You stand by the railroad tracks as a train passes by. You hear a 950-Hz frequency when the train approaches, which changes to 820 Hz as it goes away. How fast is the train moving? The speed of sound in air is 340 m/s.

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BlackTom AI · Accepted Answer

We’re given a Doppler effect scenario with a train (source) moving along a straight track and a stationary observer by the tracks. The observed frequencies are 950 Hz when the train approaches and 820 Hz as it recedes, with the speed of sound in air given as 340 m/s.

Option analysis and calculations will proceed by using the Doppler formulas for a moving source with a stationary observer. When the source moves toward the observer, the observed frequency f_on is f_on = f * v / (v - vs), where f is the emitted frequency, v is the speed of sound, and vs is the source speed. When the source moves away, the observed frequency f_off is f_off = f * v / (v + vs).

First, express the emitted frequency f in terms of the approaching observation: 950 = f * 340 / (340 - vs). Solving for f gives f = 950 * (340 - vs) / 340.

Next, use this expression for f in the receding observation: 820 = [950 * (340 - vs) / 340] * 340 / (340 + vs). The 340 terms cancel, leaving 820 = 950 * (340 - vs) / (340 + vs).

Now solve the equation 820/950 = (340 - vs)/(340 + vs). The left-hand side simplifies to 82/95 ≈ 0.863. Cross-multiplying gives: 95*(340 - vs) = 82*(340 + vs).

Compute both sides: 95*340 = 32,300 and 82*340 = 27,880. So 32,300 - 95 vs = 27,880 + 82 vs. Bring terms together: 32,300 - 27,880 = 95 vs + 82 vs, which simplifies to 4,420 = 177 vs.

Therefore vs = 4,420 / 177 ≈ 24.97 m/s, which rounds to about 25 m/s.

Now evaluate each option:
- 37 m/s: If the train were moving that fast, the ratio (340 - vs)/(340 + vs) would be (303)/(377) ≈ 0.804, corresponding to a much smaller receding frequency relative to the approaching frequency; it would not yield 820 Hz from the given 950 Hz, so this is inconsistent with the observed pair. 
- 10 m/s: This would give (340 - 10)/(340 + 10) = 330/350 ≈ 0.943, which would produce a receding frequency far closer to the emitted one and not drop to 820 Hz from 950 Hz; thus this option is not correct.
- 20 m/s: Here (340 - 20)/(340 + 20) = 320/360 ≈ 0.889, a ratio producing a receding frequency somewhat higher than observed; solving would not land precisely on 820 Hz with the given 950 Hz, so this option is not correct.
- 14 m/s: (340 - 14)/(340 + 14) = 326/354 ≈ 0.921, leading to a receding frequency that is closer to the approaching frequency than 0.863 would indicate; this does not match the 820 Hz observation, so this option is not correct.
- 25 m/s: This value makes the ratio (340 - vs)/(340 + vs) ≈ (315)/(365) ≈ 0.863, which matches the needed relationship 820/950 ≈ 0.863. The algebra above confirms vs ≈ 25 m/s is consistent with both the approaching and receding frequency observations.

In summary, the frequency-change data for approach (950 Hz) and recession (820 Hz) coherently point to a train speed of about 25 m/s, and among the provided options, 25 m/s is the one that aligns with the Doppler calculations.

2025 Fall PHYS 2205 TR 11:00 - 12:15 _ 89122 2205 Final Exam

You stand by the railroad tracks as a train passes by. You hear a 950-Hz frequency when the train approaches, which changes to 820 Hz as it goes away. How fast is the train moving? The speed of sound in air is 340 m/s.

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2025 Fall PHYS 2205 TR 11:00 - 12:15 _ 89122 2205 Final Exam

You stand by the railroad tracks as a train passes by. You hear a 950-Hz frequency when the train approaches, which changes to 820 Hz as it goes away. How fast is the train moving? The speed of sound in air is 340 m/s.

View Explanation

Log in for full answers

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