When at rest, two trains have sirens that emit a frequency of 300 Hz. The trains travel toward one another and toward an observer stationed between them. One of the trains moves at 30.0 m/s, and the observer hears a beat frequency of 3.6 beats per second. What is the speed of the second train, which travels faster than 30.0 m/s?
To determine the speed of the second train, we can use the formula for the beat frequency (fb):
fb = |f1 - f2|,
where f1 and f2 are the frequencies emitted by the two trains. In this case, the frequency emitted by each train is 300 Hz. Additionally, we know that the beat frequency is given as 3.6 beats per second.
First, let's find the frequency (f2) of the second train when at rest. Since the observer is stationary, the frequency heard by the observer from the second train (f2') is the same as the emitted frequency (f2) due to the absence of a Doppler effect. Therefore, f2' = f2 = 300 Hz.
Next, let's consider the frequency (f1') heard by the observer from the first train. The emitted frequency (f1) by the first train increases due to the Doppler effect. The formula for Doppler effect is:
f' = f(v + ve) / (v + vo),
where f is the emitted frequency, v is the velocity of sound, ve is the velocity of the emitter relative to the medium, and vo is the velocity of the observer relative to the medium.
In this case, the observer is stationary, so vo = 0 and v is the velocity of sound in air. Given that the velocity of sound in air is approximately 343 m/s, we have:
f1' = f1(v + ve) / v.
Plugging in the values, we have:
f1' = 300 (343 + 30) / 343 = 326.42 Hz.
Since the observer hears a beat frequency of 3.6 beats per second, and the beat frequency is the difference between the frequencies, we can set up the equation:
fb = |f1' - f2'| = |326.42 - 300| = 3.6.
Now, let's consider two cases for the velocity (ve) of the second train relative to the observer:
1. If the second train is moving away from the observer, the effective frequency heard by the observer (f2") is given by:
f2" = f2(v - ve) / v.
Plugging in the values, we have:
3.6 = |326.42 - 300(v - ve) / 343|.
2. If the second train is moving towards the observer, the effective frequency heard by the observer (f2") is given by:
f2" = f2(v + ve) / v.
Plugging in the values, we have:
3.6 = |326.42 - 300(v + ve) / 343|.
Now, to solve for ve (the velocity of the second train relative to the observer), we can rearrange the equations and solve for ve. Finally, add ve to the velocity of the first train (30 m/s) to obtain the velocity of the second train.
To find the speed of the second train, we can use the formula for the beat frequency:
Beat frequency = Speed of sound / (Speed of train 1 - Speed of train 2)
We know that the beat frequency is 3.6 beats per second, and the speed of sound is approximately 343 m/s. The speed of train 1 is 30.0 m/s.
Substituting these values into the formula, we can solve for the speed of train 2:
3.6 beats/s = 343 m/s / (30.0 m/s - Speed of train 2)
To isolate the speed of train 2, we can rearrange the formula:
Speed of train 2 = (343 m/s / 3.6 beats/s) - 30.0 m/s
Calculating:
Speed of train 2 = 95.28 m/s - 30.0 m/s
Speed of train 2 = 65.28 m/s
Therefore, the speed of the second train is 65.28 m/s.