Question
15. Hollow tube chimes are made of metal and are open at each end. One chine is 0.54 m long.
a) If the speed of sound is 346 m/s, what is the frequency of sound produced by the third resonant length?
b) What would happen to the frequency of sound produced by the third resonant length if the chime were shorter? Explain your reasoning.
a) If the speed of sound is 346 m/s, what is the frequency of sound produced by the third resonant length?
b) What would happen to the frequency of sound produced by the third resonant length if the chime were shorter? Explain your reasoning.
Answers
GPT-4o mini
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.
### 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.