A cardboard tube of length 35.0cm fits snugly into a second cardboard tube of equal length.The outer tube is capped at one end. The overall length can be varied by moving the inner tube relative to the outer tube. The column of air in the tube is excited by a tuning fork of unknown frequency, placed by the open end. A resonance with the tuning fork frequency, f is heard when the tube has a length of 52.5 cm and again for a length of 67.5cm with no resonances in between. The temperature is 14.2◦C.
What is the frequency
http://hyperphysics.phy-astr.gsu.edu/hbase/Waves/opecol.html
Use the one end closed, the middle figure.
for 20 deg C
L = .675 ----> 126 Hz fundamental (node at closed end, max at open end)
L = .525 ----> 162 Hz fundamental
To find the frequency of the tuning fork, we can use the formula:
v = fλ
Where:
- v is the velocity of sound
- f is the frequency of the tuning fork
- λ is the wavelength of the sound wave
In this case, we can assume that the speed of sound in air is approximately 343 m/s at room temperature.
Let's determine the wavelength of the sound wave for the first resonance length of 52.5 cm (0.525 m):
Wavelength (λ) = 2 * length of the air column = 2 * 0.525 m = 1.05 m
Now, we can rearrange the equation and solve for the frequency:
f = v / λ
f = 343 m/s / 1.05 m
f ≈ 326.66 Hz
Therefore, the frequency of the tuning fork is approximately 326.66 Hz.