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When we talk about overtones, we are discussing the frequencies at which a system, like a string, vibrates naturally. For a string fixed at both ends, its tone series starts with the fundamental frequency. This is the simplest vibration mode.
The 5th overtone, as mentioned in the exercise, actually refers to the 6th harmonic. In musical terms, each overtone corresponds to additional nodes, or points of no motion along the string.

• The fundamental frequency is the first harmonic.
• The first overtone is the second harmonic, with one additional node.
• The 5th overtone is the sixth harmonic, having five additional nodes.

Understanding overtones is crucial because they contribute to the timbre, or quality, of the sound. In our specific exercise, the string vibrates at its 5th overtone, increasing complexity and altering the sound it produces.

To determine the frequency produced by a vibrating string, we rely on a simple relationship involving wave speed and wavelength. The formula is given by:
\[v = f \lambda\]where \(v\) is the wave speed, \(f\) is the frequency, and \(\lambda\) is the wavelength.
For our scenario, the string is vibrating in its 5th overtone, which means it effectively has six complete waves known as wavelengths along its 5 m length. Hence, the wavelength of each is:
\[ \lambda = \frac{5 \text{ m}}{6} = 0.833 \text{ m}\]With the wave speed calculated as 56.74 m/s, we use the formula to find the frequency:

\[f = \frac{v}{\lambda} = \frac{56.74 \text{ m/s}}{0.833 \text{ m}} \approx 68.06 \text{ Hz}\]This frequency is the pitch of the sound wave we hear. It's crucial to note that when dealing with harmonics, larger overtones typically result in higher frequencies, emphasizing the higher pitch of the notes produced.

Wavelength in the context of sound produced by the wire is crucial because it connects with how we hear the sound in the surrounding air. The frequency of the sound wave mirrors the frequency of the wire's vibration.
However, the speed of sound in air differs from string speed. Our exercise provides the speed of sound in air as 344 m/s.
To find the wavelength of the sound in the air, we employ the wave speed formula again, solving for wavelength:
\[ \lambda = \frac{v}{f}\]Substituting the known values, we get:
\[ \lambda = \frac{344 \text{ m/s}}{68.06 \text{ Hz}} \approx 5.05 \text{ m}\]This result describes the distance the sound wave travels in air in one complete cycle. The interplay between frequency and the speed of sound elucidates why the sound depends not only on the vibrating object but also the medium through which it travels. Understanding this helps clarify why different environments can alter what we hear.