A light-rail commuter train blows its 200 Hz horn as it approaches a crossing. The speed of sound is 336 m/s.
a)An observer waiting at the crossing receives a frequency of 205 Hz. What is the speed of the train?
Fr = ((Vs+Vr)/(Vs-Vo))*Fo = 205 Hz
((336+0)/(336-Vo))*200 = 205
((336)/(336-Vo))*200 = 205
67,200)/(336-Vo) = 205
205(336-Vo) = 67200
68,880-205Vo = 67200
-205Vo = 67200-68880 = -1680
Vo = 8.2 m/s. = Speed of the train.
To solve this problem, we can use the Doppler effect equation:
f = (v + vo) / (v + vs) * fs
Where:
- f is the observed frequency
- vo is the velocity of the observer
- vs is the velocity of the source (train in this case)
- fs is the source frequency
In this case, f = 205 Hz, fs = 200 Hz, and the speed of sound (v) is 336 m/s.
We need to find the velocity of the train (vs). The observer is not moving (vo = 0), so the equation becomes:
205 Hz = 336 m/s / (336 m/s + vs) * 200 Hz
To solve for vs, we can rearrange the equation:
205 Hz * (336 m/s + vs) = 336 m/s * 200 Hz
Now we can solve for vs:
68880 Hz m/s + 205 Hz * vs = 67200 Hz m/s
205 Hz * vs = -1680 Hz m/s
vs = -1680 Hz m/s / 205 Hz
vs = -8.2 m/s
Since the speed cannot be negative, we take the magnitude of the velocity:
vs = 8.2 m/s
Therefore, the speed of the train is 8.2 m/s.
To determine the speed of the train, we can use the Doppler effect equation:
f' = f * (v+vo) / (v-vs)
Where:
- f' is the observed frequency
- f is the actual frequency of the sound
- v is the speed of sound
- vo is the velocity of the observer (in this case, 0 as the observer is stationary)
- vs is the velocity of the source (in this case, the velocity of the train)
Let's plug in the known values:
f' = 205 Hz
f = 200 Hz
v = 336 m/s
vo = 0 m/s
205 Hz = 200 Hz * (336 m/s + 0 m/s) / (336 m/s - vs)
Now, let's solve for vs:
205 Hz = 200 Hz * 336 m/s / (336 m/s - vs)
To isolate vs, we can cross-multiply:
205 Hz * (336 m/s - vs) = 200 Hz * 336 m/s
Next, we distribute and simplify:
68910 - 205 vs = 67200
Now, let's solve for vs:
205 vs = 68910 - 67200
205 vs = 1710
vs = 1710 / 205
vs ≈ 8.37 m/s
Therefore, the speed of the train is approximately 8.37 m/s.