Question 448: Waves

Question

The fundamental frequency of a closed pipe is 400 Hz . If $\frac{1}{3}$ rd pipe is filled with water, then the frequency of 2nd harmonic of the pipe will be (neglect and correction)

Accepted Answer

The fundamental frequency of a closed pipe is given as 400 Hz. Let's determine the frequency of the second harmonic when one-third of the pipe is filled with water.
Fundamental Frequency Calculation
:
The fundamental frequency of a closed pipe is represented by:
$    n = \frac{v}{4L} = 400 	ext{ Hz}    $
From this equation, the speed of sound $ v $ can be calculated as:
$    v = 400 	imes 4L    $
Adjusting for Water in the Pipe
:
When one-third of the pipe is filled with water, the remaining length of the air column becomes:
$    L - \frac{L}{3} = \frac{2L}{3}    $
New Fundamental Frequency
:
The fundamental frequency with the new air column length is calculated as:
$    	ext{New fundamental frequency} = \frac{v}{4\left(\frac{2L}{3}ight)} = \frac{3v}{8L}    $
Second Harmonic Calculation
:
The second harmonic of this configuration is three times the fundamental frequency:
$    	ext{Second harmonic} = 3 	imes 	ext{New fundamental frequency}    $
Substituting the values, we have:
$    = 3 	imes \left(\frac{3v}{8L}ight) = \frac{3 	imes 3 	imes 400 	imes 4L}{8L} = 1800 	ext{ Hz}    $
Thus, the frequency of the second harmonic of the pipe, when one-third is filled with water, is 1800 Hz.

Answer

1200 Hz

Answer

1800 Hz

Answer

600 Hz

Answer

300 Hz

Waves

The fundamental frequency of a closed pipe is 400 Hz . If \frac{1}{3} rd pipe is filled with water, then the frequency of 2nd harmonic of the pipe will be (neglect and correction)

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