Question

An air column in open pipe is in resonance in $3^{rd}$ harmonic with frequency of tuning fork '$f_1$'. By keeping the length of the pipe same, one of the pipe is closed. The frequency of tuning fork is increased such that the resonance again occurs in $n^{th}$ harmonic then [$P$ = no. of harmonic]

• A. $P = 5, f_2 = (\frac{5}{4}) f_1$
• B. $P = 2, f_2 = (\frac{4}{5}) f_1$
• C. $P = 7, f_2 = (\frac{7}{6}) f_1$
• D. $P = 7, f_2 = (\frac{6}{7}) f_1$

Answer 1

Answer: (C)

Option (c): $P = 7, f_2 = \frac{7}{6} f_1$

Answer 2

1. For an open pipe, the harmonics are given by

   $f = \frac{n v}{2L}$.

   Given the 3rd harmonic resonates,

   $f_1 = \frac{3v}{2L} = \frac{6v}{4L}$.

2. For a pipe closed at one end, the allowed harmonics are the odd multiples:

   $f = \frac{m v}{4L}$ with $m = 1, 3, 5, \ldots$.

3. Let resonance occur in the $P^{th}$ (or $m^{th}$) harmonic such that

   $f_2 = \frac{m v}{4L}$.

   Assuming $m = 7$, we have

   $f_2 = \frac{7v}{4L}$.

4. Express $f_2$ in terms of $f_1$:

   $\frac{f_2}{f_1} = \frac{\frac{7v}{4L}}{\frac{6v}{4L}} = \frac{7}{6}$

   Thus,

   $f_2 = \frac{7}{6} f_1$ and the harmonic number $P = 7$.