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.
speed of the sound from the siren of the ship
To find the frequency of the lowest note a piccolo can play, we need to determine the wavelength of the sound produced by the resonating air column.
(a) The piccolo is an open-closed pipe, which means that it has one end open and the other end closed. In this type of pipe, the fundamental frequency (lowest note) is given by:
f = v / (2L)
where f is the frequency, v is the velocity of sound in air, and L is the length of the resonating air column.
Given that the overall length of the piccolo is 30.6 cm (or 0.306 m) and the speed of sound in air is 343 m/s, we can substitute these values into the equation:
f = 343 / (2 * 0.306)
f ≈ 562.09 Hz
Therefore, the frequency of the lowest note a piccolo can play is approximately 562.09 Hz.
(b) To find the distance between adjacent antinodes for the highest note a piccolo can play, we need to determine the wavelength of the sound produced by the resonating air column.
In this case, the piccolo is considered an open-open pipe due to the presence of holes in the side, which effectively shortens the length of the resonant column. The wavelength for the highest note (fundamental frequency) in an open-open pipe is given by:
λ = 2L
where λ is the wavelength and L is the length of the resonating air column.
We know that the highest note the piccolo can sound is 41.0 kHz (or 41,000 Hz). We can use the equation:
λ = v / f
where λ is the wavelength, v is the velocity of sound in air, and f is the frequency.
Given that the speed of sound in air is 343 m/s and the frequency is 41.0 kHz (or 41,000 Hz), we can substitute these values into the equation:
λ = 343 / 41000
λ ≈ 0.00837 m
Since the wavelength is equal to twice the distance between adjacent antinodes in an open-open pipe, we divide the wavelength by 2:
Distance between adjacent antinodes = λ / 2
Distance between adjacent antinodes ≈ 0.00418 m
Therefore, the distance between adjacent antinodes for the highest note a piccolo can play is approximately 0.00418 m.