Question

A closed organ pipe (that is, open at one end) 2.5 ft long is sounded. If the velocity of sound is
1100 ft/sec, find the frequency of first overtone.
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Answer 1

Given:

A closed organ pipe (that is, open at one end) 2.5 ft long is sounded. The velocity of sound is

1100 ft/sec.

To find:

Frequency of 1st overtone ?

Calculation:

1st Overtone refers to the frequency of resonance immediately higher over the fundamental frequency.

So, now look at the diagram.

$\dfrac{ \lambda}{2} + \dfrac{ \lambda}{4} = l$

$\implies \dfrac{ 3\lambda}{4} = l$

$\implies \lambda = \dfrac{ 4l}{3}$

Now, 1st Overtone Frequency is :

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

$\implies\: f = \dfrac{v}{( \dfrac{4l}{3}) }$

$\implies\: f = \dfrac{1100}{( \dfrac{4 \times 2.5}{3}) }$

$\implies\: f = 330 \: hz$

So, frequency of 1st Overtone is 330 Hz.