A pipe which is open at both ends is $47 c m$ long and has an inner diameter $5 c m$ . If the speed of sound in air is $348 m / s$ , calculate the fundamental frequency of air column in that pipe.

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Solution

For open pipe, given
$l = 47 c m$ , inner diameter $= 5 c m$ , and $v = 348 m / s$
Fundamental frequency $n =$ ?
$∵ n = \frac{v}{2 l}$
$\Rightarrow n = \frac{v}{2 (l + 0.6 r)}$
$\Rightarrow n = \frac{348}{2 (0.47 + 0.6 \times 2.5 \times 10^{- 2})} = 358.76 H z$

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A pipe which is open at both ends is 47cm long and has an inner diameter 5cm. If the speed of sound in air is 348m/s, calculate the fundamental frequency of air column in that pipe.

For open pipe, givenl=47cm, inner diameter=5cm, and v=348m/sFundamental frequency n=?∵n=v2l⇒n=v2(l+0.6r)⇒n=3482(0.47+0.6×2.5×10−2)=358.76Hz

A pipe which is open at both ends is $47 c m$ long and has an inner diameter $5 c m$ . If the speed of sound in air is $348 m / s$ , calculate the fundamental frequency of air column in that pipe.

For open pipe, given
$l = 47 c m$ , inner diameter $= 5 c m$ , and $v = 348 m / s$
Fundamental frequency $n =$ ?
$∵ n = \frac{v}{2 l}$
$\Rightarrow n = \frac{v}{2 (l + 0.6 r)}$
$\Rightarrow n = \frac{348}{2 (0.47 + 0.6 \times 2.5 \times 10^{- 2})} = 358.76 H z$