Final answer:
A 1.80-m-long tube closed at one end will produce frequencies of 189 Hz, 567 Hz, and 945 Hz in the audible range at 20.0°C. The correct answer is option A.
Step-by-step explanation:
When a tube is closed at one end, it produces a series of harmonics with only odd multiples of the fundamental frequency present. The formula to calculate the frequencies produced by a closed tube is f = (2n-1)v/4L, where f is the frequency, v is the speed of sound, n is a positive integer representing the mode of resonance, and L is the length of the tube. In this case, the length of the tube is 1.80 m. At 20.0°C, the speed of sound in air is approximately 343 m/s. Using this information, we can calculate the frequencies produced by the tube as follows:
- Fundamental frequency (n=1): f = (2(1)-1)(343 m/s)/(4(1.80 m)) = 189.4 Hz
- Second harmonic (n=2): f = (2(2)-1)(343 m/s)/(4(1.80 m)) = 567.1 Hz
- Third harmonic (n=3): f = (2(3)-1)(343 m/s)/(4(1.80 m)) = 945.7 Hz
Therefore, the frequencies produced by the 1.80-m-long closed tube at 20.0°C are approximately 189 Hz, 567 Hz, and 945 Hz.