A flute player hears four beats per second when she compares her note to a 523 Hz tuning fork (the note C). She can match the frequency of the tuning fork by pulling out the 'tuning joint' to lengthen her flute slightly. What was her initial frequency?

Video transcript

Hey, everyone. So this problem is working with beat frequency. Let's see what it's asking us. A saxophonist detects six beats per second when playing a note through their saxophone and comparing it with a Hertz chime through adjustments in the position of the mouth piece. Specifically pulling it out. The saxophonist is able to align the frequency with the chime, determine the initial frequency of the saxophonist's played note. Our multiple choice answers here are a 822 Hertz B 560 Hertz C 870 Hertz or D 709 8 Hertz. OK. So the key to this problem is recognizing or recall our beat frequency equation. So that's F beat equals F one minus F two. Now, we are looking to solve either F one or F two. So we're gonna have to solve both of these equations um To figure out what that initial frequency is because we don't know if the um frequency went up or down just yet. So F one is equal to that frequency of the beats. That's six Hertz plus F two, which in this case would be 784 Hertz and that equals 790 Hertz or maybe we're solving for F two, which is, that's one or 784 Hertz minus F B of six Hertz and that's 778 Hertz. So the key to figuring out what the initial frequency is is how the saxophonist was able to adjust the mouthpiece to match the frequency. So when the mouthpiece is pulled out, its wavelength is increased wavelength and frequency have inversely proportional relationships, which means that the frequency decreased, which means that initially the frequency was too high. And so if we look at a frequency that was too high by six feats and then it was decreased to match 784 Hertz. That is how we get the correct answer 790 Hertz. And that aligns with answer choice D. So that's all we've got for this one. See you in the next video.