Doppler effect and echo frequency

[sfsn]

"A member of the woodwinds section is late for his concert and is practicing his flute, playing a note that has a frequency of 480 Hz, while riding in the back of the bus traveling at 100 m/s. What is the frequency he hears of the note’s echo from the canyon wall that is 200 m straight ahead on the road?"

The answer is 880 Hz.

Here is the explanation:

Explanation is copied below:
Since the bus is traveling toward the canyon wall, the frequency is increasing, eliminating choices A and B. For the period when the note is traveling toward the canyon wall, the source is traveling toward the detector and the minus sign is used in the denominator of the Doppler equation. The detected frequency by the wall is fdetected = fsource(v / (v – vsource)) = (480 Hz)((340 m/s) / (340 m/s – 100 m/s)) = (480 Hz)(340 / 240) = (2 Hz)(340) = 680 Hz. This frequency is then reflected back to the bus so the source is the wall with the detector moving toward the wall, and the plus sign is used in the numerator of the Doppler equation. The frequency detected by the member is fdetected = fsource((v + vdetector) / v) = (680 Hz)((340 m/s + 100 m/s) / 340 m/s) = (680 Hz)(440 / 340) = (2 Hz)(440) = 880 Hz.

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I kind of get it but mostly don't.

Do we first use the minus sign between the velocities because the frequency observed at the wall is greater than the actual frequency of the note, so to compensate (because the frequency is actually less), we subtract the bus's velocity of 100 m/s from the note's velocity of 340 m/s (340 m/s - 100 m/s)?

And we use the plus sign for the second half, when the note is reflected from the wall back to the bus, because...this is where I get stuck.

Here's a related thread, but I'm still confused:
http://forums.studentdoctor.net/thr...have-to-use-the-doppler-effect-twice.1060874/

Thanks to anyone who can explain this!

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[theonlytycrane]

doppler equation to use:

In the first part of the problem the canyon wall is the detector and the person is the source. So vd = 0 and vs = 100 m/s. v is the speed of sound in air = 340 m/s. I use the mnemonic "top is toward" to figure out whether to use the top or bottom sign in the numerator and denominator. vd = 0 so the numerator = 340 m/s. The person is moving toward the source so you use the negative sign in the denominator to get 680 Hz detected by the canyon wall.

Since it's an echo, you repeat the steps except for now the canyon wall emits a source frequency of 680 Hz and the person is the detector. vd = 100 m/s (in the numerator) with the + sign since the person is moving toward the source and vs = 0.

 

[sfsn]

Thanks, I hadn't seen that mnemonic before (though after doing an online search, I see it's in Kaplan).