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

(A) Since both pipes are in resonance with the same tuning fork,they must have the same frequency of vibration,$f$.
For an organ pipe closed at one end,the frequency of the $n$-th overtone is given by $f = \frac{(2n+1)v}{4l_1}$,where $n$ is the overtone number. For the first overtone $(n=1)$,the frequency is $f = \frac{3v}{4l_1}$.
For an organ pipe open at both ends,the frequency of the $n$-th overtone is given by $f = \frac{(n+1)v}{2l_2}$. For the third overtone $(n=3)$,the frequency is $f = \frac{(3+1)v}{2l_2} = \frac{4v}{2l_2} = \frac{2v}{l_2}$.
Equating the frequencies: $\frac{3v}{4l_1} = \frac{2v}{l_2}$.
Rearranging to find the ratio $\frac{l_1}{l_2}$: $\frac{l_1}{l_2} = \frac{3v}{4l_1} \cdot \frac{l_2}{2v} = \frac{3}{8}$.

(A) The frequency of the $m^{\text{th}}$ harmonic in an open organ pipe is given by $f_m = \frac{m v}{2 L}$.
For the fundamental frequency $(m=1)$, $f_1 = \frac{v}{2 L}$.
The frequency of the $n^{\text{th}}$ harmonic in a closed organ pipe is given by $f_n^{\prime} = \frac{n v}{4 L}$, where $n$ must be an odd integer.
The frequency of the $3^{\text{rd}}$ harmonic in a closed organ pipe is $f_3^{\prime} = \frac{3 v}{4 L}$.
According to the problem, $f_3^{\prime} - f_1 = 50 \,Hz$.
Substituting the expressions: $\frac{3 v}{4 L} - \frac{v}{2 L} = 50$.
$\frac{3 v - 2 v}{4 L} = 50 \Rightarrow \frac{v}{4 L} = 50 \Rightarrow \frac{v}{L} = 200$.
The fundamental frequency of the open organ pipe is $f_1 = \frac{v}{2 L} = \frac{200}{2} = 100 \,Hz$.

(A) Let the end correction be $e$. The resonance condition for a tube closed at one end is given by $l_n + e = (2n-1) \frac{\lambda}{4}$.
For the first resonance $(n=1)$: $24.1 + e = \frac{\lambda}{4} \quad ---(1)$
For the second resonance $(n=2)$: $74.1 + e = \frac{3\lambda}{4} \quad ---(2)$
Subtracting equation $(1)$ from equation $(2)$:
$(74.1 + e) - (24.1 + e) = \frac{3\lambda}{4} - \frac{\lambda}{4}$
$50 = \frac{2\lambda}{4} = \frac{\lambda}{2}$
$\lambda = 100 \ cm$
Substituting $\lambda$ in equation $(1)$:
$24.1 + e = \frac{100}{4} = 25$
$e = 25 - 24.1 = 0.9 \ cm$
The end correction $e$ is related to the diameter $D$ by the formula $e = 0.3D$.
$0.9 = 0.3D$
$D = \frac{0.9}{0.3} = 3 \ cm$.

(A) The frequency of the tuning fork is $n = 264 \,Hz$ and the speed of sound is $V = 330 \,m/s$.
For a pipe closed at one end,the resonance occurs at lengths given by $\ell = (2k-1) \frac{\lambda}{4}$,where $k = 1, 2, 3, \dots$.
The wavelength is $\lambda = \frac{V}{n} = \frac{330}{264} = 1.25 \,m = 125 \,cm$.
Thus,the possible lengths are $\ell = (2k-1) \frac{125}{4} = (2k-1) \times 31.25 \,cm$.
For $k=1$,$\ell = 31.25 \,cm$.
For $k=2$,$\ell = 3 \times 31.25 = 93.75 \,cm$.
For $k=3$,$\ell = 5 \times 31.25 = 156.25 \,cm$.
Comparing these with the given options,$62.50 \,cm$ is not an odd multiple of $31.25 \,cm$,hence it is not possible.

(D) In a pipe closed at one end, the allowed frequencies are odd multiples of the fundamental frequency $(f_0)$: $f_n = (2n-1)f_0$, where $n = 1, 2, 3, ...$.
These frequencies are $f_1 = f_0$, $f_2 = 3f_0$, $f_3 = 5f_0$, $f_4 = 7f_0$, and so on.
The difference between two consecutive harmonics in a closed pipe is $2f_0$.
Given the consecutive frequencies are $150 \,Hz$ and $250 \,Hz$, their difference is $250 \,Hz - 150 \,Hz = 100 \,Hz$.
Therefore, $2f_0 = 100 \,Hz$, which gives $f_0 = 50 \,Hz$.

(D) In the fundamental mode of a closed pipe,the length of the air column $L$ is equal to $\frac{\lambda}{4}$.
Given that the time taken for the sound wave to travel the length $L$ is $t$.
Since the wave travels a distance $L = \frac{\lambda}{4}$ in time $t$,the time taken to travel a full wavelength $\lambda$ is $T = 4t$.
Here,$T$ represents the time period of the vibration.
The frequency of vibration $n$ is the reciprocal of the time period,given by $n = \frac{1}{T}$.
Substituting $T = 4t$,we get $n = \frac{1}{4t} = \frac{0.25}{t}$.

(A) For an open organ pipe,the fundamental frequency is $f_o = \frac{v_2}{2 L_o}$. The first overtone is $f_{o1} = 2 f_o = \frac{v_2}{L_o}$.
For a closed organ pipe,the fundamental frequency is $f_c = \frac{v_1}{4 L_c}$. The first overtone is $f_{c1} = 3 f_c = \frac{3 v_1}{4 L_c}$.
Given that the frequencies are equal,$f_{o1} = f_{c1}$,so $\frac{v_2}{L_o} = \frac{3 v_1}{4 L_c}$.
Rearranging for $L_o$,we get $L_o = \frac{4 L_c}{3} \frac{v_2}{v_1}$.
The speed of sound in a gas is given by $v = \sqrt{\frac{1}{\rho \beta}}$,where $\beta$ is the compressibility. Since $\beta$ is the same for both gases,$\frac{v_2}{v_1} = \sqrt{\frac{\rho_1}{\rho_2}}$.
Substituting this into the expression for $L_o$,we get $L_o = \frac{4 L_c}{3} \sqrt{\frac{\rho_1}{\rho_2}}$.

(B) Let $L = 1.5 \,m$ be the total length of the pipe and $\ell$ be the length immersed in water. The length of the air column above the water is $L' = L - \ell = 1.5 - \ell$.
Since the pipe is now closed at one end (by water), it acts as a closed organ pipe.
The frequency of the $n^{th}$ overtone for a closed pipe is given by $f_n = \frac{(2n+1)v}{4L'}$, where $n$ is the overtone number.
For the second overtone, $n = 2$, so the frequency is $f_2 = \frac{(2(2)+1)v}{4L'} = \frac{5v}{4L'}$.
Given $f_2 = 330 \,Hz$ and $v = 330 \,m/s$, we have:
$330 = \frac{5 \times 330}{4(1.5 - \ell)}$
$1 = \frac{5}{4(1.5 - \ell)}$
$4(1.5 - \ell) = 5$
$6 - 4\ell = 5$
$4\ell = 1$
$\ell = 0.25 \,m$.

(A) The frequency of the tuning fork is $f = 500 \,Hz$ and the velocity of sound is $v = 320 \,m/s$.
The wavelength of the sound wave is $\lambda = \frac{v}{f} = \frac{320}{500} = 0.64 \,m = 64 \,cm$.
For a tube closed at one end, resonance occurs when the length of the air column $L$ satisfies $L = \frac{(2n-1)\lambda}{4}$, where $n = 1, 2, 3, \dots$.
The possible lengths for resonance are:
For $n=1: L_1 = \frac{\lambda}{4} = \frac{64}{4} = 16 \,cm$.
For $n=2: L_2 = \frac{3\lambda}{4} = 3 \times 16 = 48 \,cm$.
For $n=3: L_3 = \frac{5\lambda}{4} = 5 \times 16 = 80 \,cm$.
For $n=4: L_4 = \frac{7\lambda}{4} = 7 \times 16 = 112 \,cm$.
Since the total length of the tube is $1 \,m = 100 \,cm$, only lengths $16 \,cm, 48 \,cm,$ and $80 \,cm$ are possible.
Therefore, the total number of resonances obtained is $3$.

(C) For a tube closed at one end,the resonance lengths are given by $L = \frac{(2k-1)\lambda}{4}$,where $k = 1, 2, 3, \dots$
The first resonance occurs at $L_1 = \frac{\lambda}{4}$.
The second resonance occurs at $L_2 = \frac{3\lambda}{4}$.
Subtracting the two lengths: $L_2 - L_1 = \frac{3\lambda}{4} - \frac{\lambda}{4} = \frac{2\lambda}{4} = \frac{\lambda}{2}$.
Therefore,the wavelength is $\lambda = 2(L_2 - L_1)$.
The speed of sound $V$ is given by the formula $V = n\lambda$.
Substituting the value of $\lambda$: $V = n \times 2(L_2 - L_1) = 2n(L_2 - L_1)$.

(B) In an open pipe of length $L$,the condition for standing waves is $L = n \frac{\lambda}{2}$,where $n$ is the number of loops (or harmonics).
For a pipe open at both ends,antinodes are always formed at the open ends.
If there are $4$ antinodes and $3$ nodes,the number of loops formed is $n = 3$.
The length of the pipe is $L = 3 \frac{\lambda}{2}$.
This corresponds to the $3^{rd}$ harmonic.
The sequence of harmonics for an open pipe is $f_1, 2f_1, 3f_1, \dots$ (fundamental,$1^{st}$ overtone,$2^{nd}$ overtone,$\dots$).
Since $3f_1$ is the $3^{rd}$ harmonic,it is the $2^{nd}$ overtone.

(C) The fundamental frequency of a pipe closed at one end is given by $n = \frac{v}{4 \ell_A}$,where $v$ is the speed of sound and $\ell_A$ is the length of pipe '$A$'.
The frequencies of an open pipe are given by $n_k = \frac{k v}{2 \ell_B}$,where $k = 1, 2, 3, \dots$ and $\ell_B$ is the length of pipe '$B$'.
The first overtone corresponds to $k=2$,and the second overtone corresponds to $k=3$.
Thus,the second overtone of the open pipe is $n' = \frac{3 v}{2 \ell_B}$.
According to the problem,$n = n'$,so $\frac{v}{4 \ell_A} = \frac{3 v}{2 \ell_B}$.
Simplifying this,we get $\frac{1}{4 \ell_A} = \frac{3}{2 \ell_B}$,which implies $\frac{\ell_A}{\ell_B} = \frac{2}{4 \times 3} = \frac{2}{12} = \frac{1}{6}$.
Therefore,the ratio of the length of pipe '$A$' to that of pipe '$B$' is $1: 6$.

(D) The fundamental frequency of an open tube of length $\ell_1$ is given by $n_1 = \frac{v}{2\ell_1} = n$.
When the tube is dipped vertically in water such that one-fourth of its length is submerged, the length of the air column becomes $\ell_2 = \ell_1 - \frac{1}{4}\ell_1 = \frac{3}{4}\ell_1$.
Since the tube is now closed at one end by the water surface, it acts as a closed organ pipe.
The fundamental frequency of a closed organ pipe is $n_2 = \frac{v}{4\ell_2}$.
Substituting $\ell_2 = \frac{3}{4}\ell_1$ into the equation:
$n_2 = \frac{v}{4(\frac{3}{4}\ell_1)} = \frac{v}{3\ell_1}$.
Comparing $n_2$ with $n_1$:
$n_2 = \frac{2}{3} \times (\frac{v}{2\ell_1}) = \frac{2}{3}n$.

(D) Fundamental frequency of a closed pipe is $n_{c} = \frac{v}{4L}$.
Fundamental frequency of an open pipe is $n_{o} = \frac{v}{2L}$.
Given that they produce $2$ beats per second:
$n_{o} - n_{c} = 2$
$\frac{v}{2L} - \frac{v}{4L} = 2$
$\frac{v}{4L} = 2 \implies \frac{v}{L} = 8$.
When the length of the open pipe is halved,the new frequency is:
$n_{o}' = \frac{v}{2(L/2)} = \frac{v}{L} = 8 \text{ Hz}$.
When the length of the closed pipe is doubled,the new frequency is:
$n_{c}' = \frac{v}{4(2L)} = \frac{v}{8L} = \frac{1}{8} \times \left(\frac{v}{L}\right) = \frac{1}{8} \times 8 = 1 \text{ Hz}$.
New beat frequency $= n_{o}' - n_{c}' = 8 - 1 = 7 \text{ beats per second}$.

(B) For a pipe open at both ends, the resonant frequencies are given by the formula: $f_n = \frac{n v}{2L}$, where $n = 1, 2, 3, \dots$ is the harmonic number, $v$ is the speed of sound, and $L$ is the length of the pipe.
Given: $v = 340 \,m/s$ and $L = 1 \,m$.
Substituting these values, we get: $f_n = \frac{n \times 340}{2 \times 1} = n \times 170 \,Hz$.
This means the pipe can resonate at frequencies that are integer multiples of $170 \,Hz$ (i.e., $170 \,Hz, 340 \,Hz, 510 \,Hz, \dots$).
Comparing this with the given options, $85 \,Hz$ is not a multiple of $170 \,Hz$ and therefore the air column cannot resonate at this frequency.

(A) The frequency of the first overtone of a closed organ pipe of length $\ell$ is given by $f = \frac{3V}{4\ell}$.
The frequency of the first overtone of an open organ pipe of length $\ell^{\prime}$ is given by $f^{\prime} = \frac{2V}{2\ell^{\prime}} = \frac{V}{\ell^{\prime}}$.
Given that the frequencies are identical,we set $f = f^{\prime}$:
$\frac{3V}{4\ell} = \frac{V}{\ell^{\prime}}$.
Rearranging the terms to find the ratio of lengths $\frac{\ell}{\ell^{\prime}}$:
$\frac{\ell}{\ell^{\prime}} = \frac{3}{4}$.
Thus,the ratio of their lengths is $3: 4$.

(B) For an open organ pipe,the fundamental frequency is given by $n = \frac{V}{2\ell}$,where $V$ is the speed of sound and $\ell$ is the length of the pipe.
For the first pipe: $\ell_{1} = \frac{V}{2n_{1}}$.
For the second pipe: $\ell_{2} = \frac{V}{2n_{2}}$.
When joined in series,the total length of the new pipe is $\ell = \ell_{1} + \ell_{2}$.
The fundamental frequency of the new pipe is $n = \frac{V}{2\ell}$.
Substituting the values of $\ell_{1}$ and $\ell_{2}$ into the equation for $\ell$:
$\frac{V}{2n} = \frac{V}{2n_{1}} + \frac{V}{2n_{2}}$.
Dividing both sides by $\frac{V}{2}$,we get: $\frac{1}{n} = \frac{1}{n_{1}} + \frac{1}{n_{2}}$.
Solving for $n$: $\frac{1}{n} = \frac{n_{1} + n_{2}}{n_{1}n_{2}}$,which gives $n = \frac{n_{1}n_{2}}{n_{1} + n_{2}}$.

(B) In an organ pipe closed at one end,the standing wave pattern always has a node at the closed end and an antinode at the open end.
For the fundamental mode,there is $1$ node and $1$ antinode.
For the first overtone,there are $2$ nodes and $2$ antinodes.
Since the question specifies $2$ nodes and $2$ antinodes,this corresponds to the first overtone.

(C) In the fundamental mode of a closed pipe,the length of the pipe $L$ corresponds to one-fourth of the wavelength,i.e.,$L = \frac{\lambda}{4}$.
Given that the time taken for the sound wave to travel the length $L$ is $t$,we have $t = \frac{L}{v}$,where $v$ is the speed of sound.
Since $L = \frac{\lambda}{4}$,we can write $t = \frac{\lambda}{4v}$.
The time period $T$ of the wave is given by $T = \frac{\lambda}{v}$.
Substituting $\lambda = vT$ into the expression for $t$,we get $t = \frac{vT}{4v} = \frac{T}{4}$.
Therefore,the time period $T = 4t$.
The frequency $f$ is the reciprocal of the time period,so $f = \frac{1}{T} = \frac{1}{4t}$.

(C) Given: Length of the tube $L = 0.8 \ m$,Frequency $f = 500 \ Hz$,Speed of sound $v = 340 \ m/s$.
First,calculate the wavelength $\lambda$ of the sound wave:
$\lambda = \frac{v}{f} = \frac{340}{500} = 0.68 \ m$.
The resonance condition for a tube closed at one end is $L_n = \frac{(2n-1)\lambda}{4}$,where $n = 1, 2, 3, \dots$.
We need to find the number of values of $n$ such that $L_n \le 0.8 \ m$.
For $n=1$: $L_1 = \frac{\lambda}{4} = \frac{0.68}{4} = 0.17 \ m$.
For $n=2$: $L_2 = \frac{3\lambda}{4} = 3 \times 0.17 = 0.51 \ m$.
For $n=3$: $L_3 = \frac{5\lambda}{4} = 5 \times 0.17 = 0.85 \ m$.
Since $L_3 = 0.85 \ m > 0.8 \ m$,the third resonance cannot be formed within the tube length.
Therefore,only $2$ resonances will be heard.

(A) For a closed organ pipe of length $L_c$,the frequency of the $n^{\text{th}}$ overtone is given by $f_c = \frac{(2n+1)v}{4L_c}$,where $n$ is the overtone number.
For the $3^{\text{rd}}$ overtone $(n=3)$,$f_c = \frac{(2(3)+1)v}{4L_c} = \frac{7v}{4L_c}$.
For an open organ pipe of length $L_o$,the frequency of the $n^{\text{th}}$ overtone is given by $f_o = \frac{(n+1)v}{2L_o}$.
For the $3^{\text{rd}}$ overtone $(n=3)$,$f_o = \frac{(3+1)v}{2L_o} = \frac{4v}{2L_o} = \frac{2v}{L_o}$.
Since the frequencies are in unison,$f_c = f_o$:
$\frac{7v}{4L_c} = \frac{2v}{L_o}$.
Rearranging for the ratio $\frac{L_c}{L_o}$:
$\frac{L_c}{L_o} = \frac{7}{4 \times 2} = \frac{7}{8}$.
Thus,the correct option is $A$.

(A) Let the length of the pipe be $\ell$ and the speed of sound be $v$.
For an open pipe, the fundamental frequency is $f_0 = \frac{v}{2\ell}$. The $2^{nd}$ overtone of an open pipe is $3f_0 = \frac{3v}{2\ell}$.
For a closed pipe, the fundamental frequency is $f_c = \frac{v}{4\ell}$. The $3^{rd}$ overtone of a closed pipe is $7f_c = \frac{7v}{4\ell}$.
According to the problem, the $3^{rd}$ overtone of the closed pipe is $150 \,Hz$ higher than the $2^{nd}$ overtone of the open pipe:
$\frac{7v}{4\ell} = \frac{3v}{2\ell} + 150$
$\frac{7v}{4\ell} - \frac{6v}{4\ell} = 150$
$\frac{v}{4\ell} = 150$
Since the fundamental frequency of the open pipe is $f_0 = \frac{v}{2\ell}$, we can write:
$f_0 = 2 \times \left(\frac{v}{4\ell}\right) = 2 \times 150 = 300 \,Hz$.

(D) For a closed organ pipe of length $\ell_c$,the fundamental frequency is $f_c = \frac{V}{4\ell_c}$. The first overtone is the third harmonic,so $f_{c,1} = 3f_c = \frac{3V}{4\ell_c}$.
For an open organ pipe of length $\ell_o$,the fundamental frequency is $f_o = \frac{V}{2\ell_o}$. The first overtone is the second harmonic,so $f_{o,1} = 2f_o = \frac{2V}{2\ell_o} = \frac{V}{\ell_o}$.
Given that the frequencies of the first overtones are identical,we have $\frac{3V}{4\ell_c} = \frac{V}{\ell_o}$.
Rearranging the terms to find the ratio $\frac{\ell_o}{\ell_c}$,we get $\frac{\ell_o}{\ell_c} = \frac{4}{3}$.

(B) In a closed pipe,the fundamental mode of vibration occurs when the length of the pipe $\ell$ is equal to one-fourth of the wavelength $\lambda$ of the sound wave.
Formula: $\ell = \frac{\lambda}{4}$
Given: $\lambda = 62 \,cm$
Calculation: $\ell = \frac{62}{4} = 15.5 \,cm$
Therefore,the correct option is $B$.

(B) The fundamental frequency of a closed pipe of length $L$ is given by $f_1 = \frac{v}{4L} = 400 \,Hz$.
From this,we get $v = 1600L$.
When $1/3$ of the pipe is filled with water,the length of the air column becomes $L' = L - \frac{L}{3} = \frac{2L}{3}$.
The new fundamental frequency of the air column is $f'_1 = \frac{v}{4L'} = \frac{v}{4(2L/3)} = \frac{3v}{8L}$.
Substituting $v = 1600L$,we get $f'_1 = \frac{3(1600L)}{8L} = 3 \times 200 = 600 \,Hz$.
In a closed pipe,the harmonics are odd multiples of the fundamental frequency $(f_1, 3f_1, 5f_1, \dots)$. However,the question asks for the $2^{\text{nd}}$ harmonic of the pipe in its new state. The $2^{\text{nd}}$ harmonic (first overtone) of a closed pipe is $3f'_1$.
Therefore,the frequency of the $2^{\text{nd}}$ harmonic is $3 \times 600 \,Hz = 1800 \,Hz$.

(B) Given: Length of tube $L = 0.8 \,m$, Frequency $f = 375 \,Hz$, Speed of sound $v = 330 \,m/s$.
For the fundamental frequency in a resonance tube (closed at one end), the length of the air column $\ell$ is given by $\ell = \frac{v}{4f}$.
Substituting the values: $\ell = \frac{330}{4 \times 375} = \frac{330}{1500} = 0.22 \,m$.
This $\ell$ represents the length of the air column from the top.
The water level from the bottom is the total length of the tube minus the length of the air column: $h = L - \ell$.
$h = 0.8 \,m - 0.22 \,m = 0.58 \,m$.

(D) The fundamental frequency of an open pipe is given by $n = \frac{v}{2 \ell}$.
For a closed pipe, the harmonics are odd multiples of the fundamental frequency $n_0 = \frac{v}{4 \ell}$.
The $2^{nd}$ harmonic of a closed pipe is actually the $1^{st}$ overtone, which is the $3^{rd}$ harmonic, given by $3 n_0 = \frac{3v}{4 \ell}$.
According to the problem, the frequency of the $2^{nd}$ harmonic of the closed pipe is $200 \,Hz$ higher than $n$:
$\frac{3v}{4 \ell} - n = 200$
Since $n = \frac{v}{2 \ell}$, we have $\frac{v}{4 \ell} = \frac{n}{2}$.
Substituting this into the equation: $3(\frac{n}{2}) - n = 200$.
$1.5n - n = 200 \implies 0.5n = 200$.
$n = 400 \,Hz$.

(C) For an open organ pipe of length $L$,the frequency of the $P^{\text{th}}$ overtone is given by $f_{\text{open}} = (P+1) \frac{v}{2L}$.
For a closed organ pipe of the same length $L$,the frequency of the $P^{\text{th}}$ overtone is given by $f_{\text{closed}} = (2P+1) \frac{v}{4L}$.
Taking the ratio of the frequency of the $P^{\text{th}}$ overtone of the open pipe to that of the closed pipe:
Ratio $= \frac{(P+1) \frac{v}{2L}}{(2P+1) \frac{v}{4L}}$.
Ratio $= \frac{P+1}{2L} \times \frac{4L}{2P+1} = \frac{2(P+1)}{2P+1}$.

(D) For a closed pipe of length $l$,the fundamental frequency is given by $f_0 = \frac{v}{4l}$,where $v$ is the speed of sound.
The time $t$ taken by the sound wave to travel the length of the pipe $l$ is $t = \frac{l}{v}$,which implies $v = \frac{l}{t}$.
Substituting the value of $v$ into the frequency formula:
$f_0 = \frac{l/t}{4l} = \frac{1}{4t}$.
Thus,the frequency of vibration is $(4t)^{-1}$.

(D) For an open pipe,the fundamental frequency is given by $n = \frac{v}{2l}$,where $v$ is the speed of sound and $l$ is the length of the pipe.
For the two pipes,we have $l_1 = \frac{v}{2n_1}$ and $l_2 = \frac{v}{2n_2}$.
When the two pipes are joined,the new length becomes $L = l_1 + l_2$.
The new fundamental frequency $n'$ is given by $n' = \frac{v}{2L} = \frac{v}{2(l_1 + l_2)}$.
Substituting the values of $l_1$ and $l_2$:
$n' = \frac{v}{2(\frac{v}{2n_1} + \frac{v}{2n_2})} = \frac{v}{\frac{v}{n_1} + \frac{v}{n_2}} = \frac{1}{\frac{1}{n_1} + \frac{1}{n_2}} = \frac{n_1 n_2}{n_1 + n_2}$.

(C) For a pipe closed at one end,the natural frequencies are given by $f_k = (2k - 1) \frac{V}{4L}$,where $k = 1, 2, 3, \dots$ is the mode number.
Given $V = 332 \ m/s$ and $L = 83 \ cm = 0.83 \ m$.
The fundamental frequency $(k=1)$ is $f_1 = \frac{V}{4L} = \frac{332}{4 \times 0.83} = \frac{332}{3.32} = 100 \ Hz$.
The possible frequencies are odd multiples of the fundamental frequency: $f_k = (2k - 1) \times 100 \ Hz$.
We need to find the number of frequencies such that $f_k < 1000 \ Hz$.
$(2k - 1) \times 100 < 1000 \implies 2k - 1 < 10 \implies 2k < 11 \implies k < 5.5$.
Since $k$ must be a positive integer,$k$ can be $1, 2, 3, 4, 5$.
The frequencies are $100 \ Hz, 300 \ Hz, 500 \ Hz, 700 \ Hz, 900 \ Hz$.
There are $5$ such frequencies.

(D) For a pipe closed at one end,the fundamental frequency is given by $n_1 = \frac{V}{4L} = 100 \ Hz$.
When the same pipe is open at both ends,the new fundamental frequency $n'_1$ is given by $n'_1 = \frac{V}{2L}$.
Comparing the two expressions,we get $n'_1 = 2 \times \frac{V}{4L} = 2 \times 100 \ Hz = 200 \ Hz$.
In an open pipe,all harmonics (integer multiples of the fundamental frequency) are produced.
Therefore,the frequencies produced are $200 \ Hz, 400 \ Hz, 600 \ Hz, 800 \ Hz, .....$.

(B) For an open organ pipe,the end correction $\Delta L$ is related to the inner radius $r$ by the formula $\Delta L = 0.6 \times r$ at each end. Since an open pipe has two ends,the total end correction is $\Delta L_{total} = 2 \times (0.6 \times r) = 1.2 \times r$.
Given that the total end correction $\Delta L = 0.8 \ cm$,we have:
$0.8 = 1.2 \times r$
$r = \frac{0.8}{1.2} \ cm$
$r = \frac{8}{12} \ cm = \frac{2}{3} \ cm$.

(D) Let $f_0 = \frac{v}{2L}$ be the fundamental frequency of the open pipe.
Its second overtone is $3f_0 = \frac{3v}{2L}$.
Let $f_c = \frac{v}{4L}$ be the fundamental frequency of the closed pipe.
The frequencies of a closed pipe are given by $(2n-1)f_c$,where $n=1, 2, 3, \dots$.
The third overtone corresponds to $n=4$,so $(f_3)_{\text{closed}} = (2(4)-1)f_c = 7f_c = \frac{7v}{4L}$.
According to the problem,$(f_3)_{\text{closed}} - (3f_0) = 150 \ Hz$.
Substituting the expressions: $\frac{7v}{4L} - \frac{3v}{2L} = 150$.
$\frac{7v}{4L} - \frac{6v}{4L} = 150$.
$\frac{v}{4L} = 150$.
Since the fundamental frequency of the open pipe is $f_0 = \frac{v}{2L}$,we have $f_0 = 2 \times \frac{v}{4L} = 2 \times 150 = 300 \ Hz$.

(A) In a resonance tube (which acts as a closed pipe),the resonance lengths are given by $l_1 = \frac{\lambda}{4}$ and $l_2 = \frac{3\lambda}{4}$.
The difference between the two resonance lengths is $l_2 - l_1 = \frac{\lambda}{2}$.
Given $l_1 = 16 \ cm = 0.16 \ m$ and $l_2 = 49 \ cm = 0.49 \ m$.
Therefore,$\frac{\lambda}{2} = 0.49 \ m - 0.16 \ m = 0.33 \ m$.
This implies $\lambda = 0.66 \ m$.
Using the wave equation $v = n\lambda$,where $v = 330 \ m/s$ is the velocity of sound and $n$ is the frequency:
$n = \frac{v}{\lambda} = \frac{330}{0.66} = 500 \ Hz$.

(A) The fundamental frequency of a pipe closed at one end is $n_{1} = \frac{v}{4 l_{1}}$,which gives $l_{1} = \frac{v}{4 n_{1}}$.
The fundamental frequency of a pipe open at both ends is $n_{2} = \frac{v}{2 l_{2}}$,which gives $l_{2} = \frac{v}{2 n_{2}}$.
When both pipes are joined end to end,the new pipe is closed at one end and open at the other,with a total length $L = l_{1} + l_{2}$.
The fundamental frequency $n$ of this new closed pipe is given by $n = \frac{v}{4 L} = \frac{v}{4 (l_{1} + l_{2})}$.
Substituting the values of $l_{1}$ and $l_{2}$: $\frac{1}{4 n} = \frac{1}{4 n_{1}} + \frac{1}{2 n_{2}}$.
Multiplying by $4$: $\frac{1}{n} = \frac{1}{n_{1}} + \frac{2}{n_{2}} = \frac{n_{2} + 2 n_{1}}{n_{1} n_{2}}$.
Therefore,$n = \frac{n_{1} n_{2}}{n_{2} + 2 n_{1}}$.

(A) In the fundamental mode of a pipe closed at one end,the length of the pipe $l$ corresponds to one-fourth of the wavelength,i.e.,$l = \frac{\lambda}{4}$,which implies $\lambda = 4l$.
Given that the time taken by the wave to travel the length of the pipe $l$ is $t = 0.01 \ s$,the speed of sound $v$ is given by $v = \frac{l}{t}$.
The fundamental frequency $n$ is given by $n = \frac{v}{\lambda}$.
Substituting the values,we get $n = \frac{l/t}{4l} = \frac{1}{4t}$.
Substituting $t = 0.01 \ s$,we get $n = \frac{1}{4 \times 0.01} = \frac{1}{0.04} = 25 \ Hz$.

(C) For a pipe closed at one end,the allowed frequencies are given by $f_n = n \cdot f_1$,where $n$ is an odd integer $(n = 1, 3, 5, \dots)$.
Given the fundamental frequency $f_1 = 100 \ Hz$,the subsequent frequencies are $f_2 = 3 \times 100 \ Hz = 300 \ Hz$ and $f_3 = 5 \times 100 \ Hz = 500 \ Hz$.
Since the frequencies follow the ratio $1:3:5$,the pipe must be closed at one end and open at the other.

(A) Let the end correction be $x$. The resonance condition for a pipe closed at one end is given by $l_n + x = (2n-1) \frac{\lambda}{4}$.
For the first resonance $(n=1)$: $l_1 + x = \frac{\lambda}{4}$.
For the second resonance $(n=2)$: $l_2 + x = \frac{3\lambda}{4}$.
Dividing the two equations: $\frac{l_2 + x}{l_1 + x} = 3$.
$l_2 + x = 3l_1 + 3x$.
$2x = l_2 - 3l_1$.
$x = \frac{l_2 - 3l_1}{2}$.
Substituting the given values: $x = \frac{70.2 - 3(22.7)}{2} = \frac{70.2 - 68.1}{2} = \frac{2.1}{2} = 1.05 \,cm$.

(B) When sound waves fall normally on a perfectly reflecting wall, a stationary wave is formed due to the superposition of incident and reflected waves.
At a perfectly reflecting wall, a node is formed because the wall is a rigid boundary.
The nodes are points of zero displacement, while antinodes are points of maximum displacement (maximum amplitude).
The distance of the first antinode from the wall (node) is given by $d = \lambda/4$.
First, calculate the wavelength $\lambda$ using the formula $\lambda = v/f$, where $v = 300 \,ms^{-1}$ and $f = 600 \,Hz$.
$\lambda = 300/600 = 0.5 \,m$.
Now, calculate the distance $d = \lambda/4 = 0.5/4 = 0.125 \,m = 1/8 \,m$.
Thus, the shortest distance from the wall where particles have maximum amplitude is $1/8 \,m$.

(A) The speed of sound in a gas is given by $v = \sqrt{\frac{\gamma RT}{M}}$,which shows that the speed of sound $v$ is directly proportional to the square root of the absolute temperature $T$ $(v \propto \sqrt{T})$.
As the temperature increases,the speed of sound $v$ increases.
The frequency $n$ of an organ pipe is given by $n = \frac{v}{\lambda}$,where $\lambda$ is the wavelength determined by the length of the pipe,which remains constant.
Since $v$ increases and $\lambda$ remains constant,the frequency $n$ increases.

(C) Statement $A$: In a stationary wave,the distance between two consecutive nodes is $\frac{\lambda}{2}$ and the distance between two consecutive antinodes is also $\frac{\lambda}{2}$. Thus,statement $A$ is true.
Statement $B$: At the open end of an organ pipe,the air is free to vibrate,which means the displacement is maximum (antinode). Since pressure variation is minimum at the displacement antinode,a pressure node is formed at the open end. Thus,statement $B$ is true.
Therefore,both statements $A$ and $B$ are true.

(D) For a closed pipe of length $L$,the fundamental frequency is given by $f_c = \frac{v}{4L}$.
For an open pipe of length $L' = XL$,the frequencies are given by $f_n = \frac{n v}{2L'}$,where $n = 1, 2, 3, \dots$.
The first overtone is $n=2$ and the second overtone is $n=3$.
Therefore,the second overtone frequency of the open pipe is $f_o = \frac{3v}{2(XL)}$.
According to the problem,$f_c = f_o$,so $\frac{v}{4L} = \frac{3v}{2XL}$.
Canceling $v$ and $L$ from both sides,we get $\frac{1}{4} = \frac{3}{2X}$.
Solving for $X$,we get $2X = 12$,which implies $X = 6$.

(B) For a closed pipe of length $L_c$,the frequency of the $n^{th}$ overtone is given by $f_c = \frac{(2n+1)v}{4L_c}$,where $n$ is the overtone number. For the fourth overtone $(n=4)$,$f_c = \frac{(2(4)+1)v}{4L_c} = \frac{9v}{4L_c}$.
For an open pipe of length $L_o$,the frequency of the $m^{th}$ overtone is given by $f_o = \frac{(m+1)v}{2L_o}$,where $m$ is the overtone number. For the fifth overtone $(m=5)$,$f_o = \frac{(5+1)v}{2L_o} = \frac{6v}{2L_o} = \frac{3v}{L_o}$.
Given that the frequencies are in unison,$f_c = f_o$,so $\frac{9v}{4L_c} = \frac{3v}{L_o}$.
Simplifying the equation: $\frac{9}{4L_c} = \frac{3}{L_o} \implies \frac{L_c}{L_o} = \frac{9}{4 \times 3} = \frac{9}{12} = \frac{3}{4}$.
Thus,the ratio of the length of the closed pipe to the open pipe is $3: 4$.

(D) The speed of sound $v = 330 \ m/s$ and frequency $f = 330 \ Hz$. The wavelength $\lambda = v/f = 330/330 = 1 \ m$.
When one end of an open pipe is dipped in water,it acts as a pipe closed at one end.
The length of the air column is $L' = L - x$,where $L = 1.5 \ m$ is the total length and $x$ is the length immersed.
For a pipe closed at one end,the frequencies of harmonics are given by $f_n = (2n-1) \frac{v}{4L'}$,where $n=1$ is the fundamental,$n=2$ is the $1^{\text{st}}$ overtone,and $n=3$ is the $2^{\text{nd}}$ overtone.
For the $2^{\text{nd}}$ overtone,$n=3$,so $f_3 = 5 \frac{v}{4L'}$.
Given $f_3 = 330 \ Hz$,we have $330 = 5 \times \frac{330}{4L'}$.
This simplifies to $1 = \frac{5}{4L'}$,which gives $4L' = 5$,or $L' = 1.25 \ m$.
Since $L' = L - x$,we have $1.25 = 1.5 - x$.
Therefore,$x = 1.5 - 1.25 = 0.25 \ m$.

(C) For an air column of length $L$ closed at one end,the frequency of the $n^{\text{th}}$ overtone is given by $f_{c} = (2n + 1) \frac{v}{4L}$. For the $5^{\text{th}}$ overtone,$n = 5$,so $f_{c} = (2(5) + 1) \frac{v}{4L} = 11 \frac{v}{4L}$.
For an air column of length $L$ open at both ends,the frequency of the $n^{\text{th}}$ overtone is given by $f_{o} = (n + 1) \frac{v}{2L}$. For the $5^{\text{th}}$ overtone,$n = 5$,so $f_{o} = (5 + 1) \frac{v}{2L} = 6 \frac{v}{2L} = 12 \frac{v}{4L}$.
The ratio of the frequencies is $\frac{f_{c}}{f_{o}} = \frac{11v/4L}{12v/4L} = \frac{11}{12}$.

(C) Let the length of both pipes be $L$. The speed of sound is $v$.
For an open pipe,the frequencies are $f_{open, n} = n(v/2L)$. The first overtone $(n=2)$ is $f_{open, 1st} = 2(v/2L) = v/L$.
For a closed pipe,the frequencies are $f_{closed, n} = (2n-1)(v/4L)$. The first overtone $(n=2)$ is $f_{closed, 1st} = 3(v/4L)$.
Given that the beat frequency is $4$ Hz: $|v/L - 3v/4L| = 4 \implies v/4L = 4 \implies v/L = 16$.
Now,the new length of the open pipe is $L' = L/2$ and the new length of the closed pipe is $L'' = 2L$.
The fundamental frequency of the new open pipe is $f'_{open} = v/(2L') = v/(2(L/2)) = v/L = 16$ Hz.
The fundamental frequency of the new closed pipe is $f''_{closed} = v/(4L'') = v/(4(2L)) = v/8L = 16/8 = 2$ Hz.
The number of beats produced is $|f'_{open} - f''_{closed}| = |16 - 2| = 14$ Hz.

(D) For an open pipe of length $L$,the fundamental frequency is $f_{open, 1} = \frac{v}{2L}$. The first overtone is the second harmonic,given by $f_{open, 2} = 2 \times f_{open, 1} = 2 \times \frac{v}{2L} = \frac{v}{L}$.
For a closed pipe of length $L$,the fundamental frequency is $f_{closed, 1} = \frac{v}{4L}$. The first overtone is the third harmonic,given by $f_{closed, 2} = 3 \times f_{closed, 1} = 3 \times \frac{v}{4L} = \frac{3v}{4L}$.
The ratio of the frequency of the first overtone of the open pipe to that of the closed pipe is $\frac{f_{open, 2}}{f_{closed, 2}} = \frac{v/L}{3v/4L} = \frac{1}{1} \times \frac{4}{3} = \frac{4}{3}$.