A 425-Hz tuning fork is sounded together with an out-of-tune guitar string, and a beat frequency of 4 Hz is he

Question

A 425-Hz tuning fork is sounded together with an out-of-tune guitar string, and a beat frequency of 4 Hz is heard. When the string is tightened, the frequency at which it vibrates increases, and the beat frequency is heard to decrease. What was the original frequency of the guitar string?

Answer 1

The beat frequency produced by the interference between two waves of differing frequencies, f1 and f2 are given by the absolute value of the difference between f1 and f2:



f_beat = |f1 - f2|



which means



f_beat = f1 - f2

or

f_beat = -(f1 - f2) = f2 - f1



Here, initially, f_beat = 4Hz and let's say f1 = 425 Hz, so:



4 Hz = 425 Hz - f2

f2 = 421 Hz

or

4Hz = f2 - 425 Hz

f2 = 429 Hz



But we are told that when the string is tightened, and f2 gets larger, f_beat gets smaller. That means that the first of the above possibilities is the correct one, and the original frequency of the string was 421 Hz. (If the original frequency of the string was 429 Hz, the beat frequency would have increased when the string was tightened.)

Answer 2

For the beat note of 1 Hz the string must be either 432 or 434 Hz both of these being 1 Hz from the tuning fork. Only one of these two is ALSO 8 Hz from the 426 Hz fork.

Answer 3

Tuneguitar

Answer 4

A is 440 Hz... everyone knows that. Middle C is 256 Hz... everyone knows that too.