A train is traveling west at a velocity of 25 m/s. Another train is traveling east directly toward the west-bound train at a velocity of 15 m/s. The west-bound train blows its whistle with a frequency of 600 Hz when the two trains are 1000 m apart and then blows its whistle again 10 seconds later. For passengers on the east-bound train, how will the perceived frequency of the first whistle compare with the perceived frequency of the second whistle? (Note: Use 350 m/s for the speed of sound in air.)A.It will be 60% higher than the frequency of the second whistle.B.It will be 40% higher than the frequency of the second whistle.C.It will be approximately identical to the frequency of the second whistle.D.It will be 40% lower than the frequency of the second whistle.
Question
A train is traveling west at a velocity of 25 m/s. Another train is traveling east directly toward the west-bound train at a velocity of 15 m/s. The west-bound train blows its whistle with a frequency of 600 Hz when the two trains are 1000 m apart and then blows its whistle again 10 seconds later. For passengers on the east-bound train, how will the perceived frequency of the first whistle compare with the perceived frequency of the second whistle? (Note: Use 350 m/s for the speed of sound in air.)A.It will be 60% higher than the frequency of the second whistle.B.It will be 40% higher than the frequency of the second whistle.C.It will be approximately identical to the frequency of the second whistle.D.It will be 40% lower than the frequency of the second whistle.
Solution
To answer this question, we need to use the Doppler effect formula, which is:
f' = f * (v + vo) / (v + vs)
where: f' is the observed frequency, f is the source frequency, v is the speed of sound in air, vo is the speed of the observer, and vs is the speed of the source.
For the first whistle, the east-bound train (observer) is moving towards the west-bound train (source), so vo is positive and vs is negative. Plugging in the given values:
f'1 = 600 Hz * (350 m/s + 15 m/s) / (350 m/s - 25 m/s) = 645.16 Hz
For the second whistle, the trains have moved closer by (25 m/s + 15 m/s) * 10 s = 400 m, but their speeds haven't changed, so the observed frequency is:
f'2 = 600 Hz * (350 m/s + 15 m/s) / (350 m/s - 25 m/s) = 645.16 Hz
So, the perceived frequency of the first whistle will be approximately identical to the perceived frequency of the second whistle. Therefore, the answer is C. It will be approximately identical to the frequency of the second whistle.
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