM field around the wires The wires guide the waves Poynting vector gives the power density Studied the wave reflection Determined by the impedance matching Power is reflected or absorbed according to P reflected = V = R Z R + Z V + (R Z)2 (R + Z) 2 P input 1 2 V 0 I 0 P absorbed = 1 V 0 I 0 2 S = E H I = R Z R + Z I + c w 4RZ (R + Z) 2 P input

3 Goals for Today Reflection of mechanical waves We expect reflection in mechanical waves as well as in electromagnetic waves Where the sound hit a solid wall Where the string is tied to the door knob We expect open and shorted ends Can we define impedance for mechanical waves? Standing waves Created when sinusoidal waves are reflected Basic principle for (almos all musical instruments

4 Transverse Waves on String T ξ(x) T x Transverse wave ξ(x c w is traveling on a string The string has tension T and linear mass density ρ l From Lecture #6: Wave equation ρ l 2 ξ(x, t 2 = T 2 ξ(x, x 2 Wave velocity c w = ω k = T ρ l

5 Reflection at a Fixed End x = L Suppose one end of the string is tied to a fixed point This end cannot move ξ(l) = 0 Reflection must occur to satisfy this condition. Let s call the original and reflected waves as ξ + (x c w and ξ (x + c w At x = L, ξ(x) ξ + (L c w + ξ (L + c w = 0 Reflected waves are negative of the incoming waves Similar to the shorted end of an LC transmission line, where V + (L c w +V (L + c w = 0

6 Reflection at a Free End Imagine that an end of the string is free Let the string move vertically without any force x = L ξ(x) Vertical component of the tension is given by F = T sinθ T ξ(x, x The free end at x = L means F(x = L) = T ξ x x=l F θ = 0 i.e. the string is horizontal Not easy to make this T

7 Reflection at a Free End The boundary condition is It takes a little trick from here ξ + (x c w x 1 ξ + (x c w c w t ξ + (x c w x dξ + (L c w = dξ (L + c w dt dt Integrate with time, and assume ξ + = ξ = 0 at t = x=l x=l = ξ (x + c w x + ξ (x + c w x Reflected waves are the same as the incoming waves x=l = 1 ξ (x + c w c w t ξ + (L c w = ξ (L + c w x=l x=l = 0

8 Mechanical Impedance Reflection of electromagnetic waves on an LC transmission line was determined by the impedance matching We calculated general solutions using R and Z To analyze reflection of mechanical waves, we need mechanical impedance Consider transverse mechanical waves Vertical force is F = T ξ x = T f (x ± c w Vertical velocity is Consider the ratio: v = ξ t = ±c f (x ± c w w F v = ξ(x, = f (x ± c w proportional to each other T ±c w = ± T ρ l This is a constant: Mechanical impedance

9 Mechanical Impedance Mechanical impedance of transverse waves: Z = For longitudinal waves, Z = Kρ l For LC transmission lines Z = L m C 1/ k S The force and the velocity are related by + for forward-going waves F = ±Zv for backward-going waves Unit: [force]/[velocity] N/(m/s) = kg/s T ρ l Let s see how this helps our calculation

10 Connection Between Two Media ρ l1 ρ l2 Consider waves traversing a boundary between two media Heavy string to light string Sound transmission between different materials glass copper At the connection between two media, The force F must be equal on both sides The wave amplitude ξ(x, must be equal on both sides This can be achieved if the velocity v is equal (Assuming it was OK to at the beginning of time)

11 Connection Between Two Media Define forces and velocities for the incoming, reflected, transmitted waves F 1 (x c 1 v 1 (x c 1 F 2 (x c 2 v 2 (x c 2 Z Z 1 2 F r (x + c 1 v r (x + c 1 x = 0 Boundary conditions are v 1 ( c 1 + v r (c 1 = v 2 ( c 2 F 1 ( c 1 + F r (c 1 = F 2 ( c 2 Z 1 v 1 ( c 1 Z 1 v r (c 1 = Z 2 v 2 ( c 2

12 Connection Between Two Media v 1 ( c 1 + v r (c 1 = v 2 ( c 2 Z 1 v 1 ( c 1 Z 1 v r (c 1 = Z 2 v 2 ( c 2 v r (c 1 = Z 1 Z 2 Z 1 + Z 2 v 1 ( c 1 v 2 ( c 2 = 2Z 1 Z 1 + Z 2 v 1 ( c 1 F r (c 1 = Z 1 Z 2 Z 1 + Z 2 F 1 ( c 1 F 2 ( c 2 = 2Z 2 Z 1 + Z 2 F 1 ( c 1 Reflected and transmitted power are P r = Z Z 1 2 Z 1 + Z 2 2 P 1 P 2 = 4Z 1 Z 2 ( Z 1 + Z ) P 2 1 Add up to P 1 2

Air Water Sound waves travel from air into water inside a pipe Mechanical impedance: Z = Kρ l = A M B ρ v Z air Z water = γ air P air ρ vair M Bwater ρ vwater = 1.4 9.8 104 [N/m 2 ] 1.3[kg/m 3 ] 2.

13 Air Water Sound waves travel from air into water inside a pipe Mechanical impedance: Z = Kρ l = A M B ρ v Z air Z water = γ air P air ρ vair M Bwater ρ vwater = [N/m 2 ] 1.3[kg/m 3 ] [N/m 2 ] 10 3 [kg/m 3 ] = Fraction of power transmitted ~99.9% is reflected 4Z air Z water (Z air + Z water ) 2 = Engineering problem for the middle ear Air in the ear canal water in the inner ear Solution 1: Ear drum >> oval window Solution 2: Use levers (ossicles) to reduce velocity and increase force

14 Standing Waves Suppose a sinusoidal wave train ξ = ξ 0 cos(kx ω is being reflected at either a fixed or a free end Let s define the position of the end as x = 0 for simplicity x = 0 Reflected waves are ±ξ 0 cos(kx + ω The sum of the incoming and reflected waves is 2ξ ξ 0 cos(kx ω ± ξ 0 cos(kx + ω = 0 coskx cosωt 2ξ 0 sinkx sinωt free end fixed end

15 Standing Waves ξ 0 coskx cosωt ξ 0 sinkx sinωt free end fixed end λ = 2π / k λ = 2π / k node node node node node antinode antinode antinode antinode antinode Standing waves have nodes and antinodes There are two nodes (antinodes) in one wavelength

16 Standing Waves on String What happens when both ends of a transmission line are fixed, free, or a combination of both? Example: a string stretched between two fixed points Stringed musical instruments (piano, guitar, violin) The standing waves must have nodes on both ends L The wavelength must have particular values that match the length L of the string

17 Frequencies and Wavelengths Standing wave for one fixed end at x = 0 is ξ(x, = ξ 0 sinkx sinωt To satisfy the other fixed end at x = L, ξ(l, = ξ 0 sinkl sinωt = 0 kl = nπ any integer k = nπ L λ = 2π k = 2L n Only particular, discrete, set of frequencies and wavelengths are allowed The lowest-frequency mode is called the fundamental mode Others are called harmonics ω = c w k = nπc w L ν = ω 2π = nc w 2L Greek nu = frequency!

18 Fundamental and Harmonics Fundamental n = 1 2 nd harmonic n = 2 3 rd harmonic n = 3 4 th harmonic n = 4 λ = 2L λ = L λ = 2 3 L λ = 1 2 L ν 0 = c w 2L 2ν 0 3ν 0 4ν 0

19 Stringed Instruments Strings on piano, guitar, etc. oscillates at the fundamental frequency plus its harmonics: ν 0, 2ν 0, 3ν 0, 4ν 0, Fundamental freq. is determined by L and c w c w is determined by T and ρ l ν 0 = c w 2L = 1 2L T ρ l You hear the sum of the fundamental + harmonics, and recognize the whole sound as having a distinct pitch String instruments adjust the pitch of a note by L, T, and ρ l Relative amplitudes of higher harmonics determine timbre, i.e., the character (piano-like, guitar-like) of the sound Without harmonics, everything would sound like a tuning fork

Plucking a String Plucking a string sets up the initial condition ξ(x,t = 0) One can break it into a Fourier series Each term of the Fourier series is a harmonic Where you

20 Plucking a String Plucking a string sets up the initial condition ξ(x,t = 0) One can break it into a Fourier series Each term of the Fourier series is a harmonic Where you pluck in x determines the mixture of the harmonics You hear different timbre As the oscillation decay with time, higher harmonics disappear faster Timbre changes with time

21 Wind Instruments An open end of a pipe acts as a free end for sound Pressure of the air = 1 atm where the pipe ends This is equivalent to having no force a free end Wind instruments (pipe organ, woodwinds, etc.) are basically a pipe with at least one open end The other end is usually closed

22 Wind Instruments k = Standing wave for one free end at x = 0 is To satisfy the other fixed end at x = L, (2n 1)π 2L ξ(x, = ξ 0 cos kx cosωt ξ(l, = ξ 0 coskl cosωt = 0 kl = 2n 1 2 π λ = 2π k = 4L 2n 1 Wavelength of the fundamental is 4L ω = c w k = (2n 1)πc w 2L Only the odd-multiples of the fundamental exist any integer ν = ω 2π = (2n 1)c w 4L

23 Wind Instruments Fundamental n = 1 3 rd harmonic n = 2 5 th harmonic n = 3 7 th harmonic n = 4 λ = 4L λ = 4 3 L λ = 4 5 L λ = 4 7 L ν 0 = c w 4L 3ν 0 5ν 0 7ν 0

24 Wind Instruments Air in wind instruments oscillates at the fundamental frequency plus odd harmonics: n 0, 3n 0, 5n 0, 7n 0, Fundamental frequency is determined by L and c w c w is pretty much constant at normal temperature Pipe organ must be that big to cover the lowest audible frequency ν 0 = c w 4L = 20 L = = 4.125m Lack of even harmonics gives wind instruments their characteristic sound ν 0 = c w 4L

25 Inharmonicity Sound that contains frequencies other than integer multiples are said to be inharmonic Inharmonic sound does not have a recognizable pitch Drums and other percussion instruments are inharmonic String/wind instruments do have small inharmonicity Higher harmonics are not exactly multiples of the fundamental due to small dispersion, i.e., the wave velocity varying with the frequency Example: Strings are not perfectly flexible and resist bending Harmonics have higher frequencies This has interesting effects on tuning Our ears are accustomed to small inharmonicity in music

26 Summary Reflection of mechanical waves Similar to reflection of electromagnetic waves Mechanical impedance is defined by For transverse/longitudinal waves: Useful in analyzing reflection Standing waves Created by reflecting sinusoidal waves Oscillation pattern has nodes and antinodes F = ±Zv Z = [T or K]ρ l Musical instruments use standing waves to produce their distinct sound

PHYSICS 149: Lecture 24

PHYSICS 149: Lecture 24 Chapter 11: Waves 11.8 Reflection and Refraction 11.10 Standing Waves Chapter 12: Sound 12.1 Sound Waves 12.4 Standing Sound Waves Lecture 24 Purdue University, Physics 149 1 ILQ