The end of a stopped pipe is to be cut off so that the pipe will be open. If the stopped pipe has a total length L, what fraction of L should be cut off so that the fundamental mode of the resulting open pipe has the same frequency as the fifth harmonic (and=5) of the original stopped pipe?

4/10 of L.

Explanation:

A stopped pipe is a pipe with one closed end and one opened end. it is also called a closed pipe.

The fundamental mode of a stopped pipe is also called its fundamental frequency, and is f₁=v/4L.

Where f₁=fundamental frequency, v = velocity of sound, L = Length of pipe.

The fifth harmonic of the stopped pipe f₅ = 5v/4L ... (1)

For an open pipe,

Fundamental mode is also called fundamental frequency f₁₀=v/2l₀ ... (2)

Where f₁₀ = fundamental frequency of a closed pipe, v = velocity of sound and l₀=length of the resulting open pipe.

from the question, the fundamental mode of the resulting open pipe = The fifth harmonic of the original stopped pipe.

∴ f₅=f₁₀

⇒5v/4L = v/2l₀

Equating v from both side of the equation,

⇒ 5/4L = 1/2l₀

Cross multiplying the equation,

5*2l₀ = 4L * 1

10l₀ = 4L

Dividing both side of the equation by the coefficient of l₀ i. e 10

10l₀/10 = 4L/10

∴ l₀ = 4/10 (L)

∴ 4/10 of L must be cut off