Question

The fundamental frequency of a closed organ pipe is $220Hz$. The second overtone of this pipe has the same wavelength as the third harmonic of an open pipe. Take speed of sound in air $345m/s$. The length of this pipe is $470\times {10}^{-x}m$. Find $x$.

Answer 1

Let $l$ be the length of closed organ pipe and $l^{\prime }$ be the length of open pipe .

wavelength of second overtone in closed pipe ,

$\lambda =4l/5$ ,

wavelength of third harmonic in open pipe ,

${\lambda }^{\prime }=2l^{\prime }/3$ ,

given $\lambda ={\lambda }^{\prime }$ ,

or $4l/5=2l^{\prime }/3$ ,

or $l^{\prime }=12l/10$ ,

now given fundamental frequency of closed organ pipe , $f=220Hz$ ,

we have $f=v/4l$ ,

or $l=v/4f=345/(4\times 220)=0.392m$ ,

therefore $l^{\prime }=12l/10=1.2\times 0.392=0.470m$