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.
A. 42.2 Hz
B. 54.5 Hz
C. 61.8 Hz
D. 65.4 Hz
I used the Dopler effect formula and got B?
The shift in frequency is, for slow speeds compared to the speed of sound, f*(V/a) = 583*(16/343) = 27.2 Hz , up or down. a is the speed of sound
Double that for the difference between the upshifted and the downshifted frequencies and you get 54.4 Hz
The correct answer is B. Good work.
The exact equation, which I am too lazy to look up, would probably give you 54.5 Hz
thanks!
To solve this problem, we can use the Doppler effect formula for sound:
f' = (v + vd) / (v + vs) * f,
where f' is the observed frequency, v is the velocity of sound, vd is the velocity of the detector (observer), vs is the velocity of the source of sound (car), and f is the frequency of the source (car's horn).
In this case, the car is moving towards the observer and then receding from the observer. So, for the approaching case, vd = 0 m/s, and for the receding case, vd = 0 m/s.
Given:
v = 343 m/s (velocity of sound)
vd = 0 m/s (velocity of the observer)
vs = 16.0 m/s (velocity of the car)
f = 583 Hz (frequency of the car's horn)
For the approaching case:
f' = (v + vd) / (v + vs) * f
Substituting the given values:
f' = (343 + 0) / (343 + 16.0) * 583 Hz
f' = (343) / (359) * 583 Hz
f' = 555.4 Hz
For the receding case:
f' = (v + vd) / (v - vs) * f
Substituting the given values:
f' = (343 + 0) / (343 - 16.0) * 583 Hz
f' = (343) / (327) * 583 Hz
f' = 611.8 Hz
The difference between the frequencies is the absolute difference between the observed frequencies:
Δf = |f' - f'|
Δf = |555.4 Hz - 611.8 Hz|
Δf = 56.4 Hz
Therefore, the correct answer is not among the choices given.
To find the difference between the frequencies of sound heard when the car approaches and when it recedes from the observer, you can use the Doppler effect formula.
The formula for the observed frequency (f') is given by:
f' = f * (v + vo) / (v + vs)
Where:
f' is the observed frequency
f is the actual frequency of the sound source (the car's horn)
v is the velocity of sound (343 m/s)
vo is the velocity of the observer (0 m/s, since the observer is stationary)
vs is the velocity of the source (in this case, the velocity of the car)
When the car approaches the observer, its velocity (vs) is positive. When it recedes, its velocity (vs) is negative.
In this case, the car is moving at 16.0 m/s, so when it approaches, vs = 16.0 m/s, and when it recedes, vs = -16.0 m/s.
Let's calculate the two observed frequencies:
When the car approaches:
f' = 583 * (343 + 0) / (343 + 16)
f' = 583 * 343 / 359
f' = 558.32 Hz
When the car recedes:
f' = 583 * (343 + 0) / (343 - 16)
f' = 583 * 343 / 327
f' = 612.53 Hz
Now, let's find the difference between the frequencies:
Difference = |f' when the car approaches - f' when the car recedes|
Difference = |558.32 Hz - 612.53 Hz|
Difference = 54.21 Hz
Since the question asks for the difference between the frequencies, the answer would be B. 54.5 Hz.
So, your answer of B is correct.