Question 262: Waves

Question

An organ pipe has fundamental frequency 80 Hz . If its one end is closed, the frequencies produced will be (in Hz ) (Neglect end correction)

Accepted Answer

Step 1: Understand the problem
An
organ pipe open at both ends
has fundamental frequency = 80Ã¢ÂÂ¯Hz.
Now the same length of pipe is closed at one end (so it becomes a closed organ pipe).
We want the allowed frequencies (overtones).
Step 2: Relation of frequency and length for open and closed pipe
For an
open pipe
:
Fundamental frequency
$   f_{	ext{open}} = \frac{v}{2L}   $
For a
closed pipe
(one end closed):
Fundamental frequency
$   f_{	ext{closed}} = \frac{v}{4L}   $
And the harmonics are only odd multiples:
$ f_n = (2n - 1)\frac{v}{4L}, \quad n=1,2,3,\ldots $
Step 3: Relate the given data
Given open fundamental:
$   f_{	ext{open}} = 80 = \frac{v}{2L}.   $
So,
$   \frac{v}{L} = 160.   $
For the same pipe closed at one end:
Fundamental frequency
$   f_{	ext{closed}} = \frac{v}{4L} = \frac{160}{4} = 40 	ext{ Hz}.   $
Step 4: Find the frequencies
Thus, for closed pipe:
$ f = (2n - 1)	imes 40 \quad (n=1,2,3,\dots) $
So allowed frequencies:
$40, 120, 200, 280, \dots$
Step 5: Match with options
Option C: $40, 120, 200, 280$ Ã¢ÂÂ
Ã¢ÂÂ Final Answer:
Option C: $40, 120, 200, 280$

Answer

$40,80,120,160$

Answer

$40,80,160,240$

Answer

$40,120,200,280$

Answer

$80,160,240,320$

Waves

An organ pipe has fundamental frequency 80 Hz . If its one end is closed, the frequencies produced will be (in Hz ) (Neglect end correction)

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