You are watching a marching band practice outdoors. Near you are two trumpet players, one marching directly towards you and one away from you at the same speed. Each trumpet player is playing an A (440 Hz) and you hear a beat between the two sounds at a frequency of 2 Hz. How fast are the trumpet players marching in m/s?
Details and assumptions
The speed of sound is 340 m/s.
The trumpet players aren't moving very fast.
0.733
it's wrong
the answer is 0.77
@Anonymous is right, the correct given answer is 0.773. Given by nSolve(440/(1 - x/343) = 442, x) / 2
To solve this problem, we need to understand the concept of the beat frequency and how it is related to the relative motion and frequency of the two sources.
When two sound waves of slightly different frequencies interfere, they create a phenomenon known as beats. This is perceived as a fluctuation in the volume or intensity of the combined sound. The beat frequency is the difference between the two frequencies.
In this scenario, we have two trumpet players playing the same note, but one is marching towards you, and the other is marching away from you. Since the speed of sound is constant at 340 m/s, the difference in the frequency of the two sounds you hear is due to the Doppler effect caused by their relative motion.
The beat frequency of 2 Hz represents the difference in frequency between the two trumpet players' sounds. This means that the frequency of the sound you hear from one trumpet player is 440 + 2 = 442 Hz, and the frequency from the other trumpet player is 440 - 2 = 438 Hz.
Now, we can use the Doppler effect equation to relate the difference in frequency to the relative velocity between the sources:
f' = f * (v + v_observer) / (v + v_source)
Where:
f' = observed frequency
f = source frequency
v_observer = velocity of the observer (you)
v_source = velocity of the source (trumpet players)
Plugging in the known values:
f' = 442 Hz (for the trumpet player moving towards you)
f = 440 Hz
v_observer = 0 m/s (since you are stationary)
v_source = unknown (velocity of the trumpet players)
Since the observed frequency f' is slightly higher than the source frequency f, we know that v_source should be positive (indicating that the trumpet player is moving towards you).
Now, we can rearrange the equation to solve for v_source:
v_source = (f' - f) * v / (f' + f)
v_source = (442 - 440) * 340 / (442 + 440)
v_source = 2 * 340 / 882
v_source = 680 / 882
v_source ≈ 0.77 m/s
Therefore, the trumpet player marching towards you is moving at a speed of approximately 0.77 m/s. Since the other trumpet player is moving away from you at the same speed, their marching speed would also be 0.77 m/s.