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

(D) The tuning fork forms a resonance with the air column in a pipe closed at one end. The resonance condition is given by $n = \frac{(2N - 1)v}{4l}$,where $N = 1, 2, 3, \dots$ represents the mode of vibration.
Given $n = 340 \ Hz$ and $v = 340 \ m/s$,the length of the air column $l$ is:
$l = \frac{(2N - 1)v}{4n} = \frac{(2N - 1) \times 340}{4 \times 340} = \frac{2N - 1}{4} \ m = (2N - 1) \times 25 \ cm$.
For $N = 1, l_1 = 25 \ cm$.
For $N = 2, l_2 = 75 \ cm$.
For $N = 3, l_3 = 125 \ cm$ (not possible as the tube height is $120 \ cm$).
The height of the water column $h$ is given by $h = H_{tube} - l$.
For $l_1 = 25 \ cm$,$h_1 = 120 - 25 = 95 \ cm$.
For $l_2 = 75 \ cm$,$h_2 = 120 - 75 = 45 \ cm$.
The minimum height of water required for resonance is $45 \ cm$.

(C) The maximum frequency a normal human can hear is $20,000 \ Hz$.
For a closed organ pipe,the frequency of the $N^{th}$ mode is given by $f_N = (2N - 1)f_1$,where $f_1 = 1500 \ Hz$ is the fundamental frequency.
We require $f_N \le 20,000 \ Hz$.
$(2N - 1) \times 1500 \le 20,000$
$2N - 1 \le \frac{20,000}{1500} \approx 13.33$
$2N \le 14.33 \implies N \le 7.16$.
Since $N$ must be an integer,the maximum mode number is $N = 7$.
The number of overtones is given by $(N - 1)$.
Therefore,the number of overtones $= 7 - 1 = 6$.

(C) In a pipe open at both ends,the open ends are always displacement antinodes. Since pressure variation is inversely related to displacement,the pressure variation is minimum (zero) at the open ends. Therefore,the statement that 'Pressure change will be maximum at both ends' is false.

(C) The fundamental frequency $(f_1)$ of a pipe closed at one end is given by $f_1 = \frac{v}{4L}$.
Given: $v = 340 \, m s^{-1}$ and $L = 85 \, cm = 0.85 \, m$.
Substituting the values: $f_1 = \frac{340}{4 \times 0.85} = \frac{340}{3.4} = 100 \, Hz$.
The natural frequencies of a closed pipe are odd harmonics of the fundamental frequency: $f_n = (2n - 1)f_1$,where $n = 1, 2, 3, \dots$.
The frequencies are: $100 \, Hz, 300 \, Hz, 500 \, Hz, 700 \, Hz, 900 \, Hz, 1100 \, Hz, 1300 \, Hz, \dots$.
We need to find the number of frequencies below $1250 \, Hz$.
The frequencies are $100, 300, 500, 700, 900, 1100$.
Counting these,we get $6$ possible natural oscillations.

(A) For a string fixed at both ends,the resonant frequencies are given by $v_n = n \cdot v_0$,where $v_0$ is the fundamental frequency (lowest resonant frequency) and $n = 1, 2, 3, \dots$.
The difference between any two consecutive resonant frequencies is $\Delta v = v_{n+1} - v_n = (n+1)v_0 - nv_0 = v_0$.
Given that $420\, Hz$ and $315\, Hz$ are resonant frequencies with no other resonant frequencies between them,they must be consecutive harmonics.
Therefore,the fundamental frequency $v_0$ is the difference between these two frequencies:
$v_0 = 420\, Hz - 315\, Hz = 105\, Hz$.

(D) For an open organ pipe of length $L'$,the frequency of the $n^{th}$ harmonic is given by $f_n = n \frac{v}{2L'}$,where $n = 1, 2, 3, \dots$. The second overtone corresponds to the $3^{rd}$ harmonic $(n=3)$. Thus,$f_{open} = 3 \frac{v}{2L'}$.
For a closed organ pipe of length $L$,the frequency of the $n^{th}$ harmonic is given by $f_n = n \frac{v}{4L}$,where $n = 1, 3, 5, \dots$. The first overtone corresponds to the $3^{rd}$ harmonic $(n=3)$. Thus,$f_{closed} = 3 \frac{v}{4L}$.
According to the problem,the frequencies are equal:
$3 \frac{v}{2L'} = 3 \frac{v}{4L}$
Canceling $3v$ from both sides:
$\frac{1}{2L'} = \frac{1}{4L}$
Solving for $L'$:
$2L' = 4L \implies L' = 2L \; m$.

(B) For an air column closed at one end,the resonance occurs at odd harmonics.
The fundamental frequency (first harmonic) corresponds to the smallest length $l_1$:
$l_1 = \frac{\lambda}{4} = 50\, cm$
The next resonance (third harmonic) occurs at length $l_2$:
$l_2 = \frac{3\lambda}{4}$
Substituting $\lambda = 4 \times 50\, cm = 200\, cm$:
$l_2 = 3 \times \left(\frac{200}{4}\right) = 3 \times 50 = 150\, cm$.

(C) The fundamental frequency of an open organ pipe of length $L$ is given by $n = \frac{v}{2L}$,where $v$ is the speed of sound.
From this,the length of the pipe is $L = \frac{v}{2n}$.
For the two pipes with fundamental frequencies $n_{1}$ and $n_{2}$,their lengths are $L_{1} = \frac{v}{2n_{1}}$ and $L_{2} = \frac{v}{2n_{2}}$.
When these two pipes are joined in series,the total length of the new pipe becomes $L_{new} = L_{1} + L_{2}$.
The fundamental frequency $n$ of the new pipe is $n = \frac{v}{2L_{new}} = \frac{v}{2(L_{1} + L_{2})}$.
Substituting the values of $L_{1}$ and $L_{2}$:
$n = \frac{v}{2(\frac{v}{2n_{1}} + \frac{v}{2n_{2}})} = \frac{v}{v(\frac{1}{n_{1}} + \frac{1}{n_{2}})} = \frac{1}{\frac{n_{1} + n_{2}}{n_{1}n_{2}}}$.
Therefore,$n = \frac{n_{1}n_{2}}{n_{1} + n_{2}}$.

(B) For a tube closed at one end,the allowed frequencies (harmonics) are given by $f_n = n \cdot f_0$,where $n$ must be an odd integer $(n = 1, 3, 5, \dots)$ and $f_0$ is the fundamental frequency.
Two successive harmonics for such a tube differ by $2f_0$.
Given the two nearest harmonics are $f_1 = 220\, Hz$ and $f_2 = 260\, Hz$.
The difference between these two harmonics is $2f_0 = 260\, Hz - 220\, Hz = 40\, Hz$.
Therefore,the fundamental frequency is $f_0 = \frac{40\, Hz}{2} = 20\, Hz$.

(A) The fundamental frequency of an open organ pipe of length $l'$ is given by $f_{open} = \frac{v}{2l'}$.
The frequencies of a closed organ pipe of length $l$ are given by $f_n = \frac{nv}{4l}$,where $n = 1, 3, 5, \dots$ (odd harmonics).
The third harmonic of a closed organ pipe corresponds to $n = 3$,so $f_{closed, 3} = \frac{3v}{4l}$.
According to the problem,$f_{open} = f_{closed, 3}$.
Therefore,$\frac{v}{2l'} = \frac{3v}{4l}$.
Simplifying for $l'$,we get $l' = \frac{4l}{6} = \frac{2l}{3}$.
Given $l = 20 \ cm$,we have $l' = \frac{2 \times 20}{3} = \frac{40}{3} \approx 13.33 \ cm$.

(B) In a resonance tube experiment,the distance between two successive resonance positions is equal to half the wavelength of the sound wave,i.e.,$\frac{\lambda}{2} = L_2 - L_1$.
Given,$L_1 = 20\,cm = 0.20\,m$ and $L_2 = 73\,cm = 0.73\,m$.
Therefore,$\frac{\lambda}{2} = 0.73\,m - 0.20\,m = 0.53\,m$.
This implies $\lambda = 2 \times 0.53\,m = 1.06\,m$.
The velocity of sound $v$ is given by the formula $v = f \lambda$,where $f$ is the frequency of the tuning fork.
Given $f = 320\,Hz$.
$v = 320 \times 1.06 = 339.2\,m/s$.
Rounding to the nearest integer,the velocity of sound is $339\,m/s$.

(C) For a closed organ pipe (resonance tube),the resonance lengths are given by $l_n = (2n - 1) \frac{\lambda}{4}$,where $n = 1, 2, 3, \dots$
For the first resonance $(n=1)$,$l_1 = \frac{\lambda}{4} = 16\,cm$.
For the second resonance $(n=2)$,$l_2 = 3 \frac{\lambda}{4} = 3 \times 16 = 48\,cm$.
Thus,the next resonance is obtained at $48\,cm$.

(C) For an open pipe of length $l$,the fundamental frequency is given by $f_1 = v / (2l)$,where $v$ is the speed of sound.
When $3/4$th of the pipe is submerged in water,the effective length of the air column becomes $l' = l - (3/4)l = l/4$.
This air column acts as a pipe closed at one end.
The fundamental frequency of a pipe closed at one end is given by $f_2 = v / (4l')$.
Substituting $l' = l/4$,we get $f_2 = v / (4 \times (l/4)) = v/l$.
Comparing $f_1$ and $f_2$,we have $f_2 = 2 \times (v / (2l)) = 2f_1$.
Therefore,the ratio $f_1/f_2 = f_1 / (2f_1) = 1/2 = 0.5$.

(C) The fundamental frequency of a closed pipe of length $l$ is given by $n = \frac{v}{4l} = 220 \ Hz$.
From this,we find the speed of sound $v = 220 \times 4l = 880l \ m/s$.
When $\frac{1}{4}$ of the pipe is filled with water,the length of the air column becomes $l' = l - \frac{1}{4}l = \frac{3}{4}l$.
The new fundamental frequency of this air column is $n' = \frac{v}{4l'} = \frac{v}{4(\frac{3}{4}l)} = \frac{v}{3l}$.
The first overtone of a closed pipe is $3$ times the fundamental frequency.
Therefore,the frequency of the first overtone is $f_{1st} = 3 \times n' = 3 \times \frac{v}{3l} = \frac{v}{l}$.
Substituting $v = 880l$,we get $f_{1st} = \frac{880l}{l} = 880 \ Hz$.

(B) The tube acts as a closed organ pipe because one end is closed by the water surface and the other is open. The resonance condition for a closed pipe is given by $l = \frac{(2N - 1)v}{4f}$,where $N$ is the order of the resonance,$v = 330 \ m/s$ is the speed of sound,and $f = 660 \ Hz$ is the frequency.
Substituting the values: $1.5 = \frac{(2N - 1) \times 330}{4 \times 660}$.
Simplifying the equation: $1.5 = \frac{(2N - 1)}{8}$.
$12 = 2N - 1$.
$2N = 13$.
$N = 6.5$.
Since $N$ must be an integer,the possible values for $N$ are $1, 2, 3, 4, 5, 6$. Thus,there are $6$ resonances.

(B) For an open organ pipe,the frequency of the $n$th overtone is given by $f_n = (n+1)f_0$,where $f_0$ is the fundamental frequency.
Thus,the $5$th overtone corresponds to the $6$th harmonic mode ($n=5$,so harmonic number $p = n+1 = 6$).
In an open organ pipe vibrating in the $p$th harmonic mode:
Number of nodes $(N)$ = $p = 6$.
Number of antinodes $(A)$ = $p + 1 = 6 + 1 = 7$.
Therefore,there are $6$ nodes and $7$ antinodes.

(A) For a string fixed at both ends,the $n^{th}$ overtone corresponds to the $(n+1)^{th}$ harmonic.
The length of the string $L$ is related to the wavelength $\lambda$ by the formula: $L = (n+1) \frac{\lambda}{2} \dots (1)$.
The distance between an adjacent node and antinode is always $\frac{\lambda}{4}$. According to the problem,this distance is $d$. Therefore,$\frac{\lambda}{4} = d$,which implies $\lambda = 4d$.
Substituting $\lambda = 4d$ into equation $(1)$:
$L = (n+1) \frac{4d}{2}$
$L = 2d(n+1)$.

(A) Given frequency of sound waves $\nu = 660 \, Hz$ and velocity $v = 330 \, m/s$.
The wavelength of the sound waves is calculated as $\lambda = \frac{v}{\nu} = \frac{330}{660} = 0.5 \, m$.
When sound waves reflect from a perfectly reflecting wall,stationary waves are formed due to the interference between the incident and reflected waves.
$A$ displacement node is always formed at the reflecting surface (the wall).
The air particles have maximum amplitude of vibration at the antinodes.
The shortest distance from the wall (node) to the first point of maximum amplitude (antinode) is given by $d = \frac{\lambda}{4}$.
Substituting the value of $\lambda$,we get $d = \frac{0.5}{4} = 0.125 \, m$.

(B) For an open organ pipe,the length $L$ is given by $L = \frac{n \lambda}{2}$,where $\lambda$ is the wavelength and $n$ is an integer $(n = 1, 2, 3, \dots)$.
For the second harmonic mode,$n = 2$. Therefore,$L = \frac{2 \lambda}{2} = \lambda$.
In an open organ pipe,the ends are always antinodes $(A)$ for displacement,which correspond to nodes for pressure variation. The nodes $(N)$ for displacement are where the pressure variation is maximum.
In the second harmonic mode,the standing wave pattern consists of three antinodes and two nodes. The nodes are located at distances of $\frac{\lambda}{4}$ and $\frac{3\lambda}{4}$ from one end. Since $L = \lambda$,these positions are $\frac{L}{4}$ and $\frac{3L}{4}$ from one end.
Thus,the pressure vibration is maximum at distances of $\frac{L}{4}$ from either end inside the tube.

(C) The fundamental frequency of an open organ pipe of length $l$ is given by $f_1 = \frac{v}{2l}$.
The fundamental frequency of an open organ pipe of length $(l+x)$ is given by $f_2 = \frac{v}{2(l+x)}$.
The beat frequency is the difference between the two frequencies: $f_{beat} = |f_1 - f_2|$.
$f_{beat} = \frac{v}{2l} - \frac{v}{2(l+x)} = \frac{v}{2} \left( \frac{1}{l} - \frac{1}{l+x} \right)$.
$f_{beat} = \frac{v}{2} \left( \frac{l+x-l}{l(l+x)} \right) = \frac{v}{2} \left( \frac{x}{l(l+x)} \right)$.
Since $x << l$,we can approximate $l+x \approx l$.
Therefore,$f_{beat} \approx \frac{v}{2} \left( \frac{x}{l^2} \right) = \frac{vx}{2l^2}$.

(C) The wavelength of the sound wave is given by $\lambda = \frac{v}{\nu}$.
Substituting the values,$\lambda = \frac{340}{340} = 1 \, m = 100 \, cm$.
For a closed organ pipe (resonance tube),resonance occurs when the length of the air column $l$ satisfies $l = \frac{(2n-1)\lambda}{4}$ for $n = 1, 2, 3, \dots$.
Possible lengths of the air column are:
For $n=1$: $l_1 = \frac{\lambda}{4} = \frac{100}{4} = 25 \, cm$.
For $n=2$: $l_2 = \frac{3\lambda}{4} = \frac{300}{4} = 75 \, cm$.
For $n=3$: $l_3 = \frac{5\lambda}{4} = \frac{500}{4} = 125 \, cm$.
Since the total length of the tube is $120 \, cm$,the air column length $l_3 = 125 \, cm$ is not possible.
As water is poured,the air column length decreases. The first resonance occurs at $l_2 = 75 \, cm$ (as $l_1 = 25 \, cm$ would require more water,and we seek the minimum height of water for the first possible resonance encountered).
The height of the water column $h$ is given by $h = \text{Total length} - \text{Air column length} = 120 \, cm - 75 \, cm = 45 \, cm$.

(D) For a closed pipe of length $L = 105 \, cm$,the frequency of the $n^{th}$ overtone is given by $f_n = (2n+1) \frac{v}{4L}$.
The third overtone corresponds to $n = 3$,so the length $L$ is related to the wavelength $\lambda$ by $L = \frac{(2n+1)\lambda}{4} = \frac{7\lambda}{4}$.
Given $L = 105 \, cm$,we have $105 = \frac{7\lambda}{4}$,which gives $\lambda = \frac{105 \times 4}{7} = 60 \, cm$.
In a closed pipe,the closed end is a displacement node (pressure antinode) and the open end is a displacement antinode (pressure node).
Pressure nodes occur at distances $x = \frac{\lambda}{4}, \frac{3\lambda}{4}, \frac{5\lambda}{4}, \dots$ from the closed end.
For $\lambda = 60 \, cm$,the positions of pressure nodes are $x = 15 \, cm, 45 \, cm, 75 \, cm, \dots$
Comparing with the given options,$45 \, cm$ is a pressure node.

(D) Given: Speed of sound in air,$v = 350 \, m/s$. Frequency of tuning fork,$f = 1750 \, Hz$.
For a pipe closed at one end,the resonant frequencies are given by $f_n = \frac{(2n-1)v}{4L}$,where $n = 1, 2, 3, \dots$ and $L$ is the length of the air column.
Initially,$L_1 = 25 \, cm = 0.25 \, m$. Substituting the values: $1750 = \frac{(2n-1) \times 350}{4 \times 0.25}$.
$1750 = \frac{(2n-1) \times 350}{1} \implies 2n-1 = \frac{1750}{350} = 5 \implies 2n = 6 \implies n = 3$.
To resonate again with the same tuning fork,the next resonant length $L_2$ corresponds to $n = 4$.
$1750 = \frac{(2 \times 4 - 1) \times 350}{4 \times L_2} \implies 1750 = \frac{7 \times 350}{4 \times L_2}$.
$L_2 = \frac{7 \times 350}{4 \times 1750} = \frac{2450}{7000} = 0.35 \, m = 35 \, cm$.
The minimum distance the pipe should be raised is $\Delta L = L_2 - L_1 = 35 \, cm - 25 \, cm = 10 \, cm$.

(A) In a closed organ pipe of length $l$,the allowed frequencies are given by $\nu_n = (2n + 1) \frac{v}{4l}$,where $n = 0, 1, 2, \dots$ represents the mode of vibration.
For $n = 0$,$\nu_0 = \frac{v}{4l}$ (Fundamental frequency or first harmonic).
For $n = 1$,$\nu_1 = 3 \frac{v}{4l}$ (First overtone or third harmonic).
For $n = 2$,$\nu_2 = 5 \frac{v}{4l}$ (Second overtone or fifth harmonic).
In general,for the $p^{th}$ overtone,we set $n = p$. The frequency is $\nu_p = (2p + 1) \frac{v}{4l}$.
Since the fundamental frequency $\nu_0 = \frac{v}{4l}$,we have $\nu_p = (2p + 1) \nu_0$.
Thus,the $p^{th}$ overtone corresponds to the $(2p + 1)^{th}$ harmonic.

(C) For a closed organ pipe with end correction,the frequency of the $n^{th}$ mode is given by $\nu_{n}^{\prime} = \frac{(2n-1)v}{4(L+0.6r_1)}$,where $0.6r_1$ is the end correction.
For the fundamental mode $(n=1)$,$\nu_{1}^{\prime} = \frac{v}{4(L+0.6r_1)}$.
For an open organ pipe with end correction,the frequency of the $m^{th}$ mode is given by $\nu_{m} = \frac{mv}{2(L+2 \times 0.6r_2)}$,where $0.6r_2$ is the end correction at each end.
For the first overtone $(m=2)$,$\nu_{2} = \frac{2v}{2(L+1.2r_2)} = \frac{v}{L+1.2r_2}$.
Since both pipes resonate with the same tuning fork,$\nu_{1}^{\prime} = \nu_{2}$.
Therefore,$\frac{v}{4(L+0.6r_1)} = \frac{v}{L+1.2r_2}$.
$L + 1.2r_2 = 4L + 2.4r_1$.
$1.2r_2 - 2.4r_1 = 3L$.
Dividing by $1.2$,we get $r_2 - 2r_1 = 2.5L$.

(C) Let $L'$ and $L$ be the lengths of the closed and open organ pipes,respectively.
For a closed organ pipe,the frequency of the $n^{th}$ harmonic is given by $\nu_n' = \frac{n v}{4 L'}$,where $n$ must be an odd integer $(n = 1, 3, 5, \dots)$.
The first overtone of a closed pipe corresponds to the $3^{rd}$ harmonic $(n=3)$,so $\nu_{1st, closed} = \frac{3 v}{4 L'}$.
For an open organ pipe,the frequency of the $m^{th}$ harmonic is given by $\nu_m = \frac{m v}{2 L}$,where $m = 1, 2, 3, \dots$.
The first overtone of an open pipe corresponds to the $2^{nd}$ harmonic $(m=2)$,so $\nu_{1st, open} = \frac{2 v}{2 L} = \frac{v}{L}$.
Given $\nu_{1st, closed} = \nu_{1st, open}$,we have $\frac{3 v}{4 L'} = \frac{v}{L}$,which implies $\frac{L'}{L} = \frac{3}{4}$.
Given the $n^{th}$ harmonic of the closed pipe equals the $m^{th}$ harmonic of the open pipe: $\frac{n v}{4 L'} = \frac{m v}{2 L}$.
Substituting $\frac{L'}{L} = \frac{3}{4}$,we get $\frac{n}{4(3/4)} = \frac{m}{2} \implies \frac{n}{3} = \frac{m}{2} \implies 2n = 3m$.
Testing the options: For $n=9$ and $m=6$,$2(9) = 18$ and $3(6) = 18$. Thus,$n=9$ and $m=6$ is the correct pair.

(C) In a resonance tube,the length of the air column for the first resonance is $l_1 = \frac{\lambda}{4} - e$,where $e$ is the end correction.
For the second resonance,the length is $l_2 = \frac{3\lambda}{4} - e$.
Subtracting the two equations: $l_2 - l_1 = \frac{3\lambda}{4} - \frac{\lambda}{4} = \frac{2\lambda}{4} = \frac{\lambda}{2}$.
Thus,$\lambda = 2(l_2 - l_1)$.
Substituting $\lambda$ into the first resonance equation: $l_1 = \frac{2(l_2 - l_1)}{4} - e = \frac{l_2 - l_1}{2} - e$.
Therefore,the end correction $e = \frac{l_2 - l_1}{2} - l_1 = \frac{l_2 - 3l_1}{2}$.
The displacement antinode is located at a distance $e$ above the open end of the tube. Hence,the distance is $\frac{l_2 - 3l_1}{2}$.

(B) For a closed organ pipe,the frequencies of harmonics are given by $f_n = (2n-1)f_0$,where $n=1, 2, 3, ...$ corresponds to the fundamental,$1^{st}$ overtone,$2^{nd}$ overtone,and so on.
For the $3^{rd}$ overtone,$n=4$,so the frequency is $f_4 = (2(4)-1)f_0 = 7f_0$.
The length of the pipe $l$ in terms of wavelength $\lambda$ for the $n^{th}$ harmonic is $l = \frac{(2n-1)\lambda}{4}$.
For the $3^{rd}$ overtone $(n=4)$,$l = \frac{7\lambda}{4}$,which implies $\lambda = \frac{4l}{7}$.
The displacement amplitude $y(x)$ in a closed organ pipe at a distance $x$ from the closed end (node) is given by $y(x) = a \sin(kx)$,where $k = \frac{2\pi}{\lambda}$.
Substituting $k = \frac{2\pi}{(4l/7)} = \frac{7\pi}{2l}$,we get $y(x) = a \sin\left(\frac{7\pi x}{2l}\right)$.
At $x = l/7$,the amplitude is $y(l/7) = a \sin\left(\frac{7\pi (l/7)}{2l}\right) = a \sin\left(\frac{\pi}{2}\right) = a(1) = a$.

(D) In a resonance tube,the resonance lengths are given by $L_n = (2n-1) \frac{\lambda}{4} + e$,where $e$ is the end correction.
The difference between consecutive resonance lengths is $\Delta L = L_{n+1} - L_n = \frac{\lambda}{2}$.
Given the first resonance length $L_1 = 40\, cm$ and the second resonance length $L_2 = 122\, cm$,the difference is:
$\Delta L = 122\, cm - 40\, cm = 82\, cm$.
Since the difference between consecutive resonance lengths is constant,the third resonance length $L_3$ is:
$L_3 = L_2 + \Delta L = 122\, cm + 82\, cm = 204\, cm$.

(A) Let the length of both pipes be $L$. The fundamental frequency of an open organ pipe is $\nu_{open} = \frac{v}{2L}$.
The fundamental frequency of a closed organ pipe is $\nu_{closed} = \frac{v}{4L}$.
The beat frequency is given by $b = |\nu_{open} - \nu_{closed}| = |\frac{v}{2L} - \frac{v}{4L}| = \frac{v}{4L} = 4$.
If the length of each pipe is doubled,the new length becomes $L' = 2L$.
The new frequencies are $\nu'_{open} = \frac{v}{2(2L)} = \frac{v}{4L}$ and $\nu'_{closed} = \frac{v}{4(2L)} = \frac{v}{8L}$.
The new beat frequency is $b' = |\nu'_{open} - \nu'_{closed}| = |\frac{v}{4L} - \frac{v}{8L}| = \frac{v}{8L}$.
Since $\frac{v}{4L} = 4$,then $\frac{v}{8L} = \frac{4}{2} = 2$.
Therefore,the number of beats produced will be $2$.

(D) For an open organ pipe $A$,the frequency of the $n$-th overtone is given by $\nu_{A,n} = (n+1) \frac{v}{2 l_A}$. The second overtone $(n=2)$ is $\nu_{A,2} = \frac{3v}{2l_A}$.
For a closed organ pipe $B$,the frequency of the $n$-th overtone is given by $\nu_{B,n} = (2n+1) \frac{v}{4 l_B}$. The second overtone $(n=2)$ is $\nu_{B,2} = \frac{5v}{4l_B}$.
Given $\nu_{A,2} = \nu_{B,2}$,we have $\frac{3v}{2l_A} = \frac{5v}{4l_B}$.
Rearranging gives $\frac{l_B}{l_A} = \frac{5 \times 2}{3 \times 4} = \frac{10}{12} = \frac{5}{6}$. Thus,the ratio of lengths of $B$ to $A$ is $5:6$.
Now,the first overtone of $A$ $(n=1)$ is $\nu_{A,1} = \frac{2v}{2l_A} = \frac{v}{l_A}$.
The first overtone of $B$ $(n=1)$ is $\nu_{B,1} = \frac{3v}{4l_B}$.
The ratio of these frequencies is $\frac{\nu_{A,1}}{\nu_{B,1}} = \frac{v}{l_A} \times \frac{4l_B}{3v} = \frac{4}{3} \times \frac{l_B}{l_A} = \frac{4}{3} \times \frac{5}{6} = \frac{20}{18} = \frac{10}{9}$.
Both statements $(B)$ and $(C)$ are correct.

(C) The fundamental frequency of an open organ pipe is given by $f = \frac{v}{2L} = \frac{1}{2L} \sqrt{\frac{\gamma RT}{M}}$.
Since $T$ is constant,$f \propto \frac{\sqrt{\gamma}}{L \sqrt{M}}$.
Let $K = \frac{\sqrt{RT}}{2}$. Then $f = K \frac{\sqrt{\gamma}}{L \sqrt{M}}$.
For pipe $A$ $(H_2)$: $L_A = l$,$M_A = 2$,$\gamma_A = 1.4 = 7/5$. So,$f_A = K \frac{\sqrt{7/5}}{l \sqrt{2}}$.
For pipe $B$ $(O_2)$: $L_B = l/2$,$M_B = 32$,$\gamma_B = 1.4 = 7/5$. So,$f_B = K \frac{\sqrt{7/5}}{(l/2) \sqrt{32}} = 2K \frac{\sqrt{7/5}}{l \sqrt{32}}$.
For pipe $C$ $(N_2)$: $L_C = 2l/3$,$M_C = 28$,$\gamma_C = 1.4 = 7/5$. So,$f_C = K \frac{\sqrt{7/5}}{(2l/3) \sqrt{28}} = \frac{3}{2} K \frac{\sqrt{7/5}}{l \sqrt{28}}$.
For pipe $D$ $(CO_2)$: $L_D = l/3$,$M_D = 44$,$\gamma_D = 7/5$. So,$f_D = K \frac{\sqrt{7/5}}{(l/3) \sqrt{44}} = 3K \frac{\sqrt{7/5}}{l \sqrt{44}}$.
Calculating $f_B/f_C = \frac{2K \sqrt{7/5} / (l \sqrt{32})}{3K \sqrt{7/5} / (2l \sqrt{28})} = \frac{2}{\sqrt{32}} \times \frac{2 \sqrt{28}}{3} = \frac{4}{3} \sqrt{\frac{28}{32}} = \frac{4}{3} \sqrt{\frac{7}{8}} = \frac{4}{3} \frac{\sqrt{7}}{2\sqrt{2}} = \frac{2\sqrt{7}}{3\sqrt{2}} = \sqrt{\frac{28}{18}} = \sqrt{\frac{14}{9}}$.
Checking $f_C/f_D = \frac{3K \sqrt{7/5} / (2l \sqrt{28})}{3K \sqrt{7/5} / (l \sqrt{44})} = \frac{1}{2 \sqrt{28}} \times \sqrt{44} = \frac{1}{2} \sqrt{\frac{44}{28}} = \frac{1}{2} \sqrt{\frac{11}{7}} = \sqrt{\frac{11}{28}}$.
Thus,option $C$ is correct.

(D) The fundamental frequency of an organ pipe is given by $f = \frac{v}{2L}$ (for open pipe) or $f = \frac{v}{4L}$ (for closed pipe).
In both cases,$f \propto v$,where $v$ is the speed of sound in the gas.
The speed of sound in a gas is given by $v = \sqrt{\frac{\gamma RT}{M}}$,where $M$ is the molar mass of the gas.
Since $T$ is constant,$v \propto \frac{1}{\sqrt{M}}$. Thus,$f \propto \frac{1}{\sqrt{M}}$.
$(A)$ $M(H_2) = 2$,$M(O_2) = 32$. Since $M$ increases,$f$ decreases. Statement $(A)$ is incorrect.
$(B)$ $M(O_2) = 32$,$M(N_2) = 28$. Since $M$ decreases,$f$ increases. Statement $(B)$ is correct.
$(C)$ $M(He) = 4$,$M(CH_4) = 16$. Since $M$ increases,$f$ decreases. Statement $(C)$ is correct.
Therefore,both $(B)$ and $(C)$ are correct.

(B) For a closed organ pipe,the length $L$ is related to the wavelength $\lambda$ for the $n^{th}$ overtone as $L = (2n + 1) \frac{\lambda}{4}$.
For the first overtone,$n = 1$,so $L = \frac{3\lambda}{4}$.
Given $L = 1.2 \, m$,we have $\frac{3\lambda}{4} = 1.2 \, m$,which gives $\lambda = 1.6 \, m$.
In a standing wave,pressure variation is maximum at the displacement nodes.
In a closed organ pipe,the open end is a displacement antinode (pressure node),and the closed end is a displacement node (pressure antinode).
The displacement nodes occur at distances $x = 0, \frac{\lambda}{2}, \lambda, \dots$ from the closed end.
Alternatively,measuring from the open end,the displacement nodes occur at $x = \frac{\lambda}{4}, \frac{3\lambda}{4}, \dots$ from the open end.
For the first overtone,the first displacement node (maximum pressure variation) is at $\frac{\lambda}{4} = \frac{1.6}{4} = 0.4 \, m$ from the open end.

(C) The pressure variation in a standing wave is given by $p(x, t) = p_0 \cos(kx) \sin(\omega t)$.
At a closed end,the displacement is zero,which corresponds to a pressure antinode (maximum pressure amplitude).
At an open end,the pressure is atmospheric,which corresponds to a pressure node (zero pressure amplitude).
Given $p = p_0 \cos \frac{3\pi x}{2} \sin 300\pi t$.
At $x = 0$,$p = p_0 \cos(0) \sin(300\pi t) = p_0 \sin(300\pi t)$. Since the pressure amplitude is maximum $(p_0)$,$x = 0$ is a closed end.
For the other end to be a closed end,the pressure amplitude must be maximum,i.e.,$|\cos \frac{3\pi x}{2}| = 1$.
This occurs when $\frac{3\pi x}{2} = n\pi$ for integer $n$.
For the first possible length $L > 0$,we set $n = 3$,so $\frac{3\pi L}{2} = 3\pi$,which gives $L = 2 \ m$.
Thus,the pipe is closed at both ends with a length of $2 \ m$.

(B) In a resonance column experiment,the frequency of the tuning fork $f$ remains constant.
For the first resonance in winter,the length is $\ell_1 = 18 \ cm$. The frequency is given by $f = \frac{v}{4 \ell_1} = \frac{v}{4 \times 18}$,where $v$ is the speed of sound in winter.
For the second resonance in summer,the length is $\ell_2 = x \ cm$. The frequency is given by $f = \frac{3v'}{4 \ell_2} = \frac{3v'}{4x}$,where $v'$ is the speed of sound in summer.
Equating the frequencies: $\frac{v}{4 \times 18} = \frac{3v'}{4x}$.
Solving for $x$: $x = 54 \times \frac{v'}{v} \ cm$.
Since the temperature in summer is higher than in winter,the speed of sound $v' > v$ (because $v \propto \sqrt{T}$).
Therefore,$x > 54 \ cm$.

(B) For a pipe open at both ends with length $\ell$,the fundamental frequency is given by $f = \frac{v}{2\ell}$,where $v$ is the speed of sound in air.
When the pipe is dipped vertically in water such that half of its length is submerged,the length of the air column becomes $\ell^{\prime} = \frac{\ell}{2}$.
Since the pipe is now closed at one end (by the water surface) and open at the other,it acts as a pipe closed at one end.
The fundamental frequency of a pipe closed at one end with length $\ell^{\prime}$ is $f^{\prime} = \frac{v}{4\ell^{\prime}}$.
Substituting $\ell^{\prime} = \frac{\ell}{2}$ into the equation,we get $f^{\prime} = \frac{v}{4(\ell/2)} = \frac{v}{2\ell}$.
Comparing this with the initial frequency,we find $f^{\prime} = f$.

(D) For an open pipe of length $L$,the frequency of the $n^{th}$ harmonic is given by $f_n = \frac{n v}{2L}$,where $v$ is the speed of sound.
Given $L = 1.95 \, m$,and two successive harmonics are $f_n = 275 \, Hz$ and $f_{n+1} = 330 \, Hz$.
The difference between two successive harmonics is the fundamental frequency $f_1 = \frac{v}{2L}$.
$f_{n+1} - f_n = \frac{(n+1)v}{2L} - \frac{nv}{2L} = \frac{v}{2L} = 330 - 275 = 55 \, Hz$.
Substituting the value of $L$:
$\frac{v}{2 \times 1.95} = 55$.
$v = 55 \times 3.9 = 214.5 \, m/s$.

(A) The fundamental frequency $(n_0)$ of a string of length $d$ fixed at both ends is given by the formula: $n_0 = \frac{1}{2d} \sqrt{\frac{T}{\mu}}$.
Since the cord vibrates in its simplest standing-wave state (fundamental mode),the frequency of vibration $f$ is equal to the fundamental frequency $n_0$.
Therefore,$f = \frac{1}{2d} \sqrt{\frac{T}{\mu}}$.
Squaring both sides,we get: $f^2 = \frac{1}{4d^2} \cdot \frac{T}{\mu}$.
Rearranging to solve for tension $T$: $T = 4\mu f^2 d^2$.

(B) In a resonance column experiment,the first resonance length $l_1$ is given by the formula: $\frac{\lambda}{4} = l_1 + e$,where $e$ is the end correction.
The end correction $e$ for a tube of diameter $d$ is given by $e = 0.3d$.
The wavelength $\lambda$ is calculated as $\lambda = \frac{v}{f}$,where $v = 336 \ m/s$ and $f = 512 \ Hz$.
$\lambda = \frac{336}{512} \approx 0.65625 \ m = 65.625 \ cm$.
Therefore,$\frac{\lambda}{4} = \frac{65.625}{4} = 16.40625 \ cm$.
Given the diameter $d = 4 \ cm$,the end correction $e = 0.3 \times 4 = 1.2 \ cm$.
Substituting these values into the resonance formula: $l_1 = \frac{\lambda}{4} - e = 16.40625 - 1.2 = 15.20625 \ cm$.
Rounding to one decimal place,the reading is $15.2 \ cm$.

(C) For a string fixed at both ends,the frequency of the $2^{nd}$ harmonic is given by $f_1 = \frac{2v}{2L} = \frac{v}{L}$,where $v$ is the wave speed and $L$ is the length of the string.
When one end becomes free,the string acts like a pipe closed at one end. The resonant frequencies are given by $f_n = \frac{nv}{4L}$,where $n$ must be an odd integer $(n = 1, 3, 5, \dots)$.
We are given that the frequency is increased from $f_1$. We need to find the first resonant frequency $f_2 > f_1$ for the new configuration.
Setting $f_2 = \frac{nv}{4L} > f_1 = \frac{v}{L}$,we get $\frac{nv}{4L} > \frac{v}{L}$,which implies $n > 4$.
The smallest odd integer greater than $4$ is $n = 5$.
Thus,$f_2 = \frac{5v}{4L} = \frac{5}{4} \left( \frac{v}{L} \right) = \frac{5}{4}f_1$.

(D) For a closed organ pipe,the frequencies of the harmonics are given by $f_n = (2n - 1)f_1$,where $n = 1, 2, 3, ...$ is the harmonic number.
The fundamental frequency $(n=1)$ is the first harmonic.
The first overtone is the third harmonic $(n=2)$.
The second overtone is the fifth harmonic $(n=3)$.
The third overtone is the seventh harmonic $(n=4)$.
For the $n$-th harmonic in a closed pipe,the number of nodes is $n$ and the number of antinodes is $n$.
Since the third overtone corresponds to the $4$-th harmonic $(n=4)$,the air column in the pipe will have $4$ nodes and $4$ antinodes.

(D) The frequency of the tuning fork is $f = 340 \, Hz$ and the speed of sound is $v = 340 \, m/s$.
The wavelength is $\lambda = \frac{v}{f} = \frac{340}{340} = 1 \, m = 100 \, cm$.
For a pipe closed at one end,resonance occurs when the length of the air column $L$ satisfies $L = \frac{n \lambda}{4}$,where $n = 1, 3, 5, \ldots$.
Substituting $\lambda = 100 \, cm$,we get $L = \frac{n \times 100}{4} = 25n \, cm$.
Possible lengths of the air column are $L = 25 \, cm, 75 \, cm, 125 \, cm, \ldots$.
Since the total length of the pipe is $120 \, cm$,the air column length $L$ must be $\le 120 \, cm$. Thus,the possible values for $L$ are $25 \, cm$ and $75 \, cm$.
The height of the water column $h$ is given by $h = 120 - L$.
To find the minimum height of the water column,we must choose the maximum possible length of the air column,which is $L = 75 \, cm$.
Therefore,$h_{\min} = 120 \, cm - 75 \, cm = 45 \, cm$.

(A) For a string fixed at both ends,the resonance frequencies are given by $f_n = n f_1$,where $n = 1, 2, 3, \dots$ and $f_1$ is the fundamental frequency.
Given two consecutive resonance frequencies $f_n = 252 \ Hz$ and $f_{n+1} = 336 \ Hz$.
Taking the ratio: $\frac{f_{n+1}}{f_n} = \frac{(n+1) f_1}{n f_1} = \frac{n+1}{n} = \frac{336}{252}$.
Simplifying the fraction: $\frac{336}{252} = \frac{4}{3}$.
Thus,$\frac{n+1}{n} = \frac{4}{3}$,which implies $n = 3$.
Substituting $n = 3$ into $f_n = n f_1$,we get $252 = 3 f_1$.
Therefore,$f_1 = \frac{252}{3} = 84 \ Hz$.

(B) The fundamental frequency of a closed organ pipe of length $L$ is given by $f = \frac{v}{4L}$.
For the first pipe of length $l$,the frequency is $f_1 = \frac{v}{4l}$.
For the second pipe of length $l + x$,the frequency is $f_2 = \frac{v}{4(l + x)}$.
The beat frequency $f_b$ is the difference between these two frequencies: $f_b = f_1 - f_2 = \frac{v}{4l} - \frac{v}{4(l + x)}$.
Factoring out $\frac{v}{4}$,we get $f_b = \frac{v}{4} \left( \frac{1}{l} - \frac{1}{l + x} \right) = \frac{v}{4} \left( \frac{l + x - l}{l(l + x)} \right) = \frac{v}{4} \left( \frac{x}{l(l + x)} \right)$.
Since $x << l$,we can approximate $l + x \approx l$,so $l(l + x) \approx l^2$.
Therefore,the beat frequency is $f_b \approx \frac{vx}{4l^2}$.

(C) In a resonance tube experiment,the resonant lengths are given by the formula $\ell_n + \varepsilon = \frac{(2n-1)v}{4f_0}$,where $\varepsilon$ is the end correction,$v$ is the speed of sound,and $f_0$ is the frequency of the tuning fork.
For the first resonance $(n=1)$: $\ell_1 + \varepsilon = \frac{v}{4f_0}$.
For the second resonance $(n=2)$: $\ell_2 + \varepsilon = \frac{3v}{4f_0}$.
For the third resonance $(n=3)$: $\ell_3 + \varepsilon = \frac{5v}{4f_0}$.
Subtracting the first equation from the second: $\ell_2 - \ell_1 = \frac{3v}{4f_0} - \frac{v}{4f_0} = \frac{2v}{4f_0} = \frac{v}{2f_0}$.
Thus,$\frac{v}{4f_0} = \frac{\ell_2 - \ell_1}{2}$.
Substituting this into the third resonance equation: $\ell_3 + \varepsilon = 5 \times \frac{\ell_2 - \ell_1}{2} = 2.5(\ell_2 - \ell_1)$.
Alternatively,note that the difference between consecutive resonant lengths is constant: $\ell_2 - \ell_1 = \ell_3 - \ell_2 = \frac{v}{2f_0}$.
Therefore,$\ell_3 = \ell_2 + (\ell_2 - \ell_1) = 2\ell_2 - \ell_1$.

(C) For an open organ pipe,the frequency of the $n^{th}$ mode of vibration is given by $(v_{n})_{o} = \frac{nv}{2l}$,where $v$ is the speed of sound and $l$ is the length of the pipe.
For a closed organ pipe,the frequency of the $n^{th}$ mode of vibration is given by $(v_{n})_{c} = (2n - 1) \frac{v}{4l}$.
To find the ratio of the frequency of the $n^{th}$ mode of the open pipe to that of the closed pipe,we divide the two expressions:
$\frac{(v_{n})_{o}}{(v_{n})_{c}} = \frac{\frac{nv}{2l}}{(2n - 1) \frac{v}{4l}}$
$\frac{(v_{n})_{o}}{(v_{n})_{c}} = \frac{nv}{2l} \times \frac{4l}{(2n - 1)v}$
$\frac{(v_{n})_{o}}{(v_{n})_{c}} = \frac{2n}{2n - 1}$

(B) The fundamental frequency of a closed organ pipe is given by $f_c = \frac{v}{4\ell_c}$,where $\ell_c = 20\; cm$.
The frequencies of an open organ pipe are given by $f_n = \frac{nv}{2\ell_o}$,where $n = 1, 2, 3, \dots$.
The first overtone is $n=2$,and the second overtone is $n=3$.
Thus,the second overtone frequency of an open organ pipe is $f_{o,2} = \frac{3v}{2\ell_o}$.
According to the problem,the fundamental frequency of the closed pipe equals the second overtone of the open pipe:
$\frac{v}{4\ell_c} = \frac{3v}{2\ell_o}$
Canceling $v$ from both sides:
$\frac{1}{4\ell_c} = \frac{3}{2\ell_o}$
Rearranging to solve for $\ell_o$:
$\ell_o = \frac{3 \times 4\ell_c}{2} = 6\ell_c$
Substituting $\ell_c = 20\; cm$:
$\ell_o = 6 \times 20\; cm = 120\; cm$.