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Homework Help: Doppler Effect of a train

  1. Jun 24, 2007 #1
    1. The problem statement, all variables and given/known data

    A radio played on a moving train has frequency of 85 Hz. A second radio played at train station is at rest. What is the beat frequency heard if train approaches the station at 10 m/s?

    2. Relevant equations
    f' = [f \ (1-vsource/vsound)

    I used this equation because the source is moving towards a stationary object.

    3. The attempt at a solution

    f' = 85 \ (1 - 10\343)
    f' = 85\ 0.971
    f' = 88 Hz

    The answer is 2.5. So, I think I'm missing a second equation. However, if the source is moving towards the object I assumed the question was asking what the beat freq. would be heard at the station.
  2. jcsd
  3. Jun 24, 2007 #2


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    Staff Emeritus
    Science Advisor
    Gold Member

    No, you're not missing anything with the math, you're just either not reading the question properly or reading it but not understanding it. The question is what is the BEAT frequency? Just in case it is the latter (lack of understanding), I will explain what beats are.

    When you have two waves at very slightly different frequencies, their superposition will exhibit a very unique phenomenon known as beats. Think about it this way: the two waves start off in phase and therefore interfere constructively. As a result, the overall sound intensity is large. But due to the slight frequency difference between the two, they will not stay in phase. One will start to move slowly out of phase with the other. It will take MANY cycles before the second wave is completely out of phase with the first one, leading to destructive interference and no sound. The second wave will then start to become more and more in phase with the first again. This alternating between loud and soft is called beating. Because it takes many cycles for the superposition of the two waves to go from loud to soft to loud again, the beat frequency is LOW. In fact, the frequency of the beats is given by the DIFFERENCE between the frequencies of the two superposed waves. In this case:

    85/.971 = 87.5

    beat frequency = 87.5 - 85 = 2.5 Hz
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