An organ pipe that is closed at one end has a fundamental frequency of 128 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 217 Hz. What is the depth of the water in the pipe?

To find the depth of the water in the pipe, we can use the formula for the resonant frequency of a closed-end organ pipe:

f = (2n-1) * (v/4L)

Where:
f is the frequency of the pipe,
n is the harmonic number (1 for the fundamental frequency),
v is the speed of sound in air, and
L is the length of the pipe.

We are given that the fundamental frequency of the pipe is 128 Hz, and the frequency with the water in the pipe is 217 Hz.

Using the formula, we can set up two equations:

For the fundamental frequency:
128 = (2*1-1) * (v/4L)
128 = (v/4L)

For the frequency with the water in the pipe:
217 = (2*1-1) * (v/4(L+x))
217 = (v/4(L+x))

Notice that we introduce a variable 'x' to represent the depth of the water in the pipe.

Now we can solve these two equations simultaneously to find the value of 'x'.

We can rewrite the first equation as:
(v/4L) = 128

And the second equation as:
(v/4(L+x)) = 217

Now we can solve for v/4L and v/4(L+x) respectively.

(v/4L) = 128
Multiply both sides by 4L:
v = 512L

(v/4(L+x)) = 217
Multiply both sides by 4(L+x):
v = 868(L+x)

Now we can equate these two expressions for v:

512L = 868(L+x)

Simplify the equation:
512L = 868L + 868x

Rearrange the equation:
868x = 868L - 512L
868x = 356L

Now divide both sides by 868:
x = (356L)/868

So, the depth of the water in the pipe, 'x', is given by (356L)/868.

To find the exact value of 'x', we need the length of the pipe, 'L'. This information is not provided in the question, so we are unable to determine the exact depth of the water in the pipe without that information.