CHAPTER 2 Giancoli: Physics



CHAPTER 12 Giancoli: Physics Study Guide Dr. Lee

Summary:

Speed of Sound at 20 oC 343 m/s

Sound Intensity (I) is measured in units of W/m2

The reference intensity Io is 10-12 W /m2

Intensity level ( is measured in dB

( = 10 log (I/ Io )

Standing wave on a string:

(n = 2 L/n

The wavelength is related to the frequency by v = f (

Doppler Effect:

A source sends out sound of frequency = f

The source is moving towards the observer at speed vS

The observer is moving towards the source at speed vO

The observer will hear a frequency f’ given by:

f’ = f (V + vO) / (V - vS)

where V is the speed of sound.

Change the sign for vO and vS if they are going away.

Solution of some problems CHAPTER 12

(Note: capital X means multiplication)

3. (a) We find the extreme wavelengths from

λ1 = v/f1 = (343 m/s)/(20 Hz) = 17 m;

λ2 = v/f2 = (343 m/s)/(20,000 Hz) = 1.7 × 10–2 m = 1.7 cm.

The range of wavelengths is 1.7 cm ² λ ² 17 m.

(b) We find the wavelength from

λ = v/f = (343 m/s)/(10 × 106 Hz) = 3.4 × 10–5 m.

This is the speed through air. When ultrasound goes through

the body, its speed is higher, about 1540 m/s.

9. We find the intensity of the sound from

β1= 10 log10(I1/I0);

120 dB = 10 log(I1/10–12 W/m2), which gives I1 = 1.0 W/m2.

For a whisper we have

β2 = 10 log10(I2/I0);

20 dB = 10 log(I1/10–12 W/m2), which gives I2 = 1.0 × 10–10 W/m2.

11. We find the ratio of intensities of the sounds from

β = 10 log10(I2/I1);

2.0 dB = 10 log10(I2/I1), which gives I2/I1 = 1.58.

Because the intensity is proportional to the square of the amplitude, we have

I2/I1 = (A2/A1)2;

1.58 = (A2/A1)2, which gives A2/A1 = 1.3.

13. We find the ratio of intensities from

β = 10 log10(Isignal/Inoise);

58 dB = 10 log10(Isignal/Inoise), which gives Isignal/Inoise = 6.3 × 105.

26. The wavelength of the fundamental frequency for a string is λ = 2L, so the speed of a wave on the string is

v = λf = 2(0.32 m)(440 Hz) = 282 m/s.

Note that this is different from speed of sound in air.

We find the tension from

v = [FT/(m/L)]1/2;

282 m/s = {FT/[(0.35 × 10–3 kg)/(0.32 m)]}1/2 , which gives FT = 87 N.

29. (a) The empty soda bottle is approximately a closed pipe with a node at the bottom and an antinode at

the top. The wavelength of the fundamental frequency is λ = 4L. We find the frequency from

v = λf ;

343 m/s = 4(0.15 m)f, which gives f = 572 Hz.

(b) The length of the pipe is now 2/3 of the original length. We find the frequency from

v = λf ;

343 m/s = 4(2/3)(0.15 m)f, which gives f = 858 Hz.

31. (a) For a closed pipe the wavelength of the fundamental frequency is λ1 = 4L. We find the fundamental

frequency from

v = λ1f 1;

343 m/s = 4(1.12 m)f1 , which gives f1 = 76.6 Hz.

Only the odd harmonics are present, so we have

f3 = 3f1 = (3)(76.6 Hz) = 230 Hz;

f5 = 5f1 = (5)(76.6 Hz) = 383 Hz;

f7 = 7f1 = (7)(76.6 Hz) = 536 Hz.

(b) For an open pipe the wavelength of the fundamental frequency is λ1 = 2L. We find the fundamental

frequency from

v = λ1f 1 ;

343 m/s = 2(1.12 m)f1 , which gives f1 = 153 Hz.

All harmonics are present, so we have

f2 = 2f1 = (2)(153 Hz) = 306 Hz;

f3 = 3f1 = (3)(153 Hz) = 459 Hz;

f4 = 4f1 = (4)(153 Hz) = 612 Hz.

51. Use the Doppler effect formula:

f’ = f (V + vO) / (V - vS)

a) Here vS = 0 and vO = + 30

f’ = 1800 X (343 + 30) / (343 + 0 ) = 1957 Hz

b) Here vS = 0 and vO = -30

f’ = 1800 X (343 – 30) / (343 + 0 ) = 1590 Hz

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