1) mákes no sense at all. If it said number of waves/second (which is frequency) it would make sense.
2) D is not the answer.
3,4 The formula you cited on 2 is the operative formula. The only trick is signs, you need to review the section in your text on that.
1)As a sound source moves away from a stationary observer, the number of waves will.
decrease
2)How fast should a car move toward you for the car's horn to sound 2.88% higher in frequency than when the car is stationary? The speed of sound is 343 m/s.
fs(v-vd/v-vs)
343(1/1-987/343)
A)4.5 m/s
B)7.3 m/s
C)9.6 m/s
D)11.8 m/s
I'm doing something wrong I got D (well I guessed D)
3)A car moving at 16.0 m/s, passes an observer while its horn is pressed. Find the difference between the frequencies of sound heard when the car approaches and when it recedes from the stationary observer. The velocity of sound is 343 m/s and the frequency of the sound of the car's horn is 583 Hz.
4)A boy is blowing a whistle of frequency 536 Hz and walking toward a wall with a speed of 1.64 m/s. What frequency of the reflected sound will the boy hear if the speed of sound is 343 m/s?
I really don't have a clue on the last 3, I tried but I don't understand
4 answers
1)As the detector recedes from the source, the relative velocity is smaller, resulting in a decrease in the wave crests reaching the detector each second. (thats from the book)
A)increase
B)decrease
C)remain the same
D)need to know the speed of the source
2-4) I know how to use that formula I gave but its not working in this case and I don't know any other ones to use. I read the section but I don't know how to solve these tyypes of questions. They don't give me examples of that
A)increase
B)decrease
C)remain the same
D)need to know the speed of the source
2-4) I know how to use that formula I gave but its not working in this case and I don't know any other ones to use. I read the section but I don't know how to solve these tyypes of questions. They don't give me examples of that
3)this is everything I could come up with to solve.
fs(1/1-vs/v)
583(1/1-343/16)
583(1/1-21.4375)
583(1/-20.4735)
583(-0.048843627)
-28.47583462
583(1/1-16/343)
583(1/1-0.04664723)
583(1/0.95335277)
583(1.048929663)
611.5259937
fs(1/1-vs/v)
583(1/1-343/16)
583(1/1-21.4375)
583(1/-20.4735)
583(-0.048843627)
-28.47583462
583(1/1-16/343)
583(1/1-0.04664723)
583(1/0.95335277)
583(1.048929663)
611.5259937
<<2)How fast should a car move toward you for the car's horn to sound 2.88% higher in frequency than when the car is stationary? The speed of sound is 343 m/s.
fs(v-vd/v-vs)
343(1/1-987/343) >>
Let's take a closer look at your Doopler formula. You are applying it to sound. (For light, a different formula is required.) We need to define terms and write it as an equation. That equation is
fr = fs[(v-vd)/(v-vs)]
fs is the frequency of the source, relative to the medium carrying the waves. We assume that is also the speed with respect to the ground (no wind)
fr is the received frequency.
v is the speed of sound in the air.
vd is the speed of the detector (listener)away from the source
vs is the speed of the source towards the detector.
Now we can put numbers in the equation. vd = 0 in this case.
fr/fs = 1.0288 = 343/(343-vs)
1 - vs/343 = 0.97201
vs/343 = 0.02799
vs = 9.6 m/s
fs(v-vd/v-vs)
343(1/1-987/343) >>
Let's take a closer look at your Doopler formula. You are applying it to sound. (For light, a different formula is required.) We need to define terms and write it as an equation. That equation is
fr = fs[(v-vd)/(v-vs)]
fs is the frequency of the source, relative to the medium carrying the waves. We assume that is also the speed with respect to the ground (no wind)
fr is the received frequency.
v is the speed of sound in the air.
vd is the speed of the detector (listener)away from the source
vs is the speed of the source towards the detector.
Now we can put numbers in the equation. vd = 0 in this case.
fr/fs = 1.0288 = 343/(343-vs)
1 - vs/343 = 0.97201
vs/343 = 0.02799
vs = 9.6 m/s
1 answer