Find the fundamental, first overtone and second overtone frequencies of an open organ pipe of length 20 cm. Speed of sound in air is $340 m s^{- 1}$ .

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Solution

If V=340 m/s,

I+20 cm $\times 10^{- 2} m$

Fundamental frequency

$= (\frac{V}{2 I}) = \frac{340}{2 \times 20 \times 10^{- 2}}$

=850 Hz

We known first over tone,

$(\frac{2 V}{2 I}) = \frac{2 \times 340}{2 \times 10 \times 10^{- 2}}$ (for open pipe)

=1750 Hz

Second overtone

$= 3 (\frac{V}{2 I}) = 3 \times 850 = 2500 H z$

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Find the fundamental, first overtone and second overtone frequencies of an open organ pipe of length 20 cm. Speed of sound in air is 340ms−1.

If V=340 m/s, I+20 cm ×10−2m Fundamental frequency =(V2I)=3402×20×10−2 =850 Hz We known first over tone, (2V2I)=2×3402×10×10−2 (for open pipe) =1750 Hz Second overtone =3(V2I)=3×850=2500Hz

Find the fundamental, first overtone and second overtone frequencies of an open organ pipe of length 20 cm. Speed of sound in air is $340 m s^{- 1}$ .