Longitudinal Stationary Waves (Organ Pipes) and Resonance Tube Questions in English

(B) The frequency of the $n^{th}$ harmonic of a closed organ pipe is given by $f_{n, closed} = \frac{nv}{4L_{closed}}$,where $n$ is an odd integer $(1, 3, 5, ...)$.
For the fifth harmonic,$n = 5$,so $f_{5, closed} = \frac{5v}{4L_{closed}}$.
The frequency of the first harmonic (fundamental frequency) of an open organ pipe is given by $f_{1, open} = \frac{v}{2L_{open}}$.
Given that the fifth harmonic of the closed pipe is in unison with the first harmonic of the open pipe,we have $f_{5, closed} = f_{1, open}$.
Substituting the expressions: $\frac{5v}{4L_{closed}} = \frac{v}{2L_{open}}$.
Simplifying the equation: $\frac{5}{4L_{closed}} = \frac{1}{2L_{open}}$.
Rearranging for the ratio of lengths: $\frac{L_{closed}}{L_{open}} = \frac{5 \times 2}{4} = \frac{10}{4} = \frac{5}{2}$.
Comparing this with the given ratio $\frac{5}{x}$,we find $x = 2$.

(B) For an open organ pipe,the frequency of the $n^{\text{th}}$ harmonic is given by $f_n = n \left( \frac{v}{2L} \right)$,where $v$ is the speed of sound and $L$ is the length of the pipe.
Given $v_3 = 3 \left( \frac{v}{2L} \right)$ and $v_6 = 6 \left( \frac{v}{2L} \right)$.
According to the problem,$v_6 - v_3 = 2200 \text{ Hz}$.
Substituting the expressions: $6 \left( \frac{v}{2L} \right) - 3 \left( \frac{v}{2L} \right) = 2200$.
$3 \left( \frac{v}{2L} \right) = 2200$.
Given $v = 330 \text{ m/s}$,we have $3 \left( \frac{330}{2L} \right) = 2200$.
$990 / (2L) = 2200$.
$2L = 990 / 2200 = 0.45 \text{ m}$.
$L = 0.225 \text{ m} = 225 \text{ mm}$.