An open pipe has a length of 0.75 m. What would be the length of a closed organ pipe whose third harmonic(n=3) is the same as the fundamental frequency of the open pipe?

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Answer 1 · 2015-09-14T17:36:37

$3.4 m$ $3.4"m"$

For an open pipe the 1st harmonic looks like this:

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This refers to half a wavelength so:

$L_{1} = \frac{λ}{2}$ $L_1=lambda/2$

$λ = 2 L_{1}$ $lambda=2L_1$

Since $v = f λ$ $v=flambda$ :

$f = \frac{v}{λ} = \frac{v}{2 L_{1}}$ $f=v/lambda=(v)/(2L_1)$ $" "$ $(1)$ $color(red)((1))$

For the closed pipe the 3rd harmonic looks like this:

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In this case:

$L_{2} = \frac{3}{4} λ$ $L_2=3/4lambda$

$λ = \frac{4 L_{2}}{3}$ $lambda=(4L_2)/(3)$

So $v = f . \frac{4 L_{2}}{3}$ $v=f.(4L_2)/(3)$

So $f = \frac{3 v}{4 L_{2}}$ $f=(3v)/(4L_2)$ $" "$ $(2)$ $color(red)((2))$

These frequencies are equal so putting $(1)$ $color(red)((1))$ equal to $(2) \Rightarrow$ $color(red)((2))rArr$

$\frac{v}{2 L_{1}} = \frac{3 v}{4 L_{2}}$ $(v)/(2L_1)=(3v)/(4L_2)$

$\frac{3}{4 L_{2}} = \frac{1}{2 L_{1}}$ $(3)/(4L_2)=(1)/(2L_1)$

$L_{2} = \frac{6 L_{1}}{4}$ $L_2=(6L_1)/(4)$

$L_{2} = \frac{3 \times 6 \times 0.75}{4}$ $L_2=(3xx6xx0.75)/(4)$

$L_{2} = 3.4 m$ $L_2=3.4"m"$