# Question Video: Determining the Vibration Frequency Required to Produce a Particular Beat Frequency

The middle C hammer of a piano hits two strings, producing beats of frequency 1.50 Hz. One of the strings is tuned to 260.00 Hz. What frequencies could the other string have?

02:39

### Video Transcript

The middle C hammer of a piano hits two strings, producing beats of frequency 1.50 hertz. One of the strings is tuned to 260.00 hertz. What frequencies could the other string have?

We’re told in the statement that the beat frequency of the sounds produced by the two strings is 1.50 hertz; we’ll call that 𝑓 sub 𝐵. We’re also told that one of the strings is tuned to a frequency of 260.00 hertz; we’ll call that 𝑓 sub one. We want to know what are the possible frequencies of the other string; we’ll call those frequencies 𝑓 sub two.

To begin this problem, let’s recall the relationship for the beat frequency. Created by two overlapping frequencies, a beat frequency, 𝑓 sub 𝐵, is created when two nonidentical frequencies interfere with one another. The magnitude of the difference between those two overlapping frequencies is what produces the beat frequency, 𝑓 sub 𝐵.

In our case, 𝑓 sub 𝐵 equaling the magnitude of 𝑓 sub one minus 𝑓 sub two means that there are two possible values for 𝑓 sub two. To find them, let’s plugin for the given values of 𝑓 sub 𝐵 and 𝑓 sub one. 1.50 hertz equals the magnitude of the difference between 260.00 hertz and 𝑓 sub two.

So 𝑓 sub two could either be less than 𝑓 sub one by 1.50 hertz, which would mean that 𝑓 sub two is equal to 260.00 hertz minus 1.50 hertz, or 𝑓 sub two could be greater than 𝑓 sub one by 1.50 hertz. In this case, 𝑓 sub two is equal to the sum rather than the difference between 𝑓 sub one and 𝑓 sub 𝐵.

Calculating these values, we find that 𝑓 sub two can equal 258.50 hertz or it could also equal 261.50 hertz. These are the two possible values of the other string to create a beat frequency of 1.50 hertz.