Two trains with 112-Hz horns approach one another. The slower of the two trains has a speed of 26 m/s. What is the speed of the faster train if an observer standing near the tracks between the trains hears a beat frequency of 4.7 Hz? (Express your answer to the nearest 0.1 m/s.) I tried using the equation fo = fs((1+vo/c)/(1-vs/c)) but I think I may be plugging in the values wrong. Can anyone help me?
9 years later...
To solve this problem, we can use the Doppler effect formula for sound waves. The formula is:
f' = f * (v + vo) / (v - vs)
Where:
f' is the observed frequency
f is the actual frequency emitted by the source
v is the speed of sound in air
vo is the velocity of the observer
vs is the velocity of the source
In this case, the actual frequency emitted by each train is 112 Hz. The speed of sound in air is approximately 343 m/s. The observer is stationary, so vo = 0. The speed of the slower train is given as 26 m/s.
Using this information, we can plug in the values into the formula:
4.7 = 112 * (343 + 0) / (343 - vs)
Now, we can solve for vs, the velocity of the faster train.
4.7 * (343 - vs) = 112 * 343
1621.1 - 4.7vs = 38696
-4.7vs = 38696 - 1621.1
-4.7vs = 37074.9
vs = 37074.9 / (-4.7)
vs ≈ -7874.45 m/s
Since the speed cannot be negative, the speed of the faster train is approximately 7874.45 m/s. Rounding to the nearest 0.1 m/s, we get 7874.4 m/s.
To solve this problem, we can use the Doppler effect equation for sound waves. The equation you mentioned is correct, and I will guide you through the steps to find the speed of the faster train.
First, let's define a few variables:
- fo is the observed frequency of the beat (in this case, 4.7 Hz),
- fs is the frequency of the stationary horn (112 Hz),
- vs is the speed of the slower train (26 m/s),
- vo is the speed of the observer (who is stationary in this case),
- c is the speed of sound (approximately 343 m/s).
Now, let's substitute these values into the equation fo = fs * ((1 + vo/c) / (1 - vs/c)):
4.7 Hz = 112 Hz * ((1 + vo/343) / (1 - 26/343))
Next, let's simplify this equation:
4.7 = 112 * ((1 + vo/343) / (1 - 26/343))
To make calculations easier, let's consider the term (26/343) as a single variable, which we will call A. So,
A = 26/343
The equation now becomes:
4.7 = 112 * ((1 + vo/343) / (1 - A))
Now, let's isolate the term (1 + vo/343) by cross-multiplying:
4.7 * (1 - A) = 112 * (1 + vo/343)
4.7 - 4.7A = 112 + 112vo/343
Rearranging the equation:
112vo/343 = 4.7 - 4.7A - 112
112vo = 4.7 - 4.7A - 112) * 343
vo = ((4.7 - 4.7A - 112) * 343) / 112
Now, let's substitute the value of A:
vo = ((4.7 - 4.7 * (26/343) - 112) * 343) / 112
After performing the calculations, we find that vo ≈ 2.678 m/s.
Since the observer is stationary, the speed of the faster train is equal to the difference between the observer's speed and the slower train's speed:
vfaster = vo + vs
Substituting the values:
vfaster = 2.678 m/s + 26 m/s
Therefore, vfaster ≈ 28.678 m/s.
Rounding to the nearest 0.1 m/s, the speed of the faster train is approximately 28.7 m/s.