An open pipe is in resonance in its 2nd harmonic with tuning fork of frequency $f_1$. Now, it is closed at…
An open pipe is in resonance in its 2nd 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 $nth$ harmonic, then
$n = 3, f_1 = \frac{3}{4}f_2$
$n = 3, f_2 = \frac{5}{4}f_1$
$n = 5, f_2 = \frac{5}{4}f_1$
$n = 5, f_2 = \frac{3}{4}f_1$
Solution
For open pipe, $f_1 = 2\left(\frac{v}{2l}\right) = \frac{v}{l}$
For closed pipe, $f_2 = n\left(\frac{v}{4l}\right)$, $n = 1, 3, 5, 7, \dots$
$f_2 > f_1$
So, $f_2 = \frac{n}{4} \cdot f_1$ and $n > 4$