A guitar player plucks two strings simultaneously. One string is tuned to a (fundamental) frequency of 381 Hz. It is found that beats are audible, with a beat frequency of 5.54 Hz. What are possible values for the frequency of the other string? (Answer the larger frequency first).
F = 381 +5.54 = 386.54
F = 381 - 5.54 = 375.46
To solve this problem, we need to understand the concept of beats in sound waves. Beats occur when two sound waves of slightly different frequencies interfere with each other.
The beat frequency (fbeat) is the difference between the frequencies of the two sound waves (f1 and f2). Mathematically, it can be written as:
fbeat = |f1 - f2|
Given that the beat frequency is 5.54 Hz, we can now set up an equation using the known frequency of one string (f1 = 381 Hz) and solve for the frequency of the other string (f2).
Recall that we need to answer the larger frequency first, so let's assume the frequency of the other string is f2.
fbeat = |f1 - f2|
5.54 Hz = |381 Hz - f2|
Now, to find the possible values for the frequency of the other string (f2), we need to solve this equation.
1. Add f2 to both sides to isolate the absolute value:
5.54 Hz + f2 = 381 Hz
2. Subtract 5.54 Hz from both sides:
f2 = 381 Hz - 5.54 Hz
Now we can calculate the value of f2:
f2 = 375.46 Hz
Therefore, the possible values for the frequency of the other string are 375.46 Hz and 381 Hz. The larger frequency is 381 Hz.