Question

Two pipes are each open at one end and closed at the other. Pipe A has length L and pipe B has length 2L. Which harmonic of pipe B matches in frequency the fundamental of pipe A?

The answer is E, could anyone explain it?

A) The fundamental

B) The second

C) The third

D) The fourth

E) There are none

Answer 1

For pipes which are open at one end and closed at the other end, their fundamental frequency is given by:

fn = (2n + 1)*V/(4L)

for pipe A which has length L

fa = (2n₁ + 1)*V/(4L)

for pipe B which has length 2L

fb = (2n₂ + 1)*V/(4*2*L)

Now we need to at which harmonic of pipe B, frequency is same as pipe A, So

fb = fa

(2n₂ + 1)*V/(4*2*L) = (2n₁ + 1)*V/(4*L)

(2n₂ + 1)/2 = (2n₁ + 1)

2n₂ + 1 = 4n₁ + 2

2n₂ = 4n₁ + 1

n₂ = (4n₁ + 1)/2

Since n1 and n2 can only be integer number 1, 2, 3, ...... etc, and

4n₁ + 1 will always be an odd number, because 4n₁ will always be an even number, which means

(4n₁ + 1)/2 will never be an integer because it is odd number divide by 2, So there are no harmonic at which both pipes frequency will match.

Correct option is E.

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