An open pipe is in resonance in its 2^ nd harmonic with tuning fo… - Vidyadip

MCQ

An open pipe is in resonance in its $2^{nd}$ harmonic with tuning fork of frequency ${f_1}$. Now it is closed at one end. If the frequency of the tuning fork is increased slowly from ${f_1}$ then again a resonance is obtained with a frequency ${f_2}$. If in this case the pipe vibrates ${n^{th}}$ harmonics then

A
$n = 3,$ ${f_2} = \frac{3}{4}{f_1}$
B
$n = 3,$ ${f_2} = \frac{5}{4}{f_1}$
✓
$n = 5,$ ${f_2} = \frac{5}{4}{f_1}$
D
$n = 5,$ ${f_2} = \frac{3}{4}{f_1}$

✓

Answer

Correct option: C.

$n = 5,$ ${f_2} = \frac{5}{4}{f_1}$

c
(c) Open pipe resonance frequency ${f_1} = \frac{{2v}}{{2L}}$

Closed pipe resonance frequency ${f_2} = \frac{{nv}}{{4L}}$

${f_2} = \frac{n}{4}{f_1}$ (where n is odd and ${f_2} > {f_1}$)

$\therefore n = 5$

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