#$&* course Phy 122 3/3 8 pm 016. `Query 14
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Given Solution: `aThe beat frequency is the difference in the frequencies, in this case 277 Hz - 262 Hz = 15 Hz. One ocatave reduces frequency by half, so two octaves lower would give frequencies 1/4 as great. The difference in the frequencies would therefore also be 1/4 as great, resulting in a beat frequency of 1/4 * 15 Hz = 3.75 Hz. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: What frequency is received by a person watching an oncoming ambulance moving at 110 km/h and emitting a steady 800-Hz sound from its siren? The speed of sound on this day is 345 m/s. What frequency does she receive after the ambulance has passed? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: First I am going to convert km/h to m/s. 110 km * (1,000 m / 1 km) = 110,000 m. Now I am going to convert hours to seconds. 1 hour * (3,600 s / 1 hour) = 3,600 s Therefore the speed is 110,000 m / 2,600 s = 42.3 m/s Therefore the frequency received by a person is f` = (345 m/s + 0 m/s) / (345 m/s - 42.3 m/s) * 800 Hz = 911.8 Hz Whenever the ambulance passes, the velocity becomes negative making it f` = (345 m/s + 0 m/s) / (345 m/s - (-42.3 m/s)) * 800 Hz = 712.6 Hz confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The general formula for the Doppler shift is f ' = (c + v_r) / (c - v_s) * f where v_r is the velocity of the receiver, v_s the velocity of the source, and c the speed of sound. f is the frequency of the source, f ' the frequency observed by the receiver. v_r is considered positive if the receiver is moving in the direction of the source, and v_s is considered positive if the source is moving in the direction of the receiver. In this case, the receiver is presumably stationary, while the source (the ambulance) is moving at 110 km/h = 110 * 1000 m / (3600 s) = 30 m/s. The received frequency is thus f ' = (345 m/s + 0) / (345 m/s - 30 m/s) * 800 Hz = 876 Hz. After the ambulance has passed, it is no longer approaching the source and its velocity is negative. In this case we have f ' = (345 m/s + 0) / (345 m/s - (- 30 m/s)) * 800 Hz = 736 Hz. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I got different numbers, but I think I did it correctly, just used a different rounding technique. ------------------------------------------------ Self-critique Rating:3 36. Two eagles fly directly toward one another, the first at 15.0 m/s and the second at 20.0 m/s (both relative to the air). Both screech, the first one emitting a frequency of 3200 Hz and the second one emitting a frequency of 3800 Hz. What frequencies do they receive if the speed of sound is 330 m/s? ********************************************* Question: YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I am going to first look at the first eagle, and say that the second one is the reciever here. `f = (330 m/s + 20 m/s) / (330 m/s - 15 m/s) * 3200 Hz = 3,555.6 Hz Therefore when we flip the receiver and source, we get: `f = (330 m/s + 15 m/s) / (330 m/s - 20 m/s) * 3800 Hz = 4,229 Hz confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The general formula for the Doppler shift is f ' = (c + v_r) / (c - v_s) * f where v_r is the velocity of the receiver, v_s the velocity of the source, and c the speed of sound. f is the frequency of the source, f ' the frequency observed by the receiver. v_r is considered positive if the receiver is moving in the direction of the source, and v_s is considered positive if the source is moving in the direction of the receiver. Let's assume that the first eagle is the source and the second the receiver. In this case, the receiver is approaching the source at 20 m/s, while the source is approaching the receiver at 15 m/s, and the source is emitting frequency 3200 Hz. The received frequency is thus f ' = (330 m/s + 20 m/s) / (330 m/s - 15 m/s) * 3200 Hz = 3556 Hz, approx. In the second case the first eagle is the receiver, the second the source. The source is approaching the receiver at 20 m/s, the receiver approaching the source at 15 m/s, so f ' = (330 m/s + 15 m/s) / (330 m/s - 20 m/s) * 3800 Hz = 4230 Hz, approx. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: