Question 241: Waves

Question

In an open end organ pipe of length ' $L$ ', if the velocity of sound is ' V ', then the fundamental frequency will be (Neglect end correction)

Accepted Answer

Step 1: Behavior of open organ pipe
In an
open organ pipe
, both ends are open.
At open ends, there are
pressure nodes
(displacement antinodes).
Therefore, the fundamental mode (first harmonic) has
two open ends
, meaning:
The pipe length $L = \frac{\lambda}{2}$.
Step 2: Fundamental frequency
From wave relation:
$ f = \frac{v}{\lambda} $
For fundamental,
$ L = \frac{\lambda}{2}\quad \Rightarrow \quad \lambda = 2L $
So,
$ f_1 = \frac{v}{\lambda} = \frac{v}{2L}. $
Step 3: Harmonics present
In an open-open pipe, all harmonics are present:
Fundamental, 2nd, 3rd, etc.
Ã¢ÂÂ Final Answer
$ \boxed{\frac{V}{2L} 	ext{ and all harmonics are present.}} $
That corresponds to
Option A
.

Answer

$\frac{\mathrm{V}}{2 \mathrm{~L}}$ and all harmonics are present.

Answer

$\frac{\mathrm{V}}{4 \mathrm{~L}}$ and all harmonics are present.

Answer

$\frac{\mathrm{V}}{2 \mathrm{~L}}$ and even harmonics are present.

Answer

$\frac{\mathrm{V}}{4 \mathrm{~L}}$ and even harmonics are present.

Waves

In an open end organ pipe of length ' L ', if the velocity of sound is ' V ', then the fundamental frequency will be (Neglect end correction)

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