Question

The first overtone of an open organ pipe beats with the first overtone of a closed organ pipe with a beat frequency of 2.2 Hz. The fundamental frequency of the closed organ pipe is 110 Hz. Find the lengths of the pipes. Speed of sound in air v = 330 m/s.

Answer 1

According to problem, fundamental frequency of closed pipe,

$n_c=\frac{v}{4L_c}$

Therefore,

$L_c=\frac{v}{4n_c}=\frac{330}{4\times 110}=0.75m$

A frequency of the first overtone of closed pipe

$n_3=3\times n_c=330Hz$

The first overtone of open pipe has the frequency

$n_2=2n_0=2\left(\frac{v}{2L_0}\right)$

That is,

$\frac{2v}{2L_0}-330=+2.2or-2.2$

That is,

$L_0=\frac{330}{327.8}m$ or $\frac{330}{332.2}m$

$=1.0067m$ or $0.99m$