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.
A)42.5 Hz
B)54.5 Hz
C)61.8 Hz
D)65.4 Hz
v=16
fs=583
vs=343
(16+0)/16
Im not even gonna post the rest because quite frankly I don't get it
The choices don't even make sense. Answers given are in hz, and the problem asks for velocity. Something is major wrong.
To solve this problem, we can use the concept of the Doppler effect. The Doppler effect is the change in frequency or pitch of a sound wave perceived by an observer due to the relative motion between the source of the sound and the observer. In this case, the observer is stationary while the car is approaching.
The formula for the Doppler effect is:
f' = (v + vo) / (v + vs) * f
Where:
f' is the observed frequency
f is the actual frequency
v is the speed of sound (343 m/s)
vo is the velocity of the observer (0 m/s, assuming the observer is stationary)
vs is the velocity of the source (the car's speed towards the observer)
Now, we can plug in the values given in the question:
f' = (343 + 0) / (343 + vs) * f
f' = 1 / (1 + vs/343) * f
f' is 1.0288 times f (an increase of 2.88%) when the car is moving towards the observer. So we have:
1.0288f = 1 / (1 + vs/343) * f
Now, let's rearrange this equation to solve for the velocity of the source, vs:
1 / (1 + vs/343) = 1.0288
Now we can solve for vs:
1 + vs/343 = 1/1.0288
vs/343 = 1/1.0288 - 1
vs/343 = 0.9712 - 1
vs/343 = -0.0288
vs = -0.0288 * 343
vs ≈ -9.868 m/s
The negative sign indicates that the car is moving towards the observer. However, the question asks for the magnitude of the speed, so we take the absolute value:
|vs| ≈ 9.868 m/s
Therefore, the car should be moving towards the observer at approximately 9.868 m/s for the car's horn to sound 2.88% higher in frequency than when the car is stationary.
Now let's find the answer choice that corresponds to this change in frequency. Plugging the calculated value of vs back into the equation:
f' = (343 + 0) / (343 - 9.868) * f
f' ≈ 583 Hz
Therefore, the observed frequency would be approximately 583 Hz. Comparing this to the stationary frequency (f), we can calculate the percentage change:
Percentage change = (f' - f) / f * 100%
= (583 - f) / f * 100%
Since the percentage change is given as 2.88%, we can set up the equation:
(583 - f) / f = 2.88 / 100
Now, solve for f:
583 - f = (2.88 / 100) * f
583 = (1 + 2.88 / 100) * f
583 = (1 + 0.0288) * f
583 = 1.0288 * f
f ≈ 566.68 Hz
Now we can compare the calculated frequency (566.68 Hz) to the answer choices provided in the question:
A) 42.5 Hz
B) 54.5 Hz
C) 61.8 Hz
D) 65.4 Hz
The closest frequency to 566.68 Hz is 583 Hz, which corresponds to option:
C) 61.8 Hz
Therefore, the correct answer is option C) 61.8 Hz.
Note: The calculation involves some approximation, so the result may vary slightly.