Piano tuners tune pianos by listening to the beats between the harmonics of two different strings. When properly tuned, the note A should have a frequency of 440 Hz and the note E should be at 659 Hz. c. The tuner starts with the tension in the E string a little low, then tightens it. What is the frequency of the E string when she hears four beats per second?

Video transcript

Hello, fellow physicists today, we're gonna solve the following practice problem together. So first off, let's read the problem and highlight all the key pieces of information that we need to use in order to solve this problem, an orchestral harpist is fine tuning his harp to ensure that the notes are accurately tuned. When in tune, note C should be at a frequency of 523 Hertz and Note G should be at 784 Hertz compared with a properly tuned C string. The harpist finds the G string's tension to be somewhat low. So he tightens the string in small increments until it reaches the desired level, determine the frequency of the G string. When the harpist detects a difference of five beats per second. So that is our end goal. So we're given some multiple choice answers. They're all in the same units of Hertz. Let's read them off to see what our final answer might be. A is 782 B is 484 C is 589 and D is 684. OK. So first off, let us recall that the superposition of two slightly different frequencies produces beats. Let us also recall and use the equations for the third harmonic of the note C. And the second harmonic of the note G is the frequency subscript of three C is equal to three multiplied by F subscript. One C is equal to three multiplied by 523 Hertz, which is equal to 1500 1569 Hertz. OK. And let's make a quick little note here that C equals for the C note is equal to F subscript one C is equal to 523 Hertz. And that the G note is equal to F one G is equal to 784 Hertz. OK? Just to be clear. So let's focus on our frequency for the G note. So F two G is equal to two, F, one G is equal to two multiplied by 784 hurts, which is equal to 100. I mean sorry, 1568 hurts. OK. Thus we can go on to write, let's Boxer helpful note here in blue. OK. So we can go on to write that the beat, the frequency of the beat is equal to the absolute value of F three C minus F two G. And remember anything that we take the absolute value of will be positive. So it'll turn negative to positive. So when the harpist detects the difference of five beats per second, F two G, the second harmonic frequency of the note G can either be F two G is equal to 1000 569 hurts minus five. Hertz is equal to 1564 Hertz. And F one G is equal to 1564 Hertz divided by two which is equal to 782. Hurts. And that F two G equals 1569 shirts plus five Hertz equals 1000 574 Hertz. So F one G equals 1574 Hertz divided by two, which is equal to 787 Hertz. Ok? So therefore the fundal, therefore, the fundamental frequency of the G string will be 782 Hertz because it is lower than 787 Hertz. Awesome. So that means when we go back to look at our multiple choice answers, the correct answer has to be the letter A 782 Hertz. Thank you so much for watching. Hopefully that helped and tonight I can't wait to see you in the next video. Bye.