Chapter 17 - Waves II In this chapter we will study sound waves and concentrate on the following topics: Speed of sound waves Relation between displacement.

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Presentation transcript:

Chapter 17 - Waves II In this chapter we will study sound waves and concentrate on the following topics: Speed of sound waves Relation between displacement and pressure amplitude Interference of sound waves Sound intensity and sound level Sources of Musical sounds Beats The Doppler effect

Sound waves are mechanical longitudinal waves that propagate in solids, liquids, and gases. The figure shows an example of a point source of sound waves. We assume that the surrounding medium is isotropic, that means that sound propagates with the same speed in all directions. The sound wave spreads outward uniformly and the wave fronts are spheres centered at the point source. The single arrows indicate lines perpendicular to the wave fronts called “rays” that point along the direction in which the sound wave propagates. The double arrows indicate the motion of the molecules of the medium in which sound propagates.

The Speed of Sound The speed of any mechanical wave, transverse or longitudinal, depends on both an inertial property of the medium (to store kinetic energy) and an elastic property of the medium (to store potential energy). Thus, we can give the speed of a transverse wave along a stretched string, by writing where τ (for transverse waves) is the tension in the string and μ is the string's linear density.

If the medium is air and the wave is longitudinal, we can guess that the inertial property, μ, corresponds to the volume density, ρ of air. As a sound wave passes through air, potential energy is associated with periodic compressions and expansions of small volume elements of the air.

Bulk modulus The property that determines the extent to which an element of a medium changes in volume when the pressure (force per unit area) on it changes, is called the bulk modulus B and is defined as follows: SI-unit: pascal NB: The negative sign indicates that the volume decreases as the pressure increases. Replace τ with B μ with ρ then:

Note 1: Bulk modulus B is smaller for more compressible media Note 1: Bulk modulus B is smaller for more compressible media. Such media exhibit lower speed of sound. Nota 2: Denser mediums (larger ρ) have a smaller speed for sound.

Travelling Sound Wave

Displacement of particles because of a sound wave For transverse waves the medium’s particles oscillate vertical (y-axis) according to the horizontal (x-axis) motion of the wave. The particles’ displacement are then in the form y(x,t). For longitudinal waves, like sound waves, the particles of the medium, like air, oscillate parallel to the wave’s direction of motion, but x(x,t) is confusing. Dus eerder in die vorm s(x,t):

Δp = pf - pi < 0 → expansion; Δp > 0 → compression As the wave travels, the air pressure varies at any position x sinusiodally as given by: Phase difference = π/2 or 90˚ Δp = pf - pi < 0 → expansion; Δp > 0 → compression

Interference Two point sources S1 and S2 produce sferical sound waves that are in phase. They move past the same point P. In the figure the distance L2 , that the wave travelled from S2 , is longer than the distance L1 that the wave from S1 travelled. The path difference means that the waves are possibly out of phase when they reach P.

Interference The phase difference, φ at P depends on the difference in their path lengths: ΔL = │L2 – L1│ To relate phase difference φ and the difference in path lengths ΔL, we use the fact that a phase difference of 2π rad corresponds to a distance of one wave length. We can write the following ratio: (17-20) (17-21) Derive! Then:

Constructive Interference Takes place when φ = 0, 2π, 4π, 6π, ... Thus for φ = m(2π), for m = 0, 1, 2, ... Phase difference = integral multiples of 2π From eq.17-21 follows: m(2π) = (2π)ΔL/λ m = ΔL/λ = 0, 1, 2, ... (17-23) Thus a path length difference ΔL = 0, λ, 2λ, 3λ, ..., integral multiples of wave length gives constructive interference!

Destructive Interference Takes place when φ = π, 3π, 5π, 7π, ... Thus for φ = (2m+1)π, for m = 0, 1, 2, ... (17-24) Phase difference = odd numbered multiples π From eq.17-21 follows: (2m + 1)π = (2π)ΔL/λ ΔL/λ = 0.5, 1.5, 2.5, ... (17-25) Thus path length difference ΔL = 0.5 λ, 1.5λ, 2.5λ, ..., half-integral multiples of wave length gives destructive interference!

The intensity I of a sound wave at a surface is the average rate per unit area at which energy is transferred by the wave through or onto the surface.

Variation of Intensity with Distance In certain cases echoes can be ignored and can be assumed that the sound source is a point source that produces isotropic sound— that is sound with the same intensity in all directions. The intensity I can be given by: where 4πr2 is the area of a sphere. The Eq. indicates that the intensity of sound from an isotropic point source decreases with the square of the distance r from the source.

The Decibel Scale The displacement amplitude at the human ear ranges from about 10-5 m for the loudest tolerable sound to about 10-11 m for the faintest detectable sound, a ratio of 106. The intensity of a sound varies as the square of its amplitude, so the ratio of intensities at these two limits of the human auditory system is 1012. Humans can hear over an enormous range of intensities. Instead of speaking of the intensity I of a sound wave, it is much more convenient to speak of its sound level β, defined as: Lower than lowest range of human ear = 10-12W/m2 dB = desibel, the unit of sound level β increases with 10 dB each time the sound intensity increases with a factor 10. Standard reference level corresponds to zero decibels.

Do Sample Problems 17-4 & 17-5

Pipes with two open ends Standing waves have antinodes (maximum displacement amplitude) at both open ends. Two antinodes are always separated by a node (minimum / zero displacement) and a distance of λ/2. Fig. b shows the “fundamental mode” or “first harmonic” for the pipe. The length of pipe, L = λ/2, thus λ = 2L and f = v/λ = v/2L

(17-39) Derive!

Only odd numbered harmonics! Corresponding Frequencies

Do Sample Problem 17-5&6

Beats (17–12)

Beats (17-43)

Beats But because ω1 & ω2 are almost equal, ω >> ω’. (17-44) But because ω1 & ω2 are almost equal, ω >> ω’. Then Eq.17-45 becomes a cosine function with angular frequency, ω, and amplitude of the term in brackets’ absolute value (which varies with angular frequency, ω’). Amplitude reaches max twice , when cos function = -1 or +1.

The Doppler effect The phenomena where a change in frequency is detected when a sound source and detector move relative to each other. We will investigate sound waves in air as medium. The speed of the source S and the detector D are used. We assume that S and D move directly towards or away from each other against speeds much smaller than the speed of sound in air. Assume that the frequency of the source is equal to f and the frequency that the detector hears is f’.

If the detector D or the source S (or both) move relative to each other, the relationship between the original frequency f and the detected frequency f´ is : Derive! v = the speed of sound in air vD = the detector’s speed (relative to air) vS = the speed of the source (relative to air) The following rule can be used to determine if the plus or minus signs must be used: NB When the relative motion of source and detector is towards each other, the answer must give a higher frequency. If the two move away from each other, the answer must give a lower frequency.

Detector & Source stationary We start our investigation with both the detector & source stationary: In time t wave fronts move a distance, vt Number of waves that pass D in time t is, vt/λ Frequency of waves detected by D: (17-48) No Doppler effect!

Detector moves, Source stationary Wave moves in time t a distance vt right, while D moves a distance vDt left. D towards Source. Wave fronts move in the same time, t, relative to D, with a speed v + vD and a distance, vt + vDt. No. of waves past D: Thus detected frequency at D is (17-49) From eq.17-48, follows λ=v/f, thus: (17-50) f‘ > f

Detector moves, Source stationary If D moves away from source, to right, a distance vDt follows: Wave fronts move in the same time, t, relative to D, a distance, vt - vDt. Thus detected frequency at D is (17-51) f‘ < f Thus follows: (17-52)

Source moves, Detector stationary Detector D, is stationary, and source S, moves towards D with speed vS. In time T, wave front W1, moves a distance vT, while the source moves a distance vST. Speed of sound wave relative to detector D is thus: v - vS Distance wave travelled in time T,: λ’ = (v - vS )T Frequency detected by D: f‘ > f (17-53)

Source moves, Detector stationary Detector D, is stationary, and source S, moves away from D with speed vS. Now the speed of sound wave relative to detector D: v + vS Distance wave travelled in time T,: λ’ = (v + vS )T Frequency now detected by D: (17-54) f‘ < f Eq.’s 17-53 &17-54 together gives: (17-55)

Both source & detector move relative to each other What to do if both source & detector move, but at different speeds relative to each other? Choose the direction of the sound waves that always move from source towards detector, as positive, +. Determine the corresponding directions of the source and detector with regards to the direction of the sound waves: Easiest way is to use the following formula:

Doppler effect + - - + + + - - Sound right :+ For all situations: vS vD Doppler effect Sound right :+ + - - + For all situations: + + - - Do Sample Problems 17-5,17-6 & 17-8