Question

An open-open organ pipe is 73.5 cm long. An open-closed pipe has a fundamental frequency equal to the third harmonic of the open-open pipe. How long is the open closed pipe?

Answer 1

Answer:

Length of the open closed pipe is 196cm

Explanation:

the fundamental frequency of an open-closed

f1 = v /4L

f1 = v/4x(73.5)

and the third harmonics of an open-open pipe f2 = 2v/3L

since from the above statement in the question makes the Fundamental frequency of an open-closed pipe equal the frequency of third harmonics of an open open pipe

Hence,

f1 = f2

v/294 = 2v/3L

L = (294 x 2)/3 = 196cm

The length of the close opened pipe is 196cm

Answer 2

Answer:

The length of the  open closed pipe is 12.25 cm

Explanation:

In open - open organ pipe, third harmonic has Antinode to Node, Node to Node, Node to Node and Node to Antinode.

The length of the open-open organ pipe is equal to the sum of wavelength in Antinode to Node, Node to Node, Node to Node and Node to Antinode.

L = A→N + N→N + N→N + N→A

$L =\frac{\lambda }{4} + \frac{\lambda }{2} + \frac{\lambda }{2} + \frac{\lambda }{4} \\\\L = \frac{ 3\lambda }{2}\\\\\lambda = \frac{2L}{3} \\\\F = \frac{V}{\lambda} = \frac{3V}{2L}$

In open-closed pipe, Fundamental frequency has Antinode to Node.

Thus, length of the open-closed organ pipe is equal to the wavelength in Antinode to Node.

L = A→N

$L_o = \frac{\lambda}{4} \\\\\lambda =4L_o\\\\F_o = \frac{V}{4L_o}$

From the information in the question, fundamental frequency of open-closed pipe is to the third harmonic of the open-open pipe.

F₀ = F

$\frac{V}{4L_o} =\frac{3V}{2L} \\\\\frac{1}{4L_o} = \frac{3}{2*73.5 \ cm}\\\\\frac{1}{L_o} = \frac{3*4}{2*73.5 \ cm}\\\\\frac{1}{L_o} = \frac{12}{147 \ cm}\\\\L_o = \frac{147 \ cm}{12} = 12.25 \ cm$

Therefore, the length of the  open closed pipe is 12.25 cm