Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11

# The First Overtone Frequency of a Closed Organ Pipe P1 is Equal to the Fundamental Frequency of a Open Organ Pipe P2. - Physics

Sum

The first overtone frequency of a closed organ pipe P1 is equal to the fundamental frequency of a open organ pipe P2. If the length of the pipe P1 is 30 cm, what will be the length of P2?

#### Solution

Given:
Length of closed organ pipe L1 = 30 cm
Length of open organ pipe L2 = ?
Let $f_1$ and $f_2$ be the frequencies of the closed and open organ pipes, respectively.
The first overtone frequency of a closed organ pipe P1 is given by

$f_1 = \frac{3v}{4 L_1}$

where v is the speed of sound in air.
On substituting the respective values, we get :

$f_1 = \frac{3v}{4 \times 30}$

Fundamental frequency of an open organ pipe is given by:

$f_2 = \left( \frac{v}{2 L_2} \right)$

As per the question,

$f_1 = f_2$

$\left( \frac{3 \times v}{4 \times 30} \right) = \left( \frac{v}{2 L_2} \right)$

$\Rightarrow L_2 = 20 \text { cm }$

∴ The length of the pipe P2 will be 20 cm.

Concept: Wave Motion
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#### APPEARS IN

HC Verma Class 11, 12 Concepts of Physics 1
Chapter 16 Sound Waves
Q 42 | Page 355