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

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 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

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

<<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

1) As a sound source moves away from a stationary observer, the number of waves will decrease. This is due to the Doppler effect, which causes a change in the perceived frequency of the sound waves. When the source moves away, the wavelength of the sound waves increases, resulting in a lower frequency and fewer waves reaching the observer per unit of time.

2) To find the speed at which the car should move toward the observer for the car's horn to sound 2.88% higher in frequency, we can use the Doppler effect equation:

f' = fs((v + vd)/(v + vs))

Where f' is the observed frequency, fs is the source frequency, v is the speed of sound, vd is the velocity of the observer, and vs is the velocity of the source.

Given that the observed frequency is 2.88% higher than the source frequency, we can substitute these values into the equation:

(1 + 0.0288) * fs = fs((v + vd)/(v + vs))

Simplifying the equation:

1.0288 * fs = fs((v + vd)/(v + vs))

Cross-multiplying:

1.0288 * fs * (v + vs) = fs * (v + vd)

Canceling out the common fs factor:

1.0288 * (v + vs) = v + vd

Simplifying further:

1.0288v + 1.0288vs = v + vd

Rearranging the equation:

1.0288v - v = vd - 1.0288vs

0.0288v = vd - 1.0288vs

Solving for vd:

vd = 0.0288v + 1.0288vs

Substituting the given values:

vd = 0.0288 * 343 + 1.0288 * 0

vd ≈ 9.888 m/s

Therefore, the speed at which the car should move toward the observer is approximately 9.888 m/s.

3) To find the difference between the frequencies of sound heard when the car approaches and when it recedes from the stationary observer, we can use the Doppler effect equation again:

Δf = fs((v - vd)/(v + vs)) - fs((v + vd)/(v + vs))

Where Δf is the frequency difference.

Given the following values:

fs = 583 Hz (source frequency)
v = 343 m/s (speed of sound)
vd = 16.0 m/s (velocity of the observer, positive when the car approaches)
vs = 0 (velocity of the source, as it is stationary)

Substituting these values into the equation:

Δf = 583 * ((343 - 16)/(343 + 0)) - 583 * ((343 + 16)/(343 + 0))

Simplifying:

Δf = 583 * (327/343) - 583 * (359/343)

Calculating each term:

Δf = 583 * 0.953 - 583 * 1.047

Subtracting the two terms:

Δf ≈ -32.6 Hz

Therefore, the difference between the frequencies of sound heard when the car approaches and when it recedes from the stationary observer is approximately -32.6 Hz. The negative sign indicates a decrease in frequency, which is consistent with the observed shift when the car moves away.

4) To find the frequency of the reflected sound that the boy will hear, we can again use the Doppler effect equation:

f' = fs((v - vr)/(v - vs))

Where f' is the observed frequency, fs is the source frequency (536 Hz), v is the speed of sound (343 m/s), vr is the velocity of the wall (0 m/s, as it is stationary), and vs is the velocity of the source (1.64 m/s, as the boy is approaching the wall).

Given these values, we can substitute them into the equation:

f' = 536 * ((343 - 0)/(343 - 1.64))

Simplifying:

f' = 536 * (343/341.36)

Calculating:

f' ≈ 538.02 Hz

Therefore, the frequency of the reflected sound that the boy will hear is approximately 538.02 Hz.