College Algebra, Chapter 2, Review Exercises, Section Review Exercises, Problem 58

According to the study, the frequency $F$ of a vibrating string under constant tension $T$ is inversely proportional to the length $L$. If a string 12 inches long vibrates 440 times per second, to what length must it be shortened to vibrate 660 times per second?

Since the tension on both strings are equal

$
\begin{equation}
\begin{aligned}
F_1 &= \frac{k}{L_1}\\
\\
k &= F_1 L_1 && \Longleftarrow \text{Equation 1}
\end{aligned}
\end{equation}
$


$
\begin{equation}
\begin{aligned}
F_2 &= \frac{k}{L_2}\\
\\
k &= F_2 L_2 && \Longleftarrow \text{Equation 2}
\end{aligned}
\end{equation}
$


Using equations 1 and 2

$
\begin{equation}
\begin{aligned}
F_1 L_1 &= F_2 L_2\\
\\
\left( 440 \frac{\text{times}}{\text{second}} \right) \left( 12 \text{ inches} \right) &= \left( 660 \frac{\text{times}}{\text{second}} \right) (L_2) && \text{Solve for } L_2\\
\\
L_2 &= \frac{\left( 440 \frac{\text{times}}{\text{second}} \right) }{ \left( 660 \frac{\text{times}}{\text{second}} \right)} \left( 12 \text{ inches} \right) && \text{Cancel out like terms}\\
\\
L_2 &= 8 \text{ inches}
\end{aligned}
\end{equation}
$


It shows that the length must be shortened for about 4 inches to increase the vibration.