Question

What is the beat frequency heard when two organ pipes, each open at both ends, are sounded together at their fundamental frequencies if one pipe is 60 cm long and the other is 68 cm long?(The speed of sound is 340 m/s)

Answer 1

Answer:

The beat frequency is 33.33 Hz.

Explanation:

Given that,

Length of first pipe =60 cm

Length of other pipe = 68 cm

Speed of sound = 340 m/s

We need to calculate the frequency

We know that,

When they operate at fundamental frequency then the length is given by,

$L=\dfrac{\lambda}{2}$

The wavelength is given by

$\lambda=2L$

For first organ pipe,

Using formula of frequency

$f=\dfrac{v}{\lambda}$

$f_{1}=\dfrac{v}{2L_{1}}$...(I)

Put the value into the formula

$f_{1}=\dfrac{340}{2\times60\times10^{-2}}$

$f_{1}=283.33\ Hz$

For second organ pipe,

$f_{2}=\dfrac{v}{2L_{2}}$...(II)

Put the value in the equation (II)

$f_{2}=\dfrac{340}{2\times68\times10^{-2}}$

$f_{2}=250\ Hz$

Therefore the beat frequency

$\Delta f=f_{1}-f_{2}$

$\Delta f=283.33-250$

$\Delta f=33.33\ Hz$

Hence,  The beat frequency is 33.33 Hz.