Sound Waves. Solutions of Home Work Problems

Transcription

1 Chapter 17 Sound Waves. s of Home Work Problems 17.1 Problem 17.8 (In the text book An ultrasonic tape measure uses frequencies above 20.0 M Hz to determine dimensions of structures such as buildings. It does this by emitting a pulse of ultrasound into air and then measuring the time for an echo to return from a reflecting surface whose distance away is to be measured. The distance is displayed as a digital read-out. For a tape measure that emits a pulse of ultrasound with a frequency of 22.0 MHz, (a what is the distance to an object from which the echo pulse returns after 24.0 ms when the air temperature is 26.0 C? (b What should be the duration of the emitted pulse if it is to include 10.0 cycles of the ultrasonic wave? (c What is the spatial length of such a pulse? The speed of sound at 0 C is 331 m/s, so at 26 C it is: v = = 346 m/s 273 (a Let t represent the time for the echo to return. Then d = 1 2 vt = = 4.16 m

2 2 CHAPTER 17. SOUND WAVES. SOLUTIONS OF HOME WORK PROBLEMS (b Let t represent the duration of the pulse: t = 10λ v = 10λ fλ = 10 f = 10 = µs (c L = 10λ = 10v f = = mm

3 17.2. PROBLEM (IN THE TEXT BOOK Problem (In the text book A vacuum cleaner produces sound with a measured sound level of 70.0 db. (a What is the intensity of this sound in W/m 2? (b What is the pressure amplitude of the sound? (a Sound level β in db is given by: ( I β = 10 log I where I is the threshold of hearing. So, ( I 70 = 10 log or I = (70.0/10 = W/m (b I = P 2 max 2ρv or P max = 2ρvI = = P a

4 4 CHAPTER 17. SOUND WAVES. SOLUTIONS OF HOME WORK PROBLEMS 17.3 Problem (In the text book Two small speakers emit sound waves of different frequencies. Speaker A has an output of 1.00 mw and speaker B has an output of 1.50 mw. Determine the sound level (in db at point C (see Figure (17.31 if (a only speaker A emits sound, (b only speaker B emits sound, (c both speakers emit sound. Figure 17.31: We presume the speakers broadcast equally in all directions. (a The distance r AC between speaker A and point C is: r AC = = 5.00 m

5 17.3. PROBLEM (IN THE TEXT BOOK 5 The sound intensity I AC at point C due to speak A only, is related to the power of the speaker as: I AC = P A = 4π (5.00 = W/m 2 4πr 2 AC and the the sound level β AC is: β AC = 10 log ( IAC I ( = 10 log = 65.0 db (b Similarly for speaker B, the distance r BC between speaker B and point C is: r BC = = 4.47 m The sound intensity I BC at point C due to speak B only, is related to the power of the speaker as: I BC = P B = 4π (4.47 = W/m 2 4πr 2 BC and the the sound level β BC is: β BC = 10 log ( IBC I ( = 10 log = 67.8 db (c When the two speakers are on simultaneously, the sound intensity I and level β are: I = I AC + I BC = = W/m 2 = 9.15 µw and ( ( I β = 10 log = 10 log = 69.6 db I

6 6 CHAPTER 17. SOUND WAVES. SOLUTIONS OF HOME WORK PROBLEMS 17.4 Problem (In the text book A siren mounted on the roof of a firehouse emits sound at a frequency of 900 Hz. A steady wind is blowing with a speed of 15.0 m/s. Taking the speed of sound in calm air to be 343 m/s, find the wavelength of the sound (a upwind of the siren and (b downwind of the siren. Firefighters are approaching the siren from various directions at 15.0 m/s. What frequency does a firefighter hear (c if he or she is approaching from an upwind position, so that he is moving in the direction in which the wind is blowing? (d if he or she is approaching from a downwind position and moving against the wind? (a Sound moves upwind with speed (v v w, where v is the speed of sound in calm air and v w id wind speed. Crests pass a stationary upwind point at frequency 900 Hz. So the wave length λ is: λ = v v w = = 328 f = m (b Down wind, the sound moves with speed (v + v w and the wavelength becomes: λ = v + v w f = = = m (c We can simplify this part of the problem by making the wind be at rest. This can be done by adding a speed of to the wind in the opposite direction. To keep the total situation the same we should add a speed of in the direction opposite to the wind direction to the fire fighter and the sound source (see the left side of Figure ( The result of speed additions (shown in red in Figure (17.32 is that both the fire fighter and the wind are at rest and the source of sound moving toward the fire fighter with a speed of. The frequency heard by fire fighter is: ( ( v + f vo = f = 900 = 941 Hz v v s

7 17.4. PROBLEM (IN THE TEXT BOOK 7 Wind speed _ + Figure 17.32: (d This situation is shown in the right side of Figure ( The wind is at rest, the source is moving away from the fire fighter with a speed of and the fire fighter is moving toward the source with a speed of 30 m/s, so the frequency heard by the fire fighter in this case is: ( v + f vo = v + v s f = ( = 938 Hz

8 8 CHAPTER 17. SOUND WAVES. SOLUTIONS OF HOME WORK PROBLEMS 17.5 Problem (In the text book A train whistle (f = 400 Hz sounds higher or lower in frequency depending on whether it approaches or recedes. (a Prove that the difference in frequency between the approaching and receding train whistle is 2u/v f = 1 (u 2 /v 2 f where u is the speed of the train and v is the speed of sound. (b Calculate this difference for a train moving at a speed of 130 km/h. Take the speed of sound in air to be 340 m/s. (a Let the frequency of the whistle when the train is at rest be f, when it is approaching be f and when it is receding be f, while v be the speed of sound in air and u be speed of the train: f = v v u f and f = v v + u f The difference between the frequencies f is then: f = f f = fv (b the speed of the train is: ( 1 v u 1 v + u = fv v + u v + u v 2 u 2 = 2uv v 2 [1 (u/v 2 ] f = 2(u/v 1 (u/v 2 f and f becomes: u = 130 km/h = = 36.1 m/s f = 2(u/v 1 (u/v f = 2(36.1/ = = 86.0 Hz 2 1 (36.1/340 2