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Chapter 16 Sound and Hearing.

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Presentation on theme: "Chapter 16 Sound and Hearing."— Presentation transcript:

1 Chapter 16 Sound and Hearing

2 Goals for Chapter 16 To describe sound waves in terms of particle displacements or pressure variations To calculate the speed of sound in different materials To calculate sound intensity To find what determines the frequencies of sound from a pipe To study resonance in musical instruments To see what happens when sound waves overlap To investigate the interference of sound waves of slightly different frequencies To learn why motion affects pitch

3 Sound waves Sound is simply any longitudinal wave in a medium.
The audible range of frequency for humans is between about 20 Hz and 20,000 Hz. Ultrasonic sound waves have frequencies above human hearing and infrasonic waves are below. Figure 16.1 at the right shows sinusoidal longitudinal wave.

4 Different ways to describe a sound wave
Sound can be described by a graph of displacement versus position, or by a drawing showing the displacements of individual particles, or by a graph of the pressure fluctuation versus position. The pressure amplitude is pmax = BkA. Here B is bulk modulus, k is wavenumber, A is displacement amplitude. Bulk modulus B = Dp/DV/Vo is a measure of how incompressible a gas is.

5 Amplitude of a sound wave
Example 16.1: A sound wave of moderate loudness has pressure amplitude 3.0 x 10-2 Pa. Find the maximum displacement if the frequency is 1000 Hz. In normal air, the speed of sound is 344 m/s, and the bulk modulus is 1.42 x 105 Pa. k = 2p/l = 2p f/v A = pmax/Bk = pmax v/2pfB A = 1.16 x 10-8 m

6 Perception of sound waves
The harmonic content greatly affects our perception of sound.

7 Speed of sound waves The speed of sound depends on the characteristics of the medium. Table 16.1 gives some examples. The speed of sound:

8 Sound intensity The intensity of a sinusoidal sound wave is proportional to the square of the amplitude, the square of the frequency, and the square of the pressure amplitude. Wave intensity Avg. wave intensity Wave displacement

9 The decibel scale The sound intensity level  is  = (10 dB) log(I/I0). Table 16.2 shows examples for some common sounds.

10 Examples using decibels
For a point source of sound, sound intensity falls as 1/r2 Example 16.9, using Figure below. When you double your distance from a point source of sound, by how much does the sound intensity (in dB) decrease? If r2 = 2r1, then the intensity falls off in the ratio I2/I1 = r12/(2r1)2 =1/4 In the log space of dB, a ratio becomes a difference b2 – b1 = 10 log(1/4) = – 6.02 The intensity drops by 6 dB.

11 Standing sound waves and normal modes
The bottom figure shows displacement nodes and antinodes. A pressure node is always a displace-ment antinode, and a pressure antinode is always a displacement node, as shown in the figure at the right.

12 Organ pipes Organ pipes of different sizes produce tones with different frequencies (bottom figure). The figure at the right shows displacement nodes in two cross-sections of an organ pipe at two instants that are one-half period apart. The blue shading shows pressure variation.

13 Harmonics in an open pipe
An open pipe is open at both ends. For an open pipe n = 2L/n and fn = nv/2L (n = 1, 2, 3, …). Figure below shows some harmonics in an open pipe.

14 Harmonics in a closed pipe
A closed pipe is open at one end and closed at the other end. For a closed pipe n = 4L/n and fn = nv/4L (n = 1, 3, 5, …). Figure below shows some harmonics in a closed pipe. Follow Example

15 Resonance and sound In Figure 16.19(a) at the right, the loudspeaker provides the driving force for the air in the pipe. Part (b) shows the resulting resonance curve of the pipe. Follow Example

16 Interference The difference in the lengths of the paths traveled by the sound determines whether the sound from two sources interferes constructively or destructively, as shown in the figures below.

17 Loudspeaker interference
Example using Figure below. Two loudspeakers A and B are driven by the same amplifier in phase. (a) For what frequencies does constructive interference occur at point P? (b) For what frequencies does destructive interference occur at point P? (a) The difference in distance between AP and BP is d = 201/2 m – 171/2 m = 0.35 m. Constructive interference occurs when the difference in distance is d = 0, l, 2l, 3l, … = nl = nv/f. So the possible frequencies are fn = nv/d. Using the speed of sound in air as 350 m/s, the possible frequencies are fn = n 350 m/s / 0.35 m = 1000n Hz = 1000, 2000, 3000, … Hz. (b) Destructive interference occurs when the difference in distance is d = l/2, 3l/2, 5l/2, … = nl/2 = nv/2f (n = 1,3, 5,…). So the possible frequencies are fn = nv/2d = 500, 1500, 2500… Hz. Interference occurs for two coherent sources at the same frequency.

18 Beats Beats are heard when two tones of slightly different frequency (fa and fb) are sounded together. (See Figure below.) The beat frequency is fbeat = fa – fb. Beats occur for sources at two different frequencies.

19 The Doppler effect The Doppler effect for sound is the shift in frequency when there is motion of the source of sound, the listener, or both. Use Figure below to follow the derivation of the frequency the listener receives. Stationary source, moving listener Moving source, moving listener

20 The Doppler effect and frequencies
Follow Example using Figure below to see how the frequency of the sound is affected. = 350/380 fS (lower freq)

21 A moving listener Follow Example using Figure below to see how the motion of the listener affects the frequency of the sound. = 320/350 fS (lower freq)

22 A moving source and a moving listener
Follow Example using Figure below to see how the motion of both the listener and the source affects the frequency of the sound. = 365/395 fS (lower freq)

23 A double Doppler shift Follow Example 16.18 using Figure 16.33 below.
= 350/320 fS (higher freq) = 380/350 fS (higher freq)

24 Shock waves A “sonic boom” occurs if the source is supersonic.
Figure below shows how shock waves are generated. The angle  is given by sin = v/vS, where v/vS is called the Mach number.

25 A supersonic airplane Follow Example using Figure below.


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