Question

A tuning fork of 512 Hz is struck and a second tuning fork of an unknown lower pitch is also struck.

A series of 20 beats is subsequently heard in eight seconds-what is the frequency of the second
tuning fork?

Answer 1

Answer:

F2 = 514.5 Hz, 509.5 Hz


Steps:

We have two-component frequencies, the beat frequency can be calculated as:

Fb = | F2- F1 |

Fb = number of beats per second
F1 = the frequency of the first tuning fork
F2 = the frequency of the second tuning fork



Fb = 20/8
Fb = 2.5

2.5 = | F2 - 512 |
Note we have the absolute value. There will be 2 answers.

We cannot tell the actual frequency of the other fork (between these two values, at least) without additional information. Therefore, our two answers are:

F2 = 514.5 Hz, 509.5 Hz