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

(A) For an organ pipe closed at one end,the frequencies of the harmonics are given by $f_n = (2n - 1)f_0$,where $n = 1, 2, 3, \dots$ and $f_0 = 1500 \ Hz$.
The fundamental frequency is $f_1 = 1500 \ Hz$.
The overtones are the frequencies higher than the fundamental frequency,given by $f_n = (2n - 1)f_0$ for $n > 1$.
We need $f_n \leq 19500 \ Hz$.
$(2n - 1) \times 1500 \leq 19500$.
$2n - 1 \leq \frac{19500}{1500} = 13$.
$2n \leq 14$,so $n \leq 7$.
The possible values for $n$ are $1, 2, 3, 4, 5, 6, 7$.
Since $n=1$ is the fundamental frequency,the overtones correspond to $n = 2, 3, 4, 5, 6, 7$.
There are $7 - 1 = 6$ overtones.

(A) The fundamental resonant frequency for a pipe open at both ends of length $L$ is given by $n_1 = \frac{V}{2L}$,where $V$ is the speed of sound in air.
The fundamental resonant frequency for a pipe closed at one end of length $L$ is given by $n_2 = \frac{V}{4L}$.
Comparing the two equations:
$n_2 = \frac{V}{4L} = \frac{1}{2} \left( \frac{V}{2L} \right)$
$n_2 = \frac{n_1}{2}$
Therefore,$n_1 = 2 n_2$.

(D) The end correction $e$ for a tube of circular cross-section is given by the formula $e = 0.6r$ or $e = 0.3d$,where $r$ is the radius and $d$ is the diameter of the tube.
Since the end correction $e$ is directly proportional to the diameter $d$ of the tube $(e \propto d)$,increasing the diameter of the tube will result in a larger end correction.
Therefore,the end correction will be more if the tube is widened.

(C) For a tube of length $L$ closed at one end,the fundamental wavelength is $\lambda_1 = 4L$. The fundamental frequency is $f_A = \frac{v}{\lambda_1} = \frac{v}{4L}$.
For a tube of length $L$ open at both ends,the fundamental wavelength is $\lambda_2 = 2L$. The fundamental frequency is $f_B = \frac{v}{\lambda_2} = \frac{v}{2L}$.
The ratio of the fundamental frequency of tube $A$ to tube $B$ is $\frac{f_A}{f_B} = \frac{v/4L}{v/2L} = \frac{2L}{4L} = \frac{1}{2}$.
Thus,the ratio is $1: 2$.

(C) For a pipe closed at one end,the possible frequencies are given by $f_n = (2n + 1)f_0$,where $n = 0, 1, 2, ...$ represents the overtone number.
For the $n$-th overtone,the number of nodes is $(n + 1)$ and the number of antinodes is $(n + 1)$.
Given that the air column is vibrating in the fourth overtone,we have $n = 4$.
Therefore,the number of nodes = $4 + 1 = 5$.
The number of antinodes = $4 + 1 = 5$.
Thus,the vibrating air column has $5$ nodes and $5$ antinodes.

(B) The frequency of an open pipe of length $L_O$ is given by $f_n = \frac{n v}{2 L_O}$,where $n = 1, 2, 3, \dots$ is the harmonic number. The second overtone corresponds to the $3^{rd}$ harmonic $(n=3)$. Thus,$f_{O,2} = \frac{3 v}{2 L_O}$.
For a closed pipe of length $L_C$,the frequency is given by $f_m = \frac{(2m-1) v}{4 L_C}$,where $m = 1, 2, 3, \dots$ is the overtone number. The first overtone corresponds to $m=2$ (the $3^{rd}$ harmonic). Thus,$f_{C,1} = \frac{3 v}{4 L_C}$.
Given that $f_{O,2} = f_{C,1}$,we have:
$\frac{3 v}{2 L_O} = \frac{3 v}{4 L_C}$
$\frac{1}{2 L_O} = \frac{1}{4 L_C}$
$L_C = \frac{L_O}{2} = \frac{1.5 \ m}{2} = 0.75 \ m$.

(A) For an open organ pipe of length $L$,the fundamental frequency is given by $f = \frac{v}{2L}$,where $v$ is the speed of sound in air.
When the tube is dipped vertically into water such that half of its length is submerged,the effective length of the air column becomes $L' = \frac{L}{2}$.
The lower end of the tube is now closed by the water surface,making it a closed organ pipe of length $L' = \frac{L}{2}$.
The fundamental frequency of a closed organ pipe is given by $f' = \frac{v}{4L'}$.
Substituting $L' = \frac{L}{2}$ into the formula,we get $f' = \frac{v}{4(L/2)} = \frac{v}{2L}$.
Since $f = \frac{v}{2L}$,it follows that $f' = f$.
Therefore,the fundamental frequency remains the same.

(D) For a pipe closed at one end,the fundamental mode corresponds to a length $L = \frac{\lambda}{4}$,where $\lambda$ is the wavelength.
Thus,the wavelength is $\lambda = 4L$.
The time taken for the sound wave to travel the length of the pipe $L$ is given as $t$. Since the speed of sound $v$ is constant,we have $v = \frac{L}{t}$.
The frequency $f$ is given by $f = \frac{v}{\lambda}$.
Substituting the values,we get $f = \frac{L/t}{4L} = \frac{1}{4t} = \frac{0.25}{t}$.
Therefore,the frequency of vibration of the air column is $\frac{0.25}{t}$ Hz.
Option $(D)$ is correct.

(D) For a cylindrical tube open at both ends with length $L$,the fundamental frequency is given by $f = \frac{v}{2L}$,where $v$ is the speed of sound in air.
When the tube is dipped vertically into water such that one-third of its length is submerged,the length of the air column becomes $L' = L - \frac{L}{3} = \frac{2L}{3}$.
The tube now acts as a pipe closed at one end (the water surface) and open at the other.
For a pipe closed at one end,the fundamental frequency is $f' = \frac{v}{4L'}$.
Substituting $L' = \frac{2L}{3}$,we get $f' = \frac{v}{4(2L/3)} = \frac{3v}{8L}$.
Since $f = \frac{v}{2L}$,we can write $f' = \frac{3}{4} \times \frac{v}{2L} = \frac{3}{4}f$.

(B) In a closed pipe,sound waves travel through the air column and reflect off the closed end.
These reflected waves interfere with the incident waves to form stationary (standing) waves.
Since sound waves are pressure waves,they are longitudinal in nature.
Therefore,the waves set up in a closed pipe are longitudinal and stationary.

(D) Given: Length of pipe $L = 30 \text{ cm} = 0.3 \text{ m}$.
Frequency $f = 1.1 \text{ kHz} = 1100 \text{ Hz}$.
Speed of sound $v = 330 \text{ ms}^{-1}$.
For an open pipe,the frequency of the $n^{th}$ harmonic is given by $f_n = n \times \frac{v}{2L}$.
Substituting the values: $1100 = n \times \frac{330}{2 \times 0.3}$.
$1100 = n \times \frac{330}{0.6}$.
$1100 = n \times 550$.
$n = \frac{1100}{550} = 2$.
Therefore,the pipe resonates in the second harmonic mode.

(A) The frequency of the $n^{\text{th}}$ overtone of a closed pipe is given by $f_{c} = \frac{(2n+1)v}{4l_{1}}$. For the first overtone,$n=1$,so $f_{c} = \frac{3v}{4l_{1}}$.
The frequency of the $m^{\text{th}}$ harmonic of an open pipe is given by $f_{o} = \frac{mv}{2l_{2}}$. For the $2^{\text{nd}}$ harmonic,$m=2$,so $f_{o} = \frac{2v}{2l_{2}} = \frac{v}{l_{2}}$.
Given that $f_{c} = f_{o}$,we have $\frac{3v}{4l_{1}} = \frac{v}{l_{2}}$.
Rearranging the terms to find the ratio $\frac{l_{1}}{l_{2}}$,we get $\frac{l_{1}}{l_{2}} = \frac{3}{4}$.

(A) For an open pipe of length $L_o$,the frequency of the $n^{th}$ harmonic is $f_n = \frac{n v}{2 L_o}$. The second overtone corresponds to the $3^{rd}$ harmonic $(n=3)$,so $f_{o} = \frac{3 v}{2 L_o}$.
For a closed pipe of length $L_c$,the frequency of the $n^{th}$ harmonic is $f_n = \frac{(2n-1) v}{4 L_c}$. The first overtone corresponds to the $3^{rd}$ harmonic $(n=2)$,so $f_{c} = \frac{3 v}{4 L_c}$.
Given that $f_o = f_c$,we have $\frac{3 v}{2 L_o} = \frac{3 v}{4 L_c}$.
Simplifying this,we get $\frac{1}{2 L_o} = \frac{1}{4 L_c}$,which implies $\frac{L_o}{L_c} = \frac{4}{2} = \frac{2}{1}$.

(D) Let the length of both pipes be $L$. The fundamental frequency of an open organ pipe is $f_{o} = \frac{v}{2L}$ and that of a closed organ pipe is $f_{c} = \frac{v}{4L}$.
Given that $f_{o} - f_{c} = 2 \text{ Hz}$.
Substituting the expressions,$\frac{v}{2L} - \frac{v}{4L} = 2 \Rightarrow \frac{v}{4L} = 2 \text{ Hz}$.
Thus,$f_{c} = 2 \text{ Hz}$ and $f_{o} = 2f_{c} = 4 \text{ Hz}$.
Now,the length of the open pipe is halved $(L_{o}' = L/2)$,so its new frequency is $f_{o}' = \frac{v}{2(L/2)} = \frac{v}{L} = 2f_{o} = 2 \times 4 = 8 \text{ Hz}$.
The length of the closed pipe is doubled $(L_{c}' = 2L)$,so its new frequency is $f_{c}' = \frac{v}{4(2L)} = \frac{1}{2} \left(\frac{v}{4L}\right) = \frac{1}{2} f_{c} = \frac{1}{2} \times 2 = 1 \text{ Hz}$.
The number of beats produced per second is $|f_{o}' - f_{c}'| = |8 - 1| = 7 \text{ Hz}$.

(A) The fundamental frequency of an open cylindrical tube of length $L$ is given by $n = \frac{v}{2L} = 390 \,Hz$.
When $\frac{1}{4}$th of the tube is immersed in water, the tube acts as a closed organ pipe (closed at one end) with a new length $L' = L - \frac{1}{4}L = \frac{3}{4}L$.
The fundamental frequency of a closed organ pipe is given by $n' = \frac{v}{4L'}$.
Substituting $L' = \frac{3}{4}L$, we get $n' = \frac{v}{4(\frac{3}{4}L)} = \frac{v}{3L}$.
We can rewrite this as $n' = \frac{2}{3} \times (\frac{v}{2L})$.
Since $\frac{v}{2L} = 390 \,Hz$, we have $n' = \frac{2}{3} \times 390 \,Hz = 260 \,Hz$.

(A) For an open organ pipe of length $l$,the fundamental frequency is given by $f = \frac{v}{2l}$,where $v$ is the speed of sound in air.
When the tube is placed vertically in water such that half its length is submerged,it acts as a closed organ pipe of length $l' = \frac{l}{2}$.
The fundamental frequency of a closed organ pipe is given by $f' = \frac{v}{4l'}$.
Substituting $l' = \frac{l}{2}$ into the expression for $f'$,we get:
$f' = \frac{v}{4(l/2)} = \frac{v}{2l}$.
Comparing this with the initial frequency $f = \frac{v}{2l}$,we find that $f' = f$.

$(A)$ Given: $n_0 = 140 \,Hz$, $T_0 = 27^{\circ} C = 300 \,K$, and $T_1 = 57.75^{\circ} C = 330.75 \,K$.
Since the frequency of an organ pipe is proportional to the speed of sound $(n \propto v)$, and the speed of sound is proportional to the square root of the absolute temperature $(v \propto \sqrt{T})$, we have $n \propto \sqrt{T}$.
Therefore, $\frac{n_1}{n_0} = \sqrt{\frac{T_1}{T_0}}$.
Substituting the values: $\frac{n_1}{140} = \sqrt{\frac{330.75}{300}} = \sqrt{1.1025} = 1.05$.
So, $n_1 = 140 \times 1.05 = 147 \,Hz$.
The number of beats produced per second is the difference in frequencies: $n_1 - n_0 = 147 \,Hz - 140 \,Hz = 7 \,Hz$.
Thus, the correct option is $A$.

(B) For a closed organ pipe, the frequency of the $n^{th}$ harmonic is given by $f_n = n \cdot f_1$, where $n$ must be an odd integer $(n = 1, 3, 5, 7, 9, ...)$.
In a closed organ pipe of length $L$, the number of nodes $N$ formed for the $n^{th}$ harmonic is given by the formula $N = \frac{n+1}{2}$.
For the fifth harmonic $(n = 5)$:
$N_5 = \frac{5+1}{2} = \frac{6}{2} = 3$.
For the ninth harmonic $(n = 9)$:
$N_9 = \frac{9+1}{2} = \frac{10}{2} = 5$.
Therefore, the number of nodes formed are $3$ and $5$ respectively.

(B) For an open organ pipe,the frequency of the $n^{th}$ harmonic is given by $f_n = \frac{n v}{2L}$,where $n = 1, 2, 3, ...$ is the harmonic mode.
Given:
Length $L = 30 \ cm = 0.3 \ m$
Frequency $f = 1.65 \ kHz = 1650 \ Hz$
Velocity of sound $v = 330 \ m/s$
Substituting the values into the formula:
$1650 = \frac{n \times 330}{2 \times 0.3}$
$1650 = \frac{n \times 330}{0.6}$
$1650 = n \times 550$
$n = \frac{1650}{550} = 3$
Therefore,the pipe resonates in the $3^{rd}$ harmonic mode.

(D) For figure $(a)$,$L = \frac{\lambda_1}{4} \Rightarrow \lambda_1 = 4L$.
For figure $(b)$,$L = \frac{\lambda_2}{4} + \frac{\lambda_2}{2} + \frac{\lambda_2}{4} = \lambda_2 \Rightarrow \lambda_2 = L$.
For figure $(c)$,$L = \frac{\lambda_3}{4} + \frac{\lambda_3}{4} + \frac{\lambda_3}{2} = \lambda_3 \Rightarrow \lambda_3 = L$ (Note: Based on the diagram,it represents a pipe open at both ends with one node in the middle,so $L = \lambda_3/2 \Rightarrow \lambda_3 = 2L$).
For figure $(d)$,$L = \frac{\lambda_4}{2} + \frac{\lambda_4}{4} = \frac{3\lambda_4}{4} \Rightarrow \lambda_4 = \frac{4L}{3}$.
Since frequency $f \propto \frac{1}{\lambda}$,we have $f_1 : f_2 : f_3 : f_4 = \frac{1}{\lambda_1} : \frac{1}{\lambda_2} : \frac{1}{\lambda_3} : \frac{1}{\lambda_4}$.
Substituting the values: $f_1 : f_2 : f_3 : f_4 = \frac{1}{4L} : \frac{1}{L} : \frac{1}{2L} : \frac{3}{4L}$.
Multiplying by $4L$,we get $1 : 4 : 2 : 3$.

(C) For an open organ pipe of length $L$,the fundamental frequency is given by:
$f_0 = \frac{v}{2L} = 100 \ Hz$
When the bottom end is closed and $1/3$ of the pipe is filled with water,the effective length of the air column becomes $L' = L - \frac{L}{3} = \frac{2L}{3}$.
For a closed organ pipe of length $L'$,the fundamental frequency is:
$f' = \frac{v}{4L'}$
Substituting $L' = \frac{2L}{3}$:
$f' = \frac{v}{4(2L/3)} = \frac{3v}{8L}$
We can rewrite this as:
$f' = \frac{3}{4} \left( \frac{v}{2L} \right)$
Since $\frac{v}{2L} = 100 \ Hz$,we have:
$f' = \frac{3}{4} \times 100 = 75 \ Hz$

(A) For an open organ pipe,the frequency of the $n^{th}$ harmonic is given by $f_n = \frac{n v}{2 l_o}$.
For the third harmonic $(n=3)$,$f_{open} = \frac{3 v}{2 l_o}$.
Given $l_o = 72 \ cm$,so $f_{open} = \frac{3 v}{2 \times 72} = \frac{v}{48}$.
For a closed organ pipe,the frequency of the $n^{th}$ harmonic (where $n$ is odd) is given by $f_n = \frac{n v}{4 l_c}$.
For the fifth harmonic $(n=5)$,$f_{closed} = \frac{5 v}{4 l_c}$.
According to the problem,$f_{closed} = f_{open}$.
$\frac{5 v}{4 l_c} = \frac{3 v}{2 l_o}$.
Substituting $l_o = 72 \ cm$:
$\frac{5}{4 l_c} = \frac{3}{2 \times 72} = \frac{3}{144} = \frac{1}{48}$.
$4 l_c = 5 \times 48 = 240$.
$l_c = \frac{240}{4} = 60 \ cm$.

(D) The fundamental frequency of an open organ pipe is given by $f = \frac{v}{2l}$.
Given, beat frequency $f_b = \frac{40}{10} = 4 \,Hz$.
Let $l_1 = 50 \,cm = 0.5 \,m$ and $l_2 = 51 \,cm = 0.51 \,m$.
The difference in frequencies is $f_1 - f_2 = 4 \,Hz$.
$\frac{v}{2l_1} - \frac{v}{2l_2} = 4$
$\frac{v}{2} \left( \frac{1}{0.5} - \frac{1}{0.51} \right) = 4$
$\frac{v}{2} \left( \frac{0.51 - 0.5}{0.5 \times 0.51} \right) = 4$
$\frac{v}{2} \left( \frac{0.01}{0.255} \right) = 4$
$v \left( \frac{0.01}{0.51} \right) = 4$
$v = \frac{4 \times 0.51}{0.01} = 4 \times 51 = 204 \,ms^{-1}$.

(A) For an open pipe of length $l_O$, the frequency of the $n^{th}$ harmonic is given by $f_n = n \cdot \frac{V}{2 l_O}$.
For an open pipe, the second harmonic $(n=2)$ is $f_2 = 2 \cdot \frac{V}{2 l_O} = \frac{V}{l_O}$.
For a closed pipe of length $l_C$, the fundamental frequency is $f_1 = \frac{V}{4 l_C}$.
Given that the second harmonic of the open pipe equals the fundamental frequency of the closed pipe:
$\frac{V}{l_O} = \frac{V}{4 l_C}$.
Substituting $l_O = 80 \,cm$:
$\frac{1}{80} = \frac{1}{4 l_C}$.
$4 l_C = 80$.
$l_C = 20 \,cm$.

(B) Let the length of the tube be $L$.
The fundamental frequency for an open organ pipe is given by:
$f = \frac{V}{2L}$
When the tube is dipped vertically in water such that $60 \%$ of its length is submerged,the remaining length of the air column above the water surface acts as a closed organ pipe (closed at one end by the water surface).
The length of this air column is:
$l' = (100 \% - 60 \%) L = 40 \% L = 0.4L = \frac{2L}{5}$
The fundamental frequency of a closed organ pipe is given by:
$f' = \frac{V}{4l'}$
Substituting the value of $l'$:
$f' = \frac{V}{4(\frac{2L}{5})} = \frac{5V}{8L}$
We can rewrite this in terms of $f$:
$f' = \frac{5}{4} \left( \frac{V}{2L} \right) = \frac{5}{4} f$

(C) The frequency of an air column in a resonance tube is proportional to the speed of sound,which is proportional to the square root of the absolute temperature $(v \propto \sqrt{T})$.
Let $n$ be the frequency of the tuning fork.
At $T_1 = 273 + 51 = 324 \ K$,the frequency of the air column is $n_1 = n + 3$.
At $T_2 = 273 + 16 = 289 \ K$,the frequency of the air column is $n_2 = n - 3$.
Using the relation $\frac{n_1}{n_2} = \sqrt{\frac{T_1}{T_2}}$,we get:
$\frac{n+3}{n-3} = \sqrt{\frac{324}{289}} = \frac{18}{17}$.
Cross-multiplying gives: $17(n + 3) = 18(n - 3)$.
$17n + 51 = 18n - 54$.
$n = 51 + 54 = 105 \ Hz$.

(C) For an open organ pipe,the wavelength of the $n^{th}$ overtone is given by $\lambda_n = \frac{2l}{n+1}$.
For the $3^{rd}$ overtone,$n=3$,so $\lambda = \frac{2l}{3+1} = \frac{2l}{4} = \frac{l}{2}$.
Since the pipe is open at both ends,antinodes (maximum amplitude $A$) are formed at the open ends.
The displacement amplitude $R$ at a distance $x$ from an antinode is given by $R = A \cos(kx)$,where $k = \frac{2\pi}{\lambda}$.
Given $x = \frac{l}{16}$,we calculate the phase angle $\phi = kx = \frac{2\pi}{\lambda} \cdot \frac{l}{16}$.
Substituting $\lambda = \frac{l}{2}$,we get $\phi = \frac{2\pi}{(l/2)} \cdot \frac{l}{16} = \frac{4\pi}{l} \cdot \frac{l}{16} = \frac{\pi}{4}$.
Therefore,the amplitude $R = A \cos(\frac{\pi}{4}) = A \cdot \frac{1}{\sqrt{2}} = \frac{A}{\sqrt{2}}$.

(B) The frequency of the tuning fork $f$ remains constant. For a closed pipe, the fundamental frequency is given by $f = \frac{v}{4L}$, where $v$ is the speed of sound and $L$ is the length of the pipe.
Since $f$ is constant, $\frac{v_1}{L_1} = \frac{v_2}{L_2}$, which implies $\frac{L_2}{L_1} = \frac{v_2}{v_1}$.
The speed of sound $v$ is proportional to the square root of the absolute temperature $T$ $(v \propto \sqrt{T})$.
Thus, $\frac{L_2}{L_1} = \sqrt{\frac{T_2}{T_1}}$.
Given $T_1 = 27 + 273 = 300 \,K$ and $T_2 = 7 + 273 = 280 \,K$.
$L_1 = 20 \,cm = 200 \,mm$.
$L_2 = L_1 \sqrt{\frac{280}{300}} = 200 \times \sqrt{\frac{28}{30}} = 200 \times \sqrt{0.9333} \approx 200 \times 0.966 = 193.2 \,mm$.
The change in length $\Delta L = L_1 - L_2 = 200 \,mm - 193.2 \,mm = 6.8 \,mm \approx 7 \,mm$.

(B) For a closed organ pipe,the first overtone is the $3^{rd}$ harmonic. The frequency is given by $f = \frac{3v_1}{4L} = \frac{3}{4L} \sqrt{\frac{1}{\beta \rho_1}}$,where $\beta$ is the compressibility.
For an open organ pipe,the first overtone is the $2^{nd}$ harmonic. The frequency is given by $f = \frac{2v_2}{2L'} = \frac{v_2}{L'} = \frac{1}{L'} \sqrt{\frac{1}{\beta \rho_2}}$.
Since the compressibility $\beta$ is the same for both gases,and assuming the bulk modulus $K = 1/\beta$ is the same,we equate the frequencies:
$\frac{3}{4L} \sqrt{\frac{1}{\beta \rho_1}} = \frac{1}{L'} \sqrt{\frac{1}{\beta \rho_2}}$
$\frac{3}{4L \sqrt{\rho_1}} = \frac{1}{L' \sqrt{\rho_2}}$
$L' = \frac{4L}{3} \sqrt{\frac{\rho_1}{\rho_2}}$.

(B) The difference between successive resonance frequencies is $\Delta f = 595 - 425 = 170 \,Hz$ and $765 - 595 = 170 \,Hz$.
Since the difference between successive frequencies is $2f_0$ (where $f_0$ is the fundamental frequency), the pipe must be a closed organ pipe.
Thus, $2f_0 = 170 \,Hz$, which gives $f_0 = 85 \,Hz$.
The fundamental frequency of a closed organ pipe is given by $f_0 = \frac{v}{4l}$.
Substituting the given values: $85 = \frac{340}{4l}$.
Solving for $l$: $l = \frac{340}{4 \times 85} = \frac{340}{340} = 1 \,m$.

(A) The fundamental frequency of a closed pipe is given by $f = \frac{v}{4l}$,where $v$ is the speed of sound in the gas and $l$ is the length of the pipe.
Since $v = \sqrt{\frac{\gamma RT}{M}}$,we have $f = \frac{1}{4l} \sqrt{\frac{\gamma RT}{M}}$.
Given that the fundamental frequency $f$ and temperature $T$ are the same for both pipes,we have $l \propto \frac{1}{\sqrt{M}}$,where $M$ is the molar mass of the gas.
For oxygen $(O_2)$,$M_1 = 32 \ g/mol$. For hydrogen $(H_2)$,$M_2 = 2 \ g/mol$.
The ratio of their lengths is $\frac{l_1}{l_2} = \sqrt{\frac{M_2}{M_1}} = \sqrt{\frac{2}{32}} = \sqrt{\frac{1}{16}} = \frac{1}{4}$.
Thus,the ratio of their lengths is $1:4$.

(C) For an open pipe of length $L_1$,the fundamental frequency is $n_1 = \frac{v}{2 L_1}$,which implies $L_1 = \frac{v}{2 n_1}$.
For a closed pipe of length $L_2$,the fundamental frequency is $n_2 = \frac{v}{4 L_2}$,which implies $L_2 = \frac{v}{4 n_2}$.
The combined pipe is closed at one end and open at the other,with total length $L = L_1 + L_2$.
The fundamental frequency of the combined pipe is $n = \frac{v}{4 L} = \frac{v}{4(L_1 + L_2)}$.
Substituting the values of $L_1$ and $L_2$: $n = \frac{v}{4(\frac{v}{2 n_1} + \frac{v}{4 n_2})} = \frac{v}{4v(\frac{1}{2 n_1} + \frac{1}{4 n_2})} = \frac{1}{\frac{2}{4 n_1} + \frac{1}{4 n_2}} = \frac{1}{\frac{1}{2 n_1} + \frac{1}{4 n_2}} = \frac{1}{\frac{2 n_2 + n_1}{4 n_1 n_2}} = \frac{4 n_1 n_2}{n_1 + 2 n_2}$.
Wait,re-evaluating the denominator: $n = \frac{1}{\frac{2}{4 n_1} + \frac{1}{4 n_2}} = \frac{1}{\frac{1}{2 n_1} + \frac{1}{4 n_2}} = \frac{4 n_1 n_2}{2 n_2 + n_1}$.
Given the options,the correct expression is $\frac{2 n_1 n_2}{n_1 + 2 n_2}$ if the length relations were different,but based on standard derivation,the result is $\frac{2 n_1 n_2}{n_1 + 2 n_2}$ is not matching. Let's re-check: $L = \frac{v}{2 n_1} + \frac{v}{4 n_2} = v \frac{2 n_2 + n_1}{4 n_1 n_2}$. Then $n = \frac{v}{4 L} = \frac{v}{4 v \frac{2 n_2 + n_1}{4 n_1 n_2}} = \frac{n_1 n_2}{n_1 + 2 n_2}$.

(C) The distance between two consecutive nodes in a standing wave is equal to half the wavelength,$\lambda/2 = L$,so $\lambda = 2L$.
The speed of sound $v$ in a gas is given by $v = f \lambda = f(2L) = 2fL$.
The speed of sound in an ideal gas is also given by $v = \sqrt{\frac{\gamma R T}{M}}$.
Equating the two expressions for $v$: $2fL = \sqrt{\frac{\gamma R T}{M}}$.
Squaring both sides: $4f^2 L^2 = \frac{\gamma R T}{M}$.
Solving for $\gamma$: $\gamma = \frac{4 M f^2 L^2}{R T}$.

(B) Let $L_o$ and $L_c$ be the lengths of the open and closed pipes respectively. Given $L_o / L_c = 2 / 3$.
For an open pipe, the frequency of the $n^{th}$ harmonic is given by $f_{o,n} = n(v / 2L_o)$. The third harmonic $(n=3)$ is $f_{o,3} = 3(v / 2L_o)$.
For a closed pipe, the frequency of the $m^{th}$ harmonic is given by $f_{c,m} = m(v / 4L_c)$, where $m$ must be an odd integer. The fifth harmonic $(m=5)$ is $f_{c,5} = 5(v / 4L_c)$.
The ratio of the frequencies is $R = f_{o,3} / f_{c,5} = [3(v / 2L_o)] / [5(v / 4L_c)]$.
$R = (3v / 2L_o) \times (4L_c / 5v) = (3 \times 4 \times L_c) / (2 \times 5 \times L_o) = (12 / 10) \times (L_c / L_o)$.
Since $L_o / L_c = 2 / 3$, then $L_c / L_o = 3 / 2$.
Substituting this value: $R = (1.2) \times (1.5) = 1.8 = 18 / 10 = 9 / 5$.
Thus, the ratio is $9 : 5$.

(A) Let $v$ be the speed of sound and $l$ be the length of the pipes.
For an open organ pipe,the fundamental frequency is $f_{o} = \frac{v}{2l}$.
For a closed organ pipe,the fundamental frequency is $f_{c} = \frac{v}{4l}$.
Given that the difference between these fundamental frequencies is $100 \ Hz$:
$f_{o} - f_{c} = 100 \ Hz \implies \frac{v}{2l} - \frac{v}{4l} = 100 \ Hz \implies \frac{v}{4l} = 100 \ Hz$.
The second harmonic of the open pipe is $f_{2,o} = 2 \times f_{o} = 2 \times \frac{v}{2l} = \frac{v}{l}$.
The third harmonic of the closed pipe is $f_{3,c} = 3 \times f_{c} = 3 \times \frac{v}{4l} = \frac{3v}{4l}$.
The difference between these frequencies is:
$f_{2,o} - f_{3,c} = \frac{v}{l} - \frac{3v}{4l} = \frac{4v - 3v}{4l} = \frac{v}{4l}$.
Since $\frac{v}{4l} = 100 \ Hz$,the required difference is $100 \ Hz$.

(D) For a closed organ pipe,the frequency of the $n^{th}$ harmonic is given by $f_c = \frac{nv}{4L_c}$. For the $7^{th}$ harmonic,$n = 7$,so $f_c = \frac{7v}{4L_c}$.
For an open organ pipe,the frequency of the $n^{th}$ harmonic is given by $f_o = \frac{nv}{2L_o}$. For the $4^{th}$ harmonic,$n = 4$,so $f_o = \frac{4v}{2L_o}$.
Given that the two frequencies are in unison,$f_c = f_o$.
Therefore,$\frac{7v}{4L_c} = \frac{4v}{2L_o}$.
Simplifying the equation: $\frac{7}{4L_c} = \frac{2}{L_o}$.
Rearranging for the ratio of lengths: $\frac{L_c}{L_o} = \frac{7}{4 \times 2} = \frac{7}{8}$.

(C) An organ pipe open at both ends must have an antinode at each open end. When an extra hole is created at the center $(x = \frac{L}{2})$,the air at this point is in contact with the atmosphere,which forces an antinode to form at this position as well.
For the lowest frequency (fundamental mode),the standing wave must have antinodes at both ends and at the center hole.
The distance between two consecutive antinodes is $\frac{\lambda}{2}$.
Here,the distance between the end antinode and the center antinode is $\frac{L}{2}$.
Therefore,$\frac{\lambda}{2} = \frac{L}{2} \implies \lambda = L = 25 \,cm = 0.25 \,m$.
The frequency $f$ is given by $f = \frac{v}{\lambda}$.
Substituting the values,$f = \frac{340}{0.25} = 1360 \,Hz$.

(D) The frequency of the source is $f = 340 \,Hz$ and the velocity of sound is $v = 340 \,m/s$.
The wavelength of the sound wave is $\lambda = \frac{v}{f} = \frac{340}{340} = 1 \,m = 100 \,cm$.
For a tube closed at one end, resonance occurs when the length of the air column $l$ satisfies $l = (2n-1) \frac{\lambda}{4}$, where $n = 1, 2, 3, \dots$.
For $n=1$, $l_1 = \frac{\lambda}{4} = \frac{100}{4} = 25 \,cm$.
For $n=2$, $l_2 = \frac{3\lambda}{4} = \frac{3 \times 100}{4} = 75 \,cm$.
For $n=3$, $l_3 = \frac{5\lambda}{4} = \frac{5 \times 100}{4} = 125 \,cm$.
Since the total length of the tube is $120 \,cm$, the possible air column lengths are $25 \,cm$ and $75 \,cm$.
The height of the water level $h$ from the bottom is given by $h = L - l$, where $L = 120 \,cm$ is the total length of the tube.
For $l_1 = 25 \,cm$, $h_1 = 120 - 25 = 95 \,cm = 0.95 \,m$.
For $l_2 = 75 \,cm$, $h_2 = 120 - 75 = 45 \,cm = 0.45 \,m$.
The minimum height of the water level is $0.45 \,m$.

(C) Let the length of the pipe be $l$ and the speed of sound be $v$.
The fundamental frequency of a closed pipe is $f_c = \frac{v}{4l}$.
The $3^{rd}$ harmonic of the closed pipe is $f_{3,c} = 3 \times f_c = \frac{3v}{4l}$.
The fundamental frequency of the open pipe is $f_o = \frac{v}{2l}$.
According to the problem, the fundamental frequency of the open pipe is less than the $3^{rd}$ harmonic of the closed pipe by $55 \,Hz$:
$f_{3,c} - f_o = 55 \,Hz$
$\frac{3v}{4l} - \frac{v}{2l} = 55$
$\frac{3v - 2v}{4l} = 55$
$\frac{v}{4l} = 55 \,Hz$
Since the fundamental frequency of the closed pipe is $f_c = \frac{v}{4l}$, the value is $55 \,Hz$.

(D) The frequency of a closed organ pipe vibrating in its first harmonic is $n_1 = \frac{v_1}{4 l_1}$.
The frequency of an open organ pipe vibrating in its third harmonic is $n_3 = \frac{3 v_2}{2 l_2}$.
Since both pipes are in resonance with the same tuning fork,$n_1 = n_3$.
Therefore,$\frac{v_1}{4 l_1} = \frac{3 v_2}{2 l_2}$,which implies $\frac{l_1}{l_2} = \frac{1}{6} \left( \frac{v_1}{v_2} \right)$.
The speed of sound in a gas is given by $v = \sqrt{\frac{B}{\rho}}$,where $B$ is the bulk modulus (reciprocal of compressibility) and $\rho$ is the density.
Since the compressibility is equal,$B_1 = B_2 = B$.
Thus,$\frac{v_1}{v_2} = \sqrt{\frac{B/\rho_1}{B/\rho_2}} = \sqrt{\frac{\rho_2}{\rho_1}}$.
Substituting this into the ratio of lengths,we get $\frac{l_1}{l_2} = \frac{1}{6} \sqrt{\frac{\rho_2}{\rho_1}}$.

(A) Initially,for an open pipe of length '$L$',the fundamental frequency is given by:
$f = \frac{V}{2L}$
When the tube is dipped vertically in water such that half of its length is submerged,the remaining length of the air column above the water surface is '$L/2$'.
The water surface acts as a closed end. Thus,the tube now behaves as a pipe closed at one end with an effective length '$L' = L/2$'.
The fundamental frequency of a pipe closed at one end is given by:
$f' = \frac{V}{4L'}$
Substituting '$L' = L/2$':
$f' = \frac{V}{4(L/2)} = \frac{V}{2L} = f$
Therefore,the new fundamental frequency remains '$f$'.

(C) For a tube closed at one end,the frequencies of harmonics are given by $f_n = n \frac{v}{4L}$,where $n = 1, 3, 5, ...$ are odd integers. The fifth harmonic corresponds to $n = 5$.
The wavelength $\lambda$ is given by $\lambda = \frac{4L}{n} = \frac{4 \times 50 \ cm}{5} = 40 \ cm$.
The standing wave equation for a closed pipe (taking the open end at $x = 0$) is $y = A \sin(kx) \cos(\omega t)$,where $k = \frac{2\pi}{\lambda} = \frac{2\pi}{40} = \frac{\pi}{20} \ rad/cm$.
The particle at the open end $(x = 0)$ is at a node of the displacement wave (or antinode of pressure),but in terms of displacement amplitude,the open end is an antinode. However,the standard equation $y = A \sin(kx) \cos(\omega t)$ assumes $x=0$ is a node. For an open end at $x=0$,the displacement is $y = A \cos(kx) \cos(\omega t)$.
At $x = 0$,$y_1 = A \cos(0) \cos(\omega t) = A \cos(\omega t)$.
At $x = 42 \ cm$,$y_2 = A \cos(k \times 42) \cos(\omega t) = A \cos(\frac{\pi}{20} \times 42) \cos(\omega t) = A \cos(2.1\pi) \cos(\omega t) = A \cos(2\pi + 0.1\pi) \cos(\omega t) = A \cos(0.1\pi) \cos(\omega t)$.
Since both particles oscillate in phase (the cosine terms have the same sign),the phase difference is $0^{\circ}$.

(A) For a rod clamped at its midpoint,the midpoint acts as a node and the two ends act as antinodes for the fundamental longitudinal vibration.
Let $L$ be the length of the rod. The distance between two consecutive antinodes is $\lambda/2$. Here,the rod is clamped at the center,so the length $L$ corresponds to the distance between the two ends (antinodes),which is $\lambda/2$.
Thus,$L = \lambda/2$,which implies $\lambda = 2L$.
The fundamental frequency $f$ is given by $f = v/\lambda$,where $v$ is the speed of sound.
Substituting $\lambda = 2L$,we get $f = v/(2L)$.
Given: $L = 125 \ cm = 1.25 \ m$ and $v = 5000 \ ms^{-1}$.
$f = 5000 / (2 \times 1.25) = 5000 / 2.5 = 2000 \ Hz$.
$2000 \ Hz = 2 \ kHz$.

(D) The tube is closed at one end,so it acts as a closed organ pipe. The lowest frequency (fundamental frequency) produced in the tube corresponds to the condition where the length of the tube $l = \frac{\lambda}{4}$.
Given: $l = 1.0 \text{ m}$,$f = 84 \text{ Hz}$.
Therefore,the wavelength $\lambda = 4l = 4 \times 1.0 = 4 \text{ m}$.
The velocity of sound in air is given by $v = f \lambda = 84 \times 4 = 336 \text{ m/s}$.
The velocity of sound is also given by the formula $v = \sqrt{\frac{\gamma P}{\rho}}$,where $P$ is the atmospheric pressure and $\rho$ is the density of air.
Squaring both sides,we get $v^{2} = \frac{\gamma P}{\rho}$.
Rearranging for $\gamma$,we get $\gamma = \frac{v^{2} \rho}{P}$.
Substituting the values: $\gamma = \frac{(336)^{2} \times 1.2}{1.0 \times 10^{5}} = \frac{112896 \times 1.2}{100000} = \frac{135475.2}{100000} \approx 1.355$.
Rounding to the nearest option,we get $\gamma = 1.4$.

(C) Let the length of the closed organ pipe be $l$. The fundamental frequency of an open organ pipe of length $L = 2l$ is given by $f_{open} = \frac{v}{2L} = \frac{v}{2(2l)} = \frac{v}{4l} = 100 \ Hz$.
The frequencies of a closed organ pipe of length $l$ are given by $f_n = (2n - 1) \frac{v}{4l}$,where $n = 1, 2, 3, ...$.
The first harmonic (fundamental) is $f_1 = \frac{v}{4l} = 100 \ Hz$.
The third harmonic corresponds to $n = 2$ (since the harmonics of a closed pipe are odd multiples of the fundamental: $f_1, 3f_1, 5f_1, ...$).
Therefore,the frequency of the third harmonic is $f_3 = 3 \times f_1 = 3 \times 100 \ Hz = 300 \ Hz$.

(B) For an open pipe,the fundamental frequency $f$ is given by the formula $f = \frac{v}{2l}$,where $v$ is the speed of sound and $l$ is the length of the pipe.
Given: $f = 5100 \ Hz$,$v = 340 \ ms^{-1}$.
Substituting the values into the formula:
$5100 = \frac{340}{2l}$
$l = \frac{340}{5100 \times 2}$
$l = \frac{340}{10200} = \frac{34}{1020} = \frac{1}{30} \ m$.
To convert the length into $cm$,multiply by $100$:
$l = \frac{1}{30} \times 100 = \frac{10}{3} \ cm$.

(B) Let the length of the closed pipe be $L_{1}$ and the length of the open pipe be $L_{2}$.
The fundamental frequency of a closed pipe is given by $v_{1} = \frac{v}{4 L_{1}}$.
The frequency of the second harmonic of an open pipe is given by $v_{2} = 2 \times \frac{v}{2 L_{2}} = \frac{v}{L_{2}}$.
According to the problem,the fundamental frequency of the closed pipe is equal to the frequency of the second harmonic of the open pipe,so $v_{1} = v_{2}$.
Substituting the expressions,we get $\frac{v}{4 L_{1}} = \frac{v}{L_{2}}$.
Rearranging the terms,we find $\frac{L_{1}}{L_{2}} = \frac{1}{4}$.
Thus,the ratio of their lengths is $1: 4$.

(D) For a closed pipe,the frequency of the $n$-th overtone is given by $f_{c} = \frac{(2n+1)v}{4l_{1}}$,where $n$ is the overtone number. For the first overtone $(n=1)$,$f_{c} = \frac{3v}{4l_{1}}$.
For an open pipe,the frequency of the $n$-th overtone is given by $f_{o} = \frac{(n+1)v}{2l_{2}}$,where $n$ is the overtone number. For the first overtone $(n=1)$,$f_{o} = \frac{2v}{2l_{2}} = \frac{v}{l_{2}}$.
Given that the frequencies are equal: $\frac{3v}{4l_{1}} = \frac{v}{l_{2}}$.
Rearranging for the ratio: $\frac{l_{1}}{l_{2}} = \frac{3}{4}$.