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.
(a) To find the frequency of the lowest note a piccolo can play, we need to determine the wavelength of the corresponding sound wave. For a pipe open at both ends, the wavelength is twice the length of the pipe.
Given the overall length of the piccolo, which is 30.6 cm (or 0.306 m), the wavelength is:
Wavelength = 2 * Length = 2 * 0.306 m = 0.612 m
The speed of sound in air is given as 343 m/s. The relationship between wavelength, frequency, and speed of sound is given by the formula:
Speed of sound = Wavelength * Frequency
Rearranging the formula to solve for frequency:
Frequency = Speed of sound / Wavelength
Substituting the given values:
Frequency = 343 m/s / 0.612 m ≈ 560.457 Hz
Therefore, the frequency of the lowest note a piccolo can play is approximately 560.457 Hz.
(b) To find the distance between adjacent antinodes for the highest note a piccolo can sound, we need to determine the wavelength of the corresponding sound wave. The wavelength can be calculated using the formula:
Wavelength = Speed of sound / Frequency
Given that the highest note is 41.0 kHz (or 41,000 Hz), and the speed of sound is 343 m/s, we can substitute these values into the formula:
Wavelength = 343 m/s / 41,000 Hz ≈ 0.008366 m
Since the piccolo is open at both ends, each antinode represents a quarter of a wavelength. Therefore, the distance between adjacent antinodes can be found by:
Distance between adjacent antinodes = Wavelength / 4
Substituting the calculated wavelength:
Distance between adjacent antinodes = 0.008366 m / 4 ≈ 0.0020925 m
Therefore, the distance between adjacent antinodes for this mode of vibration is approximately 0.0020925 m.
To find the frequency of the lowest note a piccolo can play, we need to determine the length of the resonating air column when the fundamental frequency occurs.
(a) The fundamental frequency of a resonating air column that is open at both ends is given by the formula:
f = (v/2L),
where f is the frequency, v is the speed of sound in air, and L is the length of the 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 formula:
f = (343 / 2 * 0.306).
Simplifying the equation:
f = 559.8 Hz.
Therefore, the frequency of the lowest note a piccolo can play is approximately 559.8 Hz.
(b) To find the distance between adjacent antinodes for the highest note a piccolo can sound, we need to determine the length of the resonating air column when this mode of vibration occurs.
The highest note corresponds to the frequency of 41.0 kHz (or 41,000 Hz). Using the same formula as above, but solving for L:
L = v / (2f).
Substituting the given values:
L = 343 / (2 * 41000).
Simplifying the equation:
L = 0.00418 m.
Since the piccolo is open at both ends, the distance between adjacent antinodes is half the wavelength. Therefore:
Distance between adjacent antinodes = L / 2 = 0.00418 / 2 = 0.00209 m.
So, the distance between adjacent antinodes for this mode of vibration in the piccolo is approximately 0.00209 m.