AP- Waves & Sound. Ch 11-‐ 12 Giancoli. *Answers are given for the interesting
questions ☺. Ch 11. *11. Is a rattle in a car ever a resonance phenomenon?
AP- Waves & Sound. Ch 11-‐ 12 Giancoli
*Answers are given for the interesting questions
Ch 11 *11. Is a rattle in a car ever a resonance phenomenon? Explain. A rattle in a car is very often a resonance phenomenon. The car itself vibrates in many pieces, because there are many periodic motions occurring in the car – wheels rotating, pistons moving up and down, valves opening and closing, transmission gears spinning, driveshaft spinning, etc. There are also vibrations caused by irregularities in the road surface as the car is driven, such as hitting a hole in the road. If there is a loose part, and its natural frequency is close to one of the frequencies already occurring in the car’s normal operation, then that part will have a larger than usual amplitude of oscillation, and it will rattle. This is why some rattles only occur at certain speeds when driving.
*12. Is the frequency of a simple periodic wave equal to the frequency of its source? Why or why not? The frequency of a simple periodic wave is equal to the frequency of its source. The wave is created by the source moving the wave medium that is in contact with the source. If you have one end of a taut string in your hand, and you move your hand with a frequency of 2 Hz, then the end of the string in your hand will be moving at 2 Hz, because it is in contact with your hand. Then those parts of the medium that you are moving exert forces on adjacent parts of the medium and cause them to oscillate. Since those two portions of the medium stay in contact with each other, they also must be moving with the same frequency. That can be repeated all along the medium, and so the entire wave throughout the medium has the same frequency as the source.
*14. Why do the strings used for the lowest-‐frequency notes on a piano normally have wire wrapped around them? The fundamental frequency of oscillation for a string with both ends fixed is
given by
. Combining these two relationships gives
. The speed of waves on the string is
. By wrapping the string with wire, the mass of
the string can be greatly increased without changing the length or the tension of the string, and thus the string has a low fundamental frequency.
41. (II) A cord of mass 0.65 kg is stretched between two supports 28 m apart. If the tension in the cord is 150 N, how long will it take a pulse to travel from one support to the other? 53. (I) A violin string vibrates at 294 Hz when unfingered. At what frequency will it vibrate if it is fingered one-‐ third of the way down from the end? (That is, only two-‐thirds of the string vibrates as a standing wave.)
54. (I) A particular string resonates in four loops (as shown below) at a frequency of 280 Hz. Name at least three other frequencies at which it will resonate.
55. (II) The velocity of waves on a string is are two adjacent nodes?
If the frequency of standing waves is 475 Hz, how far apart
56. (II) If two successive overtones of a vibrating string are 280 Hz and 350 Hz, what is the frequency of the fundamental?
57. (II) A guitar string is 90 cm long and has a mass of 3.6 g. The distance from the bridge to the support post is and the string is under a tension of 520 N. What are the frequencies of the fundamental and first two overtones?
60. (II) The length of the string in the figure below may be adjusted by moving the pulley. If the hanging mass m is fixed at 0.080 kg, how many different standing wave patterns may be achieved by varying L between 10 cm and 1.5 m?
Ch 12 *4. When a sound wave passes from air into water, do you expect the frequency or wavelength to change? If the frequency were to change, the two media could not stay in contact with each other. If one medium vibrates with a certain frequency, and the other medium vibrates with a different frequency, then particles from the two media initially in contact could not stay in contact with each other. But particles must be in contact in order for the wave to be transmitted from one medium to the other, and so the frequency does not change. Since the wave speed changes in passing from air into water, and the frequency does not change, we expect the wavelength to change. The wave travels about four times faster in water, so we expect the wavelength in water to be about four times longer than it is in air.
*5. What evidence can you give that the speed of sound in air does not depend significantly on frequency? Listening to music while seated far away from the source of sound gives evidence that the speed of sound in air does not depend on frequency. If the speed were highly frequency dependent, then high and low sounds created at the same time at the source would arrive at your location at different times, and the music would sound very disjointed. The fact that the music “stays together” is evidence that the speed is independent of frequency.
*6. The voice of a person who has inhaled helium sounds very high-‐pitched. Why? The sound-production anatomy of a person includes various resonating cavities, such as the throat. The relatively fixed geometry of these cavities will determine the relatively fixed wavelengths of sound that a person can produce. Those wavelengths will have associated frequencies given by
. The speed of sound is determined by the gas that is filling
the resonant cavities. If the person has inhaled helium, then the speed of sound will be much higher than normal, since the speed of sound waves in helium is about 3 times that in air. Thus the person’s frequencies will go up about a factor of 3. This is about a 1.5 octave shift, and so the person sounds very high pitched.
*14. Consider the two waves shown in the figure below. Each wave can be thought of as a superposition of two sound waves with slightly different frequencies. In which of the waves, (a) or (b), are the two component frequencies farther apart? Explain. From the two waves shown, it is seen that the frequency of beating is higher in Figure (a) – the beats occur more frequently. The beat frequency is the difference between the two component frequencies, and so since (a) has a higher beat frequency, the component frequencies are further apart in (a).
15. Is there a Doppler shift if the source and observer move in the same direction, with the same velocity? Explain. *17. Figure 12–32 shows various positions of a child in motion on a swing. A monitor is blowing a whistle in front of the child on the ground. At which position, A through E, will the child hear the highest frequency for the sound of the whistle? Explain your reasoning. The highest frequency of sound will be heard at position C, while the child is swinging forward. Assuming the child is moving with SHM, then the highest speed is at the equilibrium point, point C. And to have an increased pitch, the relative motion of the source and detector must be towards each other. The child would also hear the lowest frequency of sound at point C, while swinging backwards.
24. (I) The A string on a violin has a fundamental frequency of 440 Hz. The length of the vibrating portion is 32 cm, and it has a mass of 0.35 g. Under what tension must the string be placed?
25. (I) An organ pipe is 112 cm long. What are the fundamental and first three audible overtones if the pipe is (a) closed at one end, and (b) open at both ends? 32. (II) How far from the mouthpiece of the flute in Example 12–10 should the hole be that must be uncovered to play D above middle C at 294 Hz? 39. (I) A piano tuner hears one beat every 2.0 s when trying to adjust two strings, one of which is sounding 440 Hz. How far off in frequency is the other string?
*41. (I) A certain dog whistle operates at 23.5 kHz, while another (brand X) operates at an unknown frequency. If neither whistle can be heard by humans when played separately, but a shrill whine of frequency 5000 Hz occurs when they are played simultaneously, estimate the operating frequency of brand X. The 5000 Hz shrill whine is the beat frequency generated by the combination of the two sounds. This means that the brand X whistle is either 5000 Hz higher or 5000 Hz lower than the known-frequency whistle. If it were 5000 Hz lower, then it would be in the audible range for humans. Since it cannot be heard by humans, the brand X whistle must be 5000 Hz higher than the known frequency whistle. Thus the brand X frequency is
45. (II) You have three tuning forks, A, B, and C. Fork B has a frequency of 441 Hz; when A and B are sounded together, a beat frequency of 3 Hz is heard. When B and C are sounded together, the beat frequency is 4 Hz. What are the possible frequencies of A and C? What beat frequencies are possible when A and C are sounded together?
