An organ pipe that is closed at one end has a fundamental frequency of 146 Hz. There is a leak in the church roof, and some water gets into the bottom of the pipe, as shown in the Figure. The organist then finds that this organ pipe has a frequency of 273 Hz. What is the depth of the water in the pipe?
length of pipe:1/4 wavelength=1/4 speedsound/frequency
have fun calculating.
To find the depth of the water in the pipe, we can use the concept of the effective length of the pipe. The effective length is the distance from the closed end of the pipe to the location of the leak.
First, determine the wavelength of the fundamental frequency when there is no water in the pipe. The fundamental frequency of the pipe with no water is 146 Hz. The wavelength can be found using the formula:
λ = 2L
Where λ is the wavelength and L is the length of the pipe. Since the pipe is closed at one end, the length of the pipe is a quarter of the wavelength:
L = λ/4
Substituting the given frequency of 146 Hz into the equation, we can find the length of the pipe without water.
146 = (343 m/s) / (4 × L)
Solving for L:
L = (343 m/s) / (4 × 146 Hz) ≈ 0.589 m
Next, we calculate the wavelength of the frequency with water in the pipe. The frequency with water is given as 273 Hz. Using the same formula, we find the effective length of the pipe:
L + x = λ/4
Where x is the depth of the water. Solving for the wavelength:
273 = (343 m/s) / (4 × (L + x))
Substituting the value of L from the first calculation:
273 = (343 m/s) / (4 × (0.589 m + x))
Solving for x:
x = ((343 m/s) / (4 × 273 Hz)) - 0.589 m
x ≈ 0.217 m
Therefore, the depth of the water in the pipe is approximately 0.217 meters.