What length of open pipe will produce a frequency of 1200 Hz as its first overtone on a day when the speed of sound is 340 m s⁻¹ (28.3cm)



Given:

Speed of sound = v = 340 m s⁻¹

Frequency of first overtone = f₂ = 1200 Hz

To Find:

Length of the open pipe = L =  ?


Solution:

Since for the first overtone 


f₂ = 2 f₁


But  f₁ = `\frac {v}{2 L}`


So   f₂ = 2 `\frac {v}{2 L}`


or 


f₂ =  `\frac {v}{L}`


or


L = `\frac {v}{f₂}`


by putting values


L = `\frac {340 m s⁻¹}{1200 Hz}`


L = 0.28333 m


0r 


L = 28.3 cm ----------------Ans



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