A police car, whose siren has a natural frequency of 620Hz, moves away from a wall and towards a stationary observer at a speed of 30km/h.
What is the beat frequency perceived by the observer?
If the observer stands between the car and the wall, what beat frequency does he perceive?
http://ca.answers.yahoo.com/question/index?qid=20110225140613AAXANMW
Well, let me put on my clown shoes and try to answer these questions!
When the police car is moving away from the observer, the observed frequency of the siren decreases. So, we need to find the difference between the natural frequency of the siren (620Hz) and the frequency heard by the observer.
Now, the formula for the observed frequency due to the Doppler effect is:
f' = f * (v + vo) / (v ± vs)
Where:
f' is the observed frequency
f is the natural frequency of the source (620Hz)
v is the speed of sound (which we'll assume is 343m/s)
vo is the velocity of the observer (which is 0m/s since they are stationary)
vs is the velocity of the source (which is 30km/h * 1000m/3600s = 8.33m/s since it's moving away)
So, using this formula, we get:
f' = 620 * (343 + 0) / (343 - 8.33)
≈ 623.1Hz
Therefore, the beat frequency perceived by the observer is approximately 623.1Hz.
Now, if the observer stands between the car and the wall, the situation is a little different. In this case, the police car is moving towards the observer, which means the observed frequency will be higher than the natural frequency of the siren.
Using the same formula as before, we get:
f' = 620 * (343 + 0) / (343 + 8.33)
≈ 616.0Hz
So, when the observer stands between the car and the wall, they perceive a beat frequency of approximately 616.0Hz.
I hope that brings a smile to your face! Let me know if you have any more questions.
To calculate the beat frequency perceived by the observer when the police car moves away from the wall, we need to use the Doppler effect formula:
f' = f * (v + v_obs) / (v + v_source)
Where:
f' is the perceived frequency
f is the actual frequency of the siren (620 Hz)
v is the speed of sound (approximately 343 m/s)
v_obs is the velocity of the observer (0 km/h since they are stationary)
v_source is the velocity of the source (30 km/h since the police car is moving away)
First, let's convert the velocities from km/h to m/s:
v_obs = 0 km/h = 0 m/s
v_source = 30 km/h = 8.333 m/s
Now, we can substitute the values into the formula:
f' = 620 * (343 + 0) / (343 + 8.333)
f' = 620 * 343 / 351.333
f' ≈ 605.52 Hz
Therefore, the beat frequency perceived by the observer when the police car moves away from the wall is approximately 605.52 Hz.
Now let's calculate the beat frequency perceived by the observer when they stand between the car and the wall. In this case, the observer experiences both the approaching and receding siren sound.
To calculate the beat frequency, we use the same formula as before but with the velocity of the source changed to negative:
f' = 620 * (343 + 0) / (343 - 8.333)
f' = 620 * 343 / 334.667
f' ≈ 637.96 Hz
Therefore, the beat frequency perceived by the observer when standing between the car and the wall is approximately 637.96 Hz.
To find the beat frequency perceived by the observer when the police car moves away from a wall, we need to consider the Doppler effect. The Doppler effect is the change in frequency of a wave due to the relative motion between the source of the wave and the observer.
In this case, the police car is the source of the sound wave, and the observer is stationary. The car's siren has a natural frequency (f0) of 620 Hz.
When the police car moves away from the wall, its velocity is directed away from the observer. This increases the effective wavelength of the sound waves reaching the observer. As a result, the perceived frequency of the sound decreases.
To calculate the perceived frequency, we need to use the formula:
f' = (v + vo) / (v - vs) * f0
where:
f' is the perceived frequency
v is the speed of sound in air (approximately 343 m/s)
vo is the speed of the observer (0 m/s since they are stationary)
vs is the velocity of the source (car) moving away from the observer (converted to m/s)
First, let's convert the speed of the car from km/h to m/s:
30 km/h = (30 * 1000 m) / (3600 s) ≈ 8.33 m/s
Next, we can substitute the values into the formula:
f' = (343 m/s + 0 m/s) / (343 m/s - 8.33 m/s) * 620 Hz
Calculating this expression:
f' = (343 m/s / 334.67 m/s) * 620 Hz
f' ≈ 637.17 Hz
Therefore, the beat frequency perceived by the stationary observer when the police car moves away from the wall is approximately 637.17 Hz.
Now, let's consider the scenario where the observer stands between the car and the wall. In this case, the car moves toward the observer, which results in a decrease in the effective wavelength and an increase in the perceived frequency.
Using the same formula as above, we can substitute the values:
vs = -8.33 m/s (negative sign indicates the car is moving towards the observer)
f' = (343 m/s + 0 m/s) / (343 m/s - (-8.33 m/s)) * 620 Hz
Calculating this expression:
f' = (343 m/s / 351.33 m/s) * 620 Hz
f' ≈ 605.24 Hz
Therefore, the beat frequency perceived by the observer when standing between the car and the wall is approximately 605.24 Hz.