Tippens Capitulo 22

Chapter 22. Sound

Speed of Sound Waves

22-1. Young's modulus for steel is 2.07 x 1011 Pa and its density is 7800 kg/m3. Compute the speed of sound in a steel rod.
[pic]; v = 5150 m/s

22-2. A 3-m length of copper rod has a density 8800 kg/m3, and Young's modulus for copper is 1.17 x 1011 Pa. How much time will it take for sound to travel from one end of the rod to the other?[pic]; v = 3646 m/s
[pic]; t = 1.65 ms

22-3. What is the speed of sound in air (M = 29 g/mol and ( = 1.4) on a day when the temperature is 300C? Use the approximation formula to check this result.
[pic]; v = 349 m/s
[pic] v = 349 m/s

22-4. The speed of longitudinal waves in a certain metal rod of density 7850 kg/m3 is measured to be 3380 m/s. What is theYoung's modulus for the metal?
[pic] Y = ρv2 = (7850 kg/m3)(3380 m/s)2; Y = 8.97 x 1010 Pa
22-5. If the frequency of the waves in Problem 22-4 is 312 Hz, what is the wavelength?
[pic]; λ = 10.8 m

22-6. Compare the theoretical speeds of sound in hydrogen (M = 2.0 g/mol, ( = 1.4) with helium (M = 4.0 g/mol, ( = 1.66) at 00C.
[pic]; vH = 1260 m/s
[pic][pic]; vHe = 971 m/s[pic]; vHe = 0.771 vH

*22-7. A sound wave is sent from a ship to the ocean floor, where it is reflected and returned. If the round trip takes 0.6 s, how deep is the ocean floor? Consider the bulk modulus for sea water to be 2.1 x 109 Pa and its density to be 1030 kg/m3.
[pic]; v = 1428 m/s
h = vt = (1328 m/s)(0.3 s); h = 428 m

Vibrating Air Columns
22-8. Find thefundamental frequency and the first three overtones for a 20-cm pipe at 200C if the pipe is open at both ends. v = 331 m/s + (0.6 )(300) = 343 m/s.
[pic] f1 = 858 Hz
(First overtone, n = 2) fn = nf1; f2 = 2(857.5 Hz) = 1715 Hz
22-8 (Cont.) (2nd overtone, n = 3) fn = nf1; f2 = 3(857.5 Hz) = 2573 Hz
(3rd overtone, n = 4) fn = nf1; f2 = 4(857.5 Hz) = 3430 Hz22-9. Find the fundamental frequency and the first three overtones for a 20-cm pipe at 200C if the pipe is closed at one end.
[pic] f1 = 429 Hz
(First overtone, n = 3) fn = nf1; f2 = 3(429 Hz) = 1290 Hz
(2nd overtone, n = 5) fn = nf1; f2 = 5(429 Hz) = 2140 Hz
(3rd overtone, n = 7) fn = nf1; f2 = 7(429 Hz) = 3000 Hz

22-10. What length of closed pipe will produce afundamental frequency of 256 Hz at 200C?
[pic]; l = 0.335 m; l = 33.5 cm

22-11. What length of open pipe will produce a fundamental frequency of 356 Hz at 200C?
[pic]; l = 0.482 m; l = 48.2 cm

22-12 What length of open pipe will produce a frequency of 1200 Hz as it first overtone on a day when the speed of sound is 340 m/s? [ For open pipe, first overtone is for n = 2 ][pic]; l = 28.3 cm

22-13. The second overtone of a closed pipe is 1200 Hz at 200C. What is the length of the pipe.
The second overtone for a closed pipe occurs when n = 5, and v = 343 m/s.
[pic][pic]; l = 35.7 cm
*22-14. In a resonance experiment, the air in a closed tube of variable length is found to resonate with a tuning fork when the air column is first 6 cm and then 18 cmlong. What is the frequency of the tuning fork if the temperature is 200C? [ v = 343 m/s at 200C. ]
The distance between adjacent nodes of resonance is one-half of a wavelength.
[pic]; f = 1430 Hz

*22-15. A closed pipe and an open pipe are each 3 m long. Compare the wavelength of the fourth overtone for each pipe at 200C. ? (Only odd harmonics allowed for closed pipe.)
For anopen pipe, the fourth overtone is the fifth harmonic, n = 5.
Open (fourth overtone): [pic] λ5 = 1.20 m
For closed pipe, the fourth overtone is the ninth harmonic, n = 9.
Closed (fourth overtone): [pic] λ9 = 1.33 m

Sound Intensity and Intensity Level
22-16. What is the intensity level in decibels of a sound whose intensity is 4 x 10-5 W/m2?
[pic]; β = 76.0 dB

22-17. The...