Question

How does the frequency of a guitar string change when its length is increased from 0.8 meters to 0.9 meters?

Answer 1

The frequency of a vibrating string is inversely proportional to its length. This relationship can be described by the formula for the fundamental frequency of a vibrating string, which is given by:

\[ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]

where:
- \( f \) is the frequency,
- \( L \) is the length of the string,
- \( T \) is the tension in the string, and
- \( \mu \) is the linear mass density (mass per unit length) of the string.

From this relationship, we can see that if the length \( L \) of the string increases, the frequency \( f \) decreases because they are inversely related.

To compare the frequencies at two different lengths:

1. Let \( L_1 = 0.8 \) meters and \( L_2 = 0.9 \) meters.
2. The ratio of the frequencies \( f_1 \) and \( f_2 \) at lengths \( L_1 \) and \( L_2 \) is given by:

\[
\frac{f_1}{f_2} = \frac{L_2}{L_1}
\]

Substituting the lengths:

\[
\frac{f_1}{f_2} = \frac{0.9}{0.8} = 1.125
\]

This means:

\[
f_2 = \frac{f_1}{1.125}
\]

Thus, when the length of the string increases from 0.8 meters to 0.9 meters, its frequency decreases, and specifically, the new frequency \( f_2 \) will be approximately 0.889 times the original frequency \( f_1 \). Hence, the frequency decreases when the length of the string is increased.