A 2860 Hz tuning fork is sounded at the same time that a note on a piano is struck. You hear 12 beats/s. You tighten the string slightly and now hear 20 beats/s. What is the frequency of the vibrating piano string now after the tightening?
How would you solve this without the frequency of the piano string when you hear 12 beats/s?? Some help please?
2007-02-06 10:23:21 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Physics
2680/20 or 2680*20.
They are multiples or fractions of the original. It just depends on which one you are comparing to. In your case, It would be the division. I don't know the highest or lowest frequency of piano notes
I can't remember the term off hand, but it is when the peaks of both frequecies are added to form a higher note.
It could be something just simple as that.
2007-02-06 10:33:46 · answer #1 · answered by Anonymous · 0⤊ 0⤋
The frequency of the retightened string is 4667 Hz
2007-02-06 10:28:41 · answer #2 · answered by Ronatnyu 7 · 0⤊ 0⤋
The loudness of a valid with intensity I is 10*log(I/i) dB, the area i is the intensity of a valid it really is 0 dB. Now enable the intensity I be that of the in common words hundred twenty dB sound of the completed type. Then .4I may pick to be the sound intensity of 40% of the variety. Translate those 2 "thoughts" into mathematical equations and we may have one hundred twenty = 10*log(I/i) x = 10*log(.4I/i) note that x is the dB ingredient that we are after. Now we subtract: one hundred twenty - x = 10*log(2.5) x = one hundred twenty - 10*log(2.5) x = one hundred twenty - 4 x = 116. the answer is 116 dB.
2016-11-25 21:01:18 · answer #3 · answered by jandrey 4 · 0⤊ 0⤋