The overall length of a piccolo is 30.6 cm. The resonating air column vibrates as in a pipe that is open at both ends. Assume the speed of sound is 343 m/s. (a) Find the frequency of the lowest note a piccolo can play, assuming the speed of sound in air is 343 m/s. (b) Opening holes in the side effectively shortens the length of the resonant column. If the highest note a piccolo can sound is 41.0 kHz, find the distance between adjacent antinodes for this mode of vibration. Assume the speed of sound is 343 m/s.
To find the frequency of the lowest note a piccolo can play, we need to determine the fundamental frequency of the resonating air column. In a pipe that is open at both ends, the fundamental frequency is given by the formula:
f = v / (2L)
where f is the frequency, v is the speed of sound, and L is the length of the resonating air column.
Given that the piccolo's overall length is 30.6 cm (or 0.306 m) and the speed of sound is 343 m/s, we can substitute these values into the formula to calculate the frequency:
f = 343 / (2 * 0.306)
f ≈ 559.48 Hz
Therefore, the frequency of the lowest note a piccolo can play is approximately 559.48 Hz.
To determine the distance between adjacent antinodes for the mode of vibration corresponding to the highest note a piccolo can sound (41.0 kHz), we need to first find the frequency of this highest note. We can then calculate the wavelength using the formula:
λ = v / f
Again, using the given speed of sound (343 m/s) and the frequency of the highest note (41.0 kHz = 41,000 Hz), we can calculate the wavelength:
λ = 343 / 41000 ≈ 0.0084 m
Since an antinode is formed at each end of the resonating air column, the distance between adjacent antinodes is equal to half the wavelength:
Distance between adjacent antinodes = λ/2 ≈ 0.0042 m
Therefore, the distance between adjacent antinodes for this mode of vibration is approximately 0.0042 meters.