To analyze the hollow tube chimes, we need to consider the physics of resonance in tubes. Hollow tubes that are open at both ends have frequencies determined by their length as follows:
Part (a)
For a hollow tube open at both ends, the resonant wavelengths can be calculated using the formula:
\[ \text{Wavelength} (\lambda_n) = \frac{2L}{n} \]
where \(L\) is the length of the tube and \(n\) is the harmonic number (1 for the fundamental frequency, 2 for the first overtone, 3 for the second overtone, etc.).
For the third resonant length (n = 3):
\[ \lambda_3 = \frac{2L}{3} \]
Substituting \(L = 0.54 , \text{m}\):
\[ \lambda_3 = \frac{2 \times 0.54}{3} = \frac{1.08}{3} = 0.36 , \text{m} \]
Next, we can find the frequency (f) of the sound produced using the speed of sound (v) and the wavelength (λ):
\[ f = \frac{v}{\lambda} \]
Substituting \(v = 346 , \text{m/s}\) and \(\lambda = 0.36 , \text{m}\):
\[ f = \frac{346}{0.36} \approx 961.11 , \text{Hz} \]
So, the frequency of sound produced by the third resonant length is approximately 961.11 Hz.
Part (b)
If the chime were shorter, the frequency produced by the third resonant length would increase.
Reasoning:
- The frequency is inversely proportional to the wavelength; as the length of the tube decreases, the resonant wavelength also decreases.
- Specifically, for the third harmonics (n = 3), if the length \(L\) decreases, then \(\lambda_3 = \frac{2L}{3}\) also decreases, leading to a higher frequency (since \(f = \frac{v}{\lambda}\)).
- Therefore, a shorter chime results in a shorter wavelength for the same speed of sound, which produces a higher frequency of sound.
Conclusion
As the length of the tube decreases, the frequency of the sound produced will increase, resulting in a higher pitch.