A 5.00 m, 0.732 kg wire is used to support two uniform 235 N posts of equal length (Fig. P15.55), Assume that the wire is essentially horizontal and that the speed of sound is 344 m/s. A strong wind is blowing, causing the wire to vibrate in its 5^th overtone. What are the frequency and wavelength of the sound this wire produces?

Speed of sound \(v = 344\;{\rm{m/s}}\).

The mass of wire \(m = 0.732\;{\rm{kg}}\).

Length of wire \(L = 5.00\;m\)

Tension on the wire \(F = 235\;{\rm{N}}\).

The speed is\(v = \lambda f = \sqrt {\frac{F}{\mu }} \).

Thus, the formula for frequency and wavelength is given by:

\(f = \left( {pth\;overtone + 1} \right) \times velocity\;of\;sound\;in\;air\)

Here, \(F\) is tension, \(\mu \) is linear density, \(\lambda \) is wavelength and \(f\) is frequency.

According to the question,

The frequency is calculated as follows:

\(\begin{array}{c}f = \left( {5 + 1} \right) \times 344\\ = 6 \times 344\\ = 2064\;Hz\end{array}\)

Thus, the wavelength is calculated as follows:

\(\begin{array}{c}v = \lambda f\\344 = \lambda \times 2064\;Hz\\\lambda = \frac{{344}}{{2064}}\\ = \frac{1}{6}\\ = 0.166\;m\end{array}\)

Hence, the frequency is \(2064\;Hz\) and wavelength is \(0.166\;m\).