Question

On a day when the speed of sound is $345m/s$, 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. The length of the open pipe is $470\times {10}^{-x}m$. Find $x$.

Answer 1

Let ll be the length of closed organ pipe and l′l′ be the length of open pipe .

wavelength of second overtone in closed pipe ,

$\lambda =4l/5$

λ=4l/5 ,

wavelength of third harmonic in open pipe ,

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

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

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

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

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

we have $f=v/4l$

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

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