An organ pipe that is closed at one end has a fundamental frequency of 122 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 300 Hz. What is the depth of the water in the pipe?

I can figure out the length of the pipe by 122Hz=(343m/s)/2L which gives a pipe length of 1.4m but how do you figure out how deep the water is?

To find the depth of the water in the pipe, we need to consider the change in frequency caused by the water.

When the pipe is filled with water, the effective length of the pipe will be different, leading to a change in the fundamental frequency.

The change in frequency can be calculated using the formula:
Δf = f₂ - f₁

Where:
Δf is the change in frequency
f₂ is the frequency when the pipe is filled with water (300 Hz)
f₁ is the original fundamental frequency of the pipe (122 Hz)

Now, let's solve for Δf:
Δf = 300 Hz - 122 Hz
Δf = 178 Hz

We know that the speed of sound in air is approximately 343 m/s. Therefore, the original length of the pipe can be calculated as follows:

122 Hz = (343 m/s) / (2L)
2L = (343 m/s) / (122 Hz)
2L ≈ 2.813 meters
L ≈ 1.407 meters

Now, we need to find the new length of the pipe when it is filled with water. Let's assume that the depth of the water in the pipe is x meters.

The closed end of the pipe will now be at the water level. So the length of the pipe, when filled with water, can be calculated as follows:

Length with water = x + 1.407 meters

Since the frequency is inversely proportional to the length of the pipe, we can use this relationship to find x.

122 Hz = (343 m/s) / [2(x + 1.407) meters]

Simplifying the equation:

2(x + 1.407) = (343 m/s) / 122 Hz
2x + 2(1.407) = 2.810

Simplifying further:

2x + 2.814 = 2.810
2x = 2.810 - 2.814
2x = -0.004
x = -0.002 meters

Since depth cannot be negative, the depth of the water in the pipe will be approximately 0.002 meters or 2 millimeters.

To figure out the depth of the water in the pipe, you can use the concept of effective length. The effective length of a closed organ pipe with water inside can be expressed as:

Leff = L + d

Where:
Leff is the effective length of the pipe,
L is the original length of the pipe (1.4m in this case),
d is the depth of the water in the pipe.

Now, you can relate the fundamental frequency of the pipe (300 Hz) to its effective length. The fundamental frequency can be determined using the formula:

f = (v/2) * (1/Leff)

Where:
f is the frequency (300 Hz in this case),
v is the speed of sound in air (343 m/s).

Rearranging the formula to solve for Leff gives:

Leff = (v/2) / f

Substituting the given values:

Leff = (343 m/s / 2) / 300 Hz
Leff ≈ 0.571 m

Now, we can equate the effective length of the pipe to the sum of its original length and the depth of the water:

0.571 m = 1.4 m + d

Rearranging the equation to solve for d gives:

d = 0.571 m - 1.4 m
d ≈ -0.829 m

Since the depth of water cannot be negative, we can conclude that there is no meaningful physical interpretation for a negative depth value in this case. Therefore, it is likely that the given frequencies and lengths are inaccurate or have other factors unaccounted for in the problem.