SoundPropagation and speed of longitudinal wavePeriodic sound wavesIntensity of periodic sound waves Spherical and plane wavesInterference of sound wavesBeatsDoppler Effect

Pressure Variations in Sound Waves

Speed of Sound

As sound wave passes through air, potential energy is associated with periodic compressions and expansions of small volume elements of the air. Bulk Modulus, B, determines the extent to which an element of a medium changes in volume when the pressure on it changes. B is defined as:

Here DV/V is the fractional change in volume produced by a change in pressure Dp.

But,

and if B replaces t and r replaces m,

This is the speed of sound in a medium with bulk modulus B and density r.

Speed of Sound; Derivation of result

We have:

Also,

And

Therefore,

But

Finally,

The speed of sound waves depends on the compressibility and inertia of the medium. If the medium has a bulk modulus B (see Section 12.4) and density , the speed of sound waves in that medium is

Traveling Sound Waves

Traveling Sound Waves

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Example, Pressure and Displacement Amplitudes

Intensity of Sound

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.

Therefore, I =P/A where P is the time rate of energy transfer (the power) of the sound wave and A is the area of the surface intercepting the sound.

The intensity I is related to the displacement amplitude sm of the sound wave by

Consider a thin slice of air of thickness dx, area A, and mass dm,oscillating back and forth as the sound wave passes through it.The kinetic energy dK of the slice of air is

But,

Therefore,

And,

Then the average rate at which kinetic energy is transported is

If the potential energy is carried along with the wave at this same average rate, then the wave intensity I, the average rate per unit area at which energy of both kinds is transmitted by the wave, is

Spherical and plane waves

Variation of Intensity with Distance:

If a spherical body oscillates so that its radius varies sinusoidally with time, a spherical sound wave is produced.

Power at any distance r from the source must be distributed over a spherical surface of area 4pr 2

Because is the same for any spherical surface centered at the source, we see that the intensities at distances r1 and r 2 are

Therefore, the ratio of intensities

the displacement amplitude smax of a spherical wave must vary as 1/r.

Consider a small portion of a wave front far from the source the rays passing through the wave front are nearly parallel to one another. Any small portion of a spherical wave farfrom its source can be considered a plane wave.

The plane wave function depends only on x and t and has the form:

Interference of sound waves

Phase difference f can be related to path length difference DL, by noting that a phase difference of 2 p rad corresponds to one wavelength.

Therefore,

Fully constructive interference occurs when f is zero, 2p, or any integer multipleof 2p.

Fully destructive interference occurs when f is an odd multiple of p:

Example, Interference:

Example, Interference:

Two identical loudspeakers placed 3.00 m apart are driven by the same oscillator (Fig. 18.5). A listener is originally at point O, located 8.00 m from the center of the line connecting the two speakers. The listener then moves to point P, which is a perpendicular distance 0.350 m from O, and she experiences the first minimum in sound intensity. What is the frequency of the oscillator?Analyze Figure 18.5 shows the physical arrangement of the speakers, along with two shaded right triangles that can be drawn on the basis of the lengths described in the problem. The first minimum occurs when the two waves reaching the listener at point P are 180° out of phase, in other words, when their path difference Dr equals l/2.From the shaded triangles, find the path lengths from the speakers to the listener:

Hence, the path difference is r2 - r1 = 0.13 m. Because this path difference must equal l/2 for the first minimum, l = 0.26 m.To obtain the oscillator frequency, v = lf, where v is the speed of sound in air, 343 m/s:

Example:

BeatsWhen two sound waves whose frequencies are close, but not the same, are superimposed, a striking variation in the intensity of the resultant sound wave is heard. This is the beat phenomenon. The wavering of intensity occurs at a frequency which is the difference between the two combining frequencies.

Example, Beat Frequencies:

Doppler EffectWhen the motion of detector or source is toward the other, the sign on its speed must give an upward shift in frequency. When the motion of detector or source is away from the other, the sign on its speed must give a downward shift in frequency.

Here the emitted frequency is f, the detected frequency f’ , v is the speed of sound through the air, vD is the detector’s speed relative to the air, and vS is the source’s speed relative to the air.

No Motion

Fig. 17-19 The planar wavefronts (a) reach and (b) pass a stationary detector D; they move a distance vt to the right in time t.

In time t, the wavefronts move to the right a distance vt. The number of wavelengths in that distance vt is the number of wavelengths intercepted by D in time t, and that number is vt/l. The rate at which D intercepts wavelengths, which is the frequency f detected by D, is

In this situation, with D stationary, there is no Doppler effect—the frequency detected by D is the frequency emitted by S.

Doppler Effect; Detector Moving, Source Stationary

If D moves in the direction opposite the wavefront velocity, in time t, the wavefronts move to the right a distance vt, but now D moves to the left a distance vDt.

Thus, in this time t, the distance moved by the wavefronts relative to D is vt +vDt. The number of wavelengths in this relative distance vt +vDt is (vt +vDt)/l.

The rate at which D intercepts wavelengths in this situation is the frequency f’ , given by

Similarly, we can find the frequency detected by D if D moves away from the source. In this situation, the wavefronts move a distance vt -vDt relative to D in time t, and f’ is given by

Doppler Effect; Source Moving, Detector StationaryDetector D is stationary with respect to the body of air, and source S move toward D at speed vS.

If T ( =1/f ) is the time between the emission of any pair of successive wavefronts W1 and W2, during T, wavefront W1 moves a distance vT and the source moves a distance vST.

At the end of T, wavefront W2 is emitted.

In the direction in which S moves, the distance between W1 and W2, which is the wavelength l of the waves moving in that direction, is (vT –vST).

If D detects those waves, it detects frequency f given by

In the direction opposite that taken by S, the wavelength l of the waves is again the distance between successive waves but now that distance is (vT vST). D detects frequency f given by

Eg.The siren of a police car at rest emits at a predominant frequency of 1600 Hz. What frequency will you hear if you are at rest and the police car moves at 25.0 m/s (a) toward you, and (b) away from you?

a) 1726 Hz, b) 1491 Hz

Eg. Loud music was being played at a roadside restaurant. A person, driving a car at a steady speed of 30 m/s passes by the restaurant. If the apparent predominant frequency heard by him as he moves toward the restaurant is 1250 Hz, a) Calculate the actual predominant frequency of the music.b) the frequency heard as he moves away from the restaurant.

Eg. A 5000-Hz sound wave is emitted by a stationary source. This sound wave reflects from an object moving toward the source. What is the frequency of the wave reflected by the moving object as detected by a detector at rest near the source?

Ans: First Doppler effect, f’ = 5051 Hz, Second Doppler effect, f’’ = 5103 Hz