Two Standing Waves on a Vibrating String

Write the frequency for the n-th and the (n+1)-th harmonic numbers, and combine them to find the length.

For the m-th harmonic, the frequency can be written as:

\begin{equation*}
f_m = \frac{m v}{ 2L},
\end{equation*}

and for the (m+1) harmonic we have:

\begin{equation*}
f_{m+1} = \frac{(m+1) v}{2L}.
\end{equation*}

As a \(2 \times 2\) equation system, solve for \(m\) in the first one and replace it in the second one. After some algebra, we get:

\begin{equation*}
L = \frac{v}{2(f_{m+1} - f_m)},
\end{equation*}

which, with numerical values, becomes:

\begin{equation*}
L = 0.5 \ \text{m}.
\end{equation*}

For a more detailed explanation of any of these steps, click on “Detailed Solution”.

In order to find the length of the string, we should use the equation that relates the standing-wave frequencies with the length of the string, the speed of the waves, and the harmonic number. Both the harmonic number and the length of the string are unknown. Hence, if we write these equations for the two successive frequencies that were given to us, we can solve for these two variables, and in particular for the length of the string.

The standing-wave frequencies \(f_n\) of a string are given by

\begin{equation}
\label{EQ:FN}
f_n = \frac{n v}{2L}, \text{ for } n = 1,2,3,...,
\end{equation}

where \(v\) is the speed of the waves, \(L\) is the length of the string, and \(n\) is a positive integer. In our case, we do not know to which harmonic number do the frequencies that were given to us correspond to, but we know that they are successive. Therefore, if we let \(n = m\) be the harmonic number for the frequency \(f_m = 500\) Hz, the frequency of \(550\) Hz will correspond to harmonic \(m+1\) i.e. \(f_{m+1} = 550\) Hz. Therefore, using eq. \eqref{EQ:FN} for both frequencies gives

\begin{equation}
\label{EQ:FM}
f_m = \frac{m v}{ 2L},
\end{equation}

and

\begin{equation}
\label{EQ:FM1}
f_{m+1} = \frac{(m+1) v}{2L}.
\end{equation}

We need to solve this system of equations for the length of the string \(L\). Let us first solve eq. \eqref{EQ:FM} for m. Multiplying eq. \eqref{EQ:FM} by \(2L\) gives

\begin{equation}
2Lf_m = m v,
\end{equation}

and dividing by \(v\) yields

\begin{equation}
m = \frac{2Lf_m}{v}.
\end{equation}

Now, if we substitute this eq. in eq. \eqref{EQ:FM1} we obtain

\begin{equation}
f_{m+1} = \frac{\left(\frac{2Lf_m}{v} + 1\right)v}{2L}.
\end{equation}

We should solve this equation for \(L\). If we multiply on both sides by \(2L\) we get

\begin{equation}
2Lf_{m+1} = \left(\frac{2Lf_m}{v} + 1\right)v,
= 2Lf_m + v.
\end{equation}

We can rewrite this equation as

\begin{equation}
2Lf_{m+1} - 2Lf_m = v,
\end{equation}

and after factoring out the term \(2L\) we obtain

\begin{equation}
2L(f_{m+1} - f_m) = v.
\end{equation}

Finally, dividing by \(2(f_{m+1} - f_m)\) gives

\begin{equation}
L = \frac{v}{2(f_{m+1} - f_m)}
= \frac{300 \ \frac{\text{m}}{\text{s}}}{2(550 \ \text{s}^{-1}) - (500\ \text{s}^{-1})}
= 0.5 \ \text{m}.
\end{equation}