Question 245: Waves

Question

The fundamental frequency of a closed pipe is 400 Hz . If $\left(\frac{1}{3}ight)^{	ext {rd }}$ length of the pipe is filled with water, the frequency of the $2^{	ext {nd }}$ harmonic of the pipe will be (Neglect end correction)

Accepted Answer

Step 1: Recall fundamental frequency of a closed pipe
For a closed organ pipe:
$   f_1 = \frac{v}{4L}   $
where $L$ = length of the pipe, $v$ = speed of sound.
Given: $ f_1 = 400 $.
Ã¢ÂÂ $\dfrac{v}{4L} = 400$
Ã¢ÂÂ $ v = 1600 L $.
So relationship established: $v$ is proportional to $L$.
Step 2: Pipe partially filled with water
When $	frac{1}{3}$ of length filled with water, air column length becomes:
$   L' = L - \frac{L}{3} = \frac{2L}{3}.   $
Step 3: Frequency of a closed pipe
For closed pipe, harmonics are
odd multiples
:
$ f_n = (2n-1) \frac{v}{4L'} $.
So,
Fundamental with new air column:
$   f_1' = \frac{v}{4 L'}.   $
Second harmonic (i.e., 3rd harmonic actually):
$   f_2' = 3 \frac{v}{4 L'}.   $
Step 4: Compute
Use $v = 1600 L$, $L' = 	frac{2L}{3}$.
So,
$ f_2' = \frac{3v}{4L'} = \frac{3v}{4 \cdot 	frac{2L}{3}}. $
Simplify denominator:
$ 4 \cdot \frac{2L}{3} = \frac{8L}{3}. $
So,
$ f_2' = \frac{3v}{	frac{8L}{3}} = \frac{3v \cdot 3}{8L} = \frac{9v}{8L}. $
Step 5: Substitute $v = 1600 L$
$ f_2' = \frac{9 \cdot 1600L}{8L} = \frac{14400}{8} = 1800 \, 	ext{Hz}. $
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Answer: Option D (1800 Hz)

Answer

1500 Hz

Answer

1200 Hz

Answer

600 Hz

Answer

1800 Hz

Waves

The fundamental frequency of a closed pipe is 400 Hz . If \left(\frac{1}{3}\right)^{\text {rd }} length of the pipe is filled with water, the frequency of the 2^{\text {nd }} harmonic of the pipe will be (Neglect end correction)

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