# Problem: An open organ pipe (i.e., a pipe open at both ends) of length L0 has a fundamental frequency f0.Part A. If the organ pipe is cut in half, what is the new fundamental frequency?A. 4f0 B. 2f0 C. f0 D. f0/2E. f0/4Part B. After being cut in half in Part A, the organ pipe is closed off at one end. What is the new fundamental frequency?A. 4f0 B. 2f0 C. f0 D. f0/2E. f0/4Part C. The air from the pipe in Part B (i.e., the original pipe after being cut in half and closed off at one end) is replaced with helium. (The speed of sound in helium is about three times faster than in air). What is the appropriate new fundamental frequency?A. 3f0 B. 2f0 C. f0 D. f0/2E. f0/3

###### FREE Expert Solution

Part A

Frequency:

$\overline{){\mathbf{f}}{\mathbf{=}}\frac{\mathbf{v}}{\mathbf{2}\mathbf{L}}}$

f is inversely proportional to L

80% (452 ratings) ###### Problem Details

An open organ pipe (i.e., a pipe open at both ends) of length L0 has a fundamental frequency f0.

Part A. If the organ pipe is cut in half, what is the new fundamental frequency?

A. 4f0
B. 2f0
C. f0
D. f0/2
E. f0/4

Part B. After being cut in half in Part A, the organ pipe is closed off at one end. What is the new fundamental frequency?

A. 4f0
B. 2f0
C. f0
D. f0/2
E. f0/4

Part C. The air from the pipe in Part B (i.e., the original pipe after being cut in half and closed off at one end) is replaced with helium. (The speed of sound in helium is about three times faster than in air). What is the appropriate new fundamental frequency?

A. 3f0
B. 2f0
C. f0
D. f0/2
E. f0/3

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