A tuning fork, which has a frequency of 288 Hz, is resonating over the mouth of a tube closed at one end and which has an inner diameter of 3.50 cm. What is the length of the tube? 13. What is the lowest frequency tuning fork which will resonate over a closed tube which has an inner diameter of 3.00 cm and is 65.0 cm long? An open tube [tube open at both ends!], which has an inner diameter of 4.0 cm and is 38.0 cm long, is placed in front of a speaker which is attached to an audio oscillator. The audio oscillator begins at the lowest possible frequency setting and is slowly turned up to higher frequencies: [A = 2 ( + 0.8d)] Audio Oscillator Tube open at both ends 14. What is the lowest frequency that will cause this open tube to resonate? 15. What is the next higher frequency which will cause this open tube to resonate? This open tube is replaced by a closed tube [a tube closed at one end and open at the other!] which has the same length and inner diameter as the open tube. 16. What is the lowest frequency that will cause this closed tube to resonate? 17. What is the next higher frequency which will cause this closed tube to resonate
The Correct Answer and Explanation is:
Question 12: A tube closed at one end resonates with a 288 Hz tuning fork. The fundamental wavelength in a closed tube fits one quarter of the wave, so:
f=v4L⇒L=v4ff = \frac{v}{4L} \quad \Rightarrow \quad L = \frac{v}{4f}
L=3434×288≈0.297 m=29.7 cmL = \frac{343}{4 \times 288} \approx 0.297 \text{ m} = 29.7 \text{ cm}
Question 13: For a closed tube of 65.0 cm (0.65 m):
f=v4L=3434×0.65≈132 Hzf = \frac{v}{4L} = \frac{343}{4 \times 0.65} \approx 132 \text{ Hz}
Question 14: For an open tube with end correction, use:
λ=2(l+0.8d)\lambda = 2(l + 0.8d)
Here, l=0.38 ml = 0.38 \text{ m}, d=0.04 md = 0.04 \text{ m}, so:
λ=2(0.38+0.032)=2(0.412)=0.824 m\lambda = 2(0.38 + 0.032) = 2(0.412) = 0.824 \text{ m}
f=vλ=3430.824≈416 Hzf = \frac{v}{\lambda} = \frac{343}{0.824} \approx 416 \text{ Hz}
Question 15: The second harmonic of an open tube is just double the fundamental:
f2=2×416=832 Hzf_2 = 2 \times 416 = 832 \text{ Hz}
Question 16: Now the same tube is closed at one end. The fundamental wavelength is:
λ=4(l+0.8d)=4(0.38+0.032)=4(0.412)=1.648 m\lambda = 4(l + 0.8d) = 4(0.38 + 0.032) = 4(0.412) = 1.648 \text{ m}
f=3431.648≈208 Hzf = \frac{343}{1.648} \approx 208 \text{ Hz}
Question 17: Next higher resonance for a closed tube is the third harmonic (odd multiples only):
f3=3×208=624 Hzf_3 = 3 \times 208 = 624 \text{ Hz}
Summary Explanation: Resonance in tubes depends on how standing waves fit within the tube’s boundaries. Closed tubes support odd harmonics only because of the displacement node at the closed end and an antinode at the open end. Open tubes support all harmonics since both ends are antinodes. The diameter of the tube affects the end correction, which accounts for the fact that the wave does not end exactly at the physical opening. This correction is modeled by 0.8d0.8d and is added to the length of the tube before calculating the wavelength. The speed of sound connects frequency and wavelength directly through f=v/λf = v / \lambda. These resonance relationships form the acoustic basis for instruments like flutes and clarinets, making them not only mathematically elegant but musically impactful.
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