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Wave problem solving.

I was trying to do this problem over on this quiz:

"You are watching a marching band practice outdoors. Near you are two trumpet players, one marching directly towards you and one away from you at the same speed. Each trumpet player is playing an A (440 Hz) and you hear a beat between the two sounds at a frequency of 2 Hz. How fast are the trumpet players marching in m/s?

Details and assumptions

The speed of sound is 340 m/s. The trumpet players aren't moving very fast."

I first made the expression for the shifted wavelengths by thinking about 2 peaks on the original wave and noted that the second peak would be off by the velocity of the source times the period, so I can up with this

W-V*T

W-\(\frac{V}{f}\)

\(\frac{340}{440}\)-\(\frac{V}{440}\)

\(\frac{340-V}{440}\)

I would like to know if I did this correctly.

After that, I got lost (and consequently frustrated; if you can, please give tips on how to deal with that as well.) because I didn't understand what the 2 Hz part means: does it mean that a peak from either waves occurs every half-second or does it mean that the peaks coincide every half-second?

Please tell me.

Note by Guy Alves
1 month ago

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Hey Guy, you were pretty close. I'm going to keep things in symbols so it's clear what the various quantities are referring to.

Like you say the natural wavelength of the signal is shifted up or down depending on whether the trumpet player is moving toward or away from the observer. I find it simpler to think about this in terms of the frequencies, so I'll do that and switch back at the end.

The speed of sound of course depends only on the medium (air) and therefore will not change. It's all about the time \(\Delta T_\textrm{observed}\) between signals arriving at the observer, who will observe the frequency \(f_\textrm{observed}= 1/\Delta T_\textrm{observed}.\)

If the source were stationary relative to the medium, then the observer would hear one cycle every \(T = f^{-1} = c/\lambda\) seconds, but things are more involved for the moving source. If the source is moving at speed \(v_\textrm{source}\) relative to the medium, we can track the point in space at which it starts and finishes one cycle.

Suppose the trumpet is a distance \(\ell\) from the observer when it begins cycle \(n\) (or equivalently, finishes cycle \(n-1\)). This signal will immediately begin traveling toward the observer at speed \(v_\textrm{sound},\) arriving at time \(t_{n-1} = \ell/v_\textrm{sound}.\)

Meanwhile, the trumpet will move a distance \(v_\textrm{source}/f\) before finishing cycle \(n.\) As before, this signal will take off toward the observer, now a distance \(\ell - v_\textrm{source}/f\) away, and arrive at time \(\displaystyle t_n = f^{-1} + \frac{\ell - v_\textrm{source}/f}{v_\textrm{sound}}.\)

According to the observer, the time between these two signals is \[\begin{align}\Delta T_\textrm{observed}&= t_n - t_{n-1} \\ &= f^{-1} - f^{-1} \frac{v_\textrm{source}}{v_\textrm{sound}} \\ &= f^{-1}\frac{v_\textrm{sound} - v_\textrm{source}}{v_\textrm{sound}}\end{align}.\]

Inserting this to the relation from above, we find \[\Delta T_\textrm{observed}^{-1} = f_\textrm{observed}(\textrm{toward}) = f_\textrm{source} \frac{v_\textrm{sound}}{v_\textrm{sound} - v_\textrm{source}}\]

If you do the same analysis for the case when the trumpet is moving away, you find \[f_\textrm{observed}(\textrm{toward}) = f_\textrm{source} \frac{v_\textrm{sound}}{v_\textrm{sound} + v_\textrm{source}}.\]

From there you should be able to find the difference in the frequency of the sound from the approaching and receding trumpeteers, which leads us to beats (below).

Obviously you can get the equivalent result in terms of frequencies by making the replacement \(f\rightarrow v_\textrm{sound}/\lambda.\)

Now, beat phenomena arise when two signals arrive at an observer with different frequencies. Due to superposition, the observer doesn't hear each signal independently, but instead hears their combined signal. If one signal has frequency \(f_A\) and another has \(f_B,\) the audible signal will go from maximum amplitude to minimum amplitude with frequency \(f_\textrm{beat} = f_A - f_B\) as signals A and B go in and out of phase with each other.

If you want to understand beats in more detail, I encourage you to do some research but I'll leave an example with \(\cos 55t\) (frequency \(f_A = 55/2\pi\)), \(\cos 50t\) (frequency \(f_B = 55/2\pi\)), and \(\cos 55t - \cos 50t\) which has beat frequency \(f_\textrm{beat} = f_A - f_B = 5/2\pi.\)

Josh Silverman Staff - 3 weeks, 4 days ago

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Thanks for replying. I'm still trying to understand everything you wrote. The frequency for cos50t should be 50/2π by the way. Just wanted you to know ^_^.

Guy Alves - 2 days, 4 hours ago

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