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

(C) For an open organ pipe of length $L$,the fundamental frequency is given by $n = \frac{v}{2L}$.
For the first pipe of length $l$,the frequency is $n_1 = \frac{v}{2l}$.
For the second pipe of length $(l + \Delta l)$,the frequency is $n_2 = \frac{v}{2(l + \Delta l)}$.
The beat frequency is the difference between the two frequencies: $f_{beat} = n_1 - n_2 = \frac{v}{2l} - \frac{v}{2(l + \Delta l)}$.
Taking $\frac{v}{2}$ as a common factor: $f_{beat} = \frac{v}{2} \left( \frac{1}{l} - \frac{1}{l + \Delta l} \right)$.
Simplifying the expression: $f_{beat} = \frac{v}{2} \left( \frac{l + \Delta l - l}{l(l + \Delta l)} \right) = \frac{v \Delta l}{2l(l + \Delta l)}$.
Since $\Delta l$ is very small compared to $l$,we can approximate $l + \Delta l \approx l$.
Therefore,$f_{beat} \approx \frac{v \Delta l}{2l^2}$.

(A) For a tube closed at one end,the fundamental frequency is given by $f_{closed} = \frac{v}{4L}$,where $v$ is the speed of sound and $L$ is the length of the tube.
Given $f_{closed} = 512 \ Hz$.
For a tube open at both ends,the fundamental frequency is given by $f_{open} = \frac{v}{2L}$.
Comparing the two,we get $f_{open} = 2 \times f_{closed}$.
Therefore,$f_{open} = 2 \times 512 \ Hz = 1024 \ Hz$.

(C) For a closed pipe of length $l_1$,the fundamental frequency is $f_{c1} = \frac{v}{4l_1}$. The first overtone is the third harmonic,$f_{c,over1} = 3 \times \frac{v}{4l_1} = \frac{3v}{4l_1}$.
For an open pipe of length $l_2$,the fundamental frequency is $f_{o1} = \frac{v}{2l_2}$. The first overtone is the second harmonic,$f_{o,over1} = 2 \times \frac{v}{2l_2} = \frac{v}{l_2}$.
Given that the first overtones are identical in frequency:
$\frac{3v}{4l_1} = \frac{v}{l_2}$
$\frac{3}{4l_1} = \frac{1}{l_2}$
$\frac{l_1}{l_2} = \frac{3}{4}$
Therefore,the ratio of their lengths is $3:4$.

(D) For a closed pipe of length $l$,the fundamental frequency is $f_0 = \frac{v}{4l}$. The first overtone is the next harmonic,which is $f_1 = \frac{3v}{4l}$.
For an open pipe of the same length $l$,the fundamental frequency is $f_{open,0} = \frac{v}{2l}$.
The first overtone of an open pipe is $f_{open,1} = \frac{2v}{2l} = \frac{v}{l}$.
Comparing $f_1 = \frac{3v}{4l}$ with the options:
- Option $A$: $f_{open,0} = \frac{v}{2l} = \frac{2v}{4l} \neq \frac{3v}{4l}$.
- Option $B$: $2 \times f_{open,0} = 2 \times \frac{v}{2l} = \frac{v}{l} = \frac{4v}{4l} \neq \frac{3v}{4l}$.
- Option $C$: $f_{open,1} = \frac{v}{l} = \frac{4v}{4l} \neq \frac{3v}{4l}$.
Therefore,none of the given options are correct.

(C) For a closed pipe (vessel),the fundamental frequency $n$ is given by $n = \frac{v}{4l}$,where $v$ is the speed of sound and $l$ is the length of the air column.
From the formula,we can see that $n \propto \frac{1}{l}$.
When the vessel is partially filled with water,the length of the air column $l$ decreases.
Since the frequency $n$ is inversely proportional to the length $l$,a decrease in $l$ leads to an increase in the frequency $n$.
Therefore,the frequency of vibration of the air column increases.

(D) The fundamental frequency of a closed pipe is given by $n = \frac{v}{4l}$,where $v$ is the speed of sound and $l$ is the length of the tube.
$1$. Replacing air with hydrogen gas: Since $v = \sqrt{\frac{\gamma RT}{M}}$,the speed of sound is inversely proportional to the square root of the molar mass $(v \propto \frac{1}{\sqrt{M}})$. Because the molar mass of hydrogen $(M_{H_2})$ is less than that of air $(M_{air})$,the speed of sound in hydrogen is higher,which increases the frequency $n$. Thus,option $(A)$ is correct.
$2$. Decreasing the length of the tube: Since $n \propto \frac{1}{l}$,decreasing the length $l$ increases the fundamental frequency $n$. Thus,option $(B)$ is correct.
$3$. Opening the closed end of the tube: An open pipe has a fundamental frequency $n_{open} = \frac{v}{2l}$,which is twice the fundamental frequency of a closed pipe $(n_{closed} = \frac{v}{4l})$. Thus,opening the end increases the frequency. Option $(C)$ is correct.
Since all three methods increase the fundamental frequency,the correct answer is $(D)$.

(D) For a pipe closed at one end,the fundamental frequency is given by $n = \frac{v}{4l}$.
Here,$n = 166 \,Hz$ and $v = 332 \,m/s$.
Rearranging the formula for length $l$,we get $l = \frac{v}{4n}$.
Substituting the values: $l = \frac{332}{4 \times 166} = \frac{332}{664} = 0.5 \,m$.
Thus,the length of the air column is $0.5 \,m$.

(A) The fundamental frequency $(n)$ of an open organ pipe is given by the formula: $n = \frac{v}{2l}$.
Given:
Velocity of sound $(v)$ = $350 \ m/s$.
Length of the pipe $(l)$ = $50 \ cm = 0.5 \ m$.
Substituting the values into the formula:
$n = \frac{350}{2 \times 0.5} = \frac{350}{1} = 350 \ Hz$.
Therefore,the fundamental frequency is $350 \ Hz$.

(B) For a closed organ pipe of length $l$,the fundamental frequency (first note) is given by $n_1 = \frac{v}{4l}$.
Given $v = 330 \ m/s$ and $l = 1 \ m$,we have $n_1 = \frac{330}{4 \times 1} = \frac{330}{4} \ Hz$.
The frequencies of the notes in a closed pipe are given by odd harmonics: $n_k = (2k - 1)n_1$,where $k = 1, 2, 3, ...$.
The first note is $n_1 = 1 \times n_1$.
The second note (first overtone) is $n_2 = 3 \times n_1 = 3 \times \frac{330}{4} \ Hz$.

(C) The fundamental frequency of a closed organ pipe of length $l$ is given by $n_{\text{closed}} = \frac{v}{4l} = f$.
The fundamental frequency of an open organ pipe of the same length $l$ is given by $n_{\text{open}} = \frac{v}{2l}$.
By comparing the two expressions,we get $n_{\text{open}} = 2 \times \left( \frac{v}{4l} \right) = 2f$.
Therefore,the fundamental frequency of the open organ pipe is $2f$.

(B) The minimum frequency that a human can hear is $f_{min} = 20 \ Hz$.
For a closed pipe of length $l$,the fundamental frequency is given by $f = \frac{v}{4l}$.
To find the maximum length $l$ that produces a just audible sound,we use the minimum audible frequency:
$20 = \frac{336}{4l}$
$l = \frac{336}{4 \times 20}$
$l = \frac{336}{80}$
$l = 4.2 \ m$.
Thus,the maximum length of the pipe is $4.2 \ m$.

(C) For a closed organ pipe of length $L_1$,the frequency of the $k$-th overtone is given by $n = \frac{(2k+1)v}{4L_1}$. For the first overtone,$k=1$,so $n_1 = \frac{3v}{4L_1}$.
For an open organ pipe of length $L_2$,the frequency of the $m$-th overtone is given by $n = \frac{(m+1)v}{2L_2}$. For the third overtone,$m=3$,so $n_2 = \frac{4v}{2L_2} = \frac{2v}{L_2}$.
Since both pipes are in resonance with the same tuning fork,$n_1 = n_2$.
Equating the two: $\frac{3v}{4L_1} = \frac{2v}{L_2}$.
Rearranging for the ratio $\frac{L_1}{L_2}$: $\frac{L_1}{L_2} = \frac{3v}{4} \times \frac{1}{2v} = \frac{3}{8}$.
Thus,the ratio is $3:8$.

(B) For a closed pipe (resonance air column),the fundamental frequency is given by the formula $n = \frac{v}{4l}$,where $n$ is the frequency,$v$ is the speed of sound,and $l$ is the length of the air column.
Given: $n = 250 \ Hz$ and $l = 20 \ cm = 0.2 \ m$.
Substituting the values into the formula:
$250 = \frac{v}{4 \times 0.2}$
$250 = \frac{v}{0.8}$
$v = 250 \times 0.8 = 200 \ m/s$.
Therefore,the speed of sound in air is $200 \ m/s$.

(B) The fundamental frequency of an open tube of length $L$ is given by $f_0 = \frac{v}{2L}$,where $v$ is the speed of sound in air.
When the tube is dipped vertically into water such that half of its length is inside the water,the tube effectively becomes a closed pipe (closed at one end by the water surface) with a new length $L' = L/2$.
The fundamental frequency of a closed pipe of length $L'$ is given by $f' = \frac{v}{4L'}$.
Substituting $L' = L/2$ into the formula,we get $f' = \frac{v}{4(L/2)} = \frac{v}{2L}$.
Comparing this with the original frequency,we find $f' = f_0$.

(B) For a closed organ pipe,the frequencies of the notes are given by the formula $f_n = \frac{(2n-1)v}{4L}$,where $n = 1, 2, 3, \dots$ represents the $n^{th}$ note.
Here,$L = 1.5 \, m$ and $v = 330 \, m/s$.
The first note (fundamental frequency) corresponds to $n = 1$.
The second note corresponds to $n = 2$.
Substituting $n = 2$ into the formula:
$f_2 = \frac{(2 \times 2 - 1)v}{4L} = \frac{3v}{4L}$.
$f_2 = \frac{3 \times 330}{4 \times 1.5} = \frac{990}{6} = 165 \, Hz$.

(B) The fundamental frequency $(n_1)$ of an open pipe is given by the formula: $n_1 = \frac{v}{2l}$.
Given,$v = 330 \ ms^{-1}$ and $l = 30 \ cm = 0.3 \ m$.
Substituting the values: $n_1 = \frac{330}{2 \times 0.3} = \frac{330}{0.6} = 550 \ Hz$.
The frequencies of the harmonics in an open pipe are given by $n_k = k \times n_1$,where $k = 1, 2, 3, \dots$ is the harmonic number.
We are given the source frequency $f = 1.1 \ kHz = 1100 \ Hz$.
Setting $n_k = f$,we get: $k \times 550 = 1100$.
Solving for $k$: $k = \frac{1100}{550} = 2$.
Thus,the $2^{nd}$ harmonic mode is resonantly excited.

(B) The fundamental frequency of a closed organ pipe is given by $n = \frac{v}{4l}$.
Let the lengths of the two pipes be $l_1 = 1 \ m$ and $l_2$ (where $l_2 < l_1$).
The frequencies are $n_1 = \frac{v}{4l_1}$ and $n_2 = \frac{v}{4l_2}$.
Since $l_2 < l_1$,$n_2 > n_1$. The beat frequency is $n_2 - n_1 = 4 \ Hz$.
Substituting the values: $\frac{v}{4l_2} - \frac{v}{4l_1} = 4$.
$\frac{300}{4} \left( \frac{1}{l_2} - \frac{1}{1} \right) = 4$.
$75 \left( \frac{1}{l_2} - 1 \right) = 4$.
$\frac{1}{l_2} - 1 = \frac{4}{75} = 0.0533$.
$\frac{1}{l_2} = 1.0533$.
$l_2 = \frac{1}{1.0533} \approx 0.9493 \ m = 94.93 \ cm$.
Thus,the length of the shorter pipe is approximately $94.9 \ cm$.

(A) In a resonance column,the closed end acts as a displacement node and a pressure antinode.
At the closed end,the pressure variation due to the acoustic wave reaches its maximum amplitude ${\rho _0}$.
The total pressure at any point is the sum of the atmospheric pressure ${\rho _A}$ and the pressure variation due to the sound wave.
Therefore,the maximum pressure at the closed end is ${\rho _{\max }} = {\rho _A} + {\rho _0}$.
The minimum pressure at the closed end is ${\rho _{\min }} = {\rho _A} - {\rho _0}$.
The ratio of maximum to minimum pressure is given by $\frac{{{\rho _{\max }}}}{{{\rho _{\min }}}} = \frac{{{\rho _A} + {\rho _0}}}{{{\rho _A} - {\rho _0}}}$.

(C) The fundamental frequency of a closed pipe is given by $n = \frac{v}{4l}$.
Since $v$ is constant,$n \propto \frac{1}{l}$,which implies $\frac{n_1}{n_2} = \frac{l_2}{l_1}$.
Given the ratio of lengths $l_1 : l_2 = 25 : 26$,we have $\frac{n_1}{n_2} = \frac{26}{25}$,or $n_1 = \frac{26}{25}n_2$.
The beat frequency is $|n_1 - n_2| = 10$.
Substituting $n_1$,we get $\frac{26}{25}n_2 - n_2 = 10$.
$\frac{1}{25}n_2 = 10 \implies n_2 = 250 \text{ Hz}$.
Then $n_1 = \frac{26}{25} \times 250 = 260 \text{ Hz}$.
Thus,the frequencies are $260 \text{ Hz}$ and $250 \text{ Hz}$.

(A) Let $l_1$ be the length of the closed organ pipe and $l_2$ be the length of the open organ pipe.
For a closed organ pipe,the fundamental frequency is $n_1 = \frac{v}{4l_1}$.
For an open organ pipe,the fundamental frequency is $n_2 = \frac{v}{2l_2}$.
Given that the fundamental frequencies are the same,$n_1 = n_2$.
Therefore,$\frac{v}{4l_1} = \frac{v}{2l_2}$.
Simplifying this,we get $\frac{1}{4l_1} = \frac{1}{2l_2}$,which implies $\frac{l_1}{l_2} = \frac{2}{4} = \frac{1}{2}$.
Thus,the ratio of the lengths of the closed pipe to the open pipe is $1:2$.

(B) The distance between two consecutive nodes in a stationary wave is given by $\frac{\lambda}{2}$.
Given that the nodes are at $16 \ cm$ and $46 \ cm$ from the open end,the distance between them is:
$\frac{\lambda}{2} = 46 \ cm - 16 \ cm = 30 \ cm$.
Therefore,the wavelength $\lambda = 2 \times 30 \ cm = 60 \ cm = 0.6 \ m$.
The speed of sound $v$ is given by the formula $v = f \lambda$,where $f$ is the frequency.
Given $f = 500 \ Hz$ and $\lambda = 0.6 \ m$:
$v = 500 \times 0.6 = 300 \ m/s$.

(A) For a closed pipe,the fundamental frequency $n$ is given by the formula: $n = \frac{v}{4l}$.
Here,the speed of sound $v = 332 \, m/s$ and the length of the air column $l = 42 \, m$.
Substituting these values into the formula:
$n = \frac{332}{4 \times 42}$
$n = \frac{332}{168}$
$n \approx 1.976 \, Hz$,which is approximately $2 \, Hz$.

(A) For a closed pipe,the resonance occurs when the length of the pipe $l$ corresponds to the odd harmonics of the fundamental frequency.
The fundamental frequency (first harmonic) of a closed pipe is given by the formula $n = \frac{v}{4l}$,where $v$ is the speed of sound and $l$ is the length of the pipe.
To find the shortest length $l$ for a given frequency $n$,we rearrange the formula:
$l = \frac{v}{4n}$.
Therefore,the shortest length of the closed pipe is $\frac{v}{4n}$.

(B) For an open organ pipe,the fundamental frequency is given by $n_{open} = \frac{v}{2L} = 320 \ Hz$.
For a closed organ pipe of the same length $L$,the fundamental frequency is given by $n_{closed} = \frac{v}{4L}$.
Comparing the two expressions,we get $n_{closed} = \frac{1}{2} \times n_{open}$.
Substituting the given value: $n_{closed} = \frac{1}{2} \times 320 \ Hz = 160 \ Hz$.

(C) For a closed pipe,the fundamental frequency is given by $n_1 = \frac{v}{4L}$.
The frequencies of the harmonics are given by $n_k = (2k-1)n_1$,where $k = 1, 2, 3, \dots$.
The $1^{st}$ overtone is the $3^{rd}$ harmonic $(3n_1)$.
The $2^{nd}$ overtone is the $5^{th}$ harmonic $(5n_1)$.
Given the fundamental frequency $n_1 = 50\,Hz$.
Therefore,the frequency of the $2^{nd}$ overtone is $n_3 = 5 \times n_1 = 5 \times 50\,Hz = 250\,Hz$.

(A) The fundamental frequency of an open organ pipe is given by $n = \frac{v}{2l}$.
Given lengths are $l_1 = 25 \ cm = 0.25 \ m$ and $l_2 = 25.5 \ cm = 0.255 \ m$.
The beat frequency is $n_1 - n_2 = 10 \ \text{beats/sec}$.
Substituting the values: $\frac{v}{2l_1} - \frac{v}{2l_2} = 10$.
$\frac{v}{2} \left( \frac{1}{0.25} - \frac{1}{0.255} \right) = 10$.
$\frac{v}{2} \left( \frac{0.255 - 0.25}{0.25 \times 0.255} \right) = 10$.
$\frac{v}{2} \left( \frac{0.005}{0.06375} \right) = 10$.
$v \left( \frac{0.005}{0.1275} \right) = 10$.
$v = \frac{10 \times 0.1275}{0.005} = 2000 \times 0.1275 = 255 \ m/s$.

(A) For a tube open at both ends,the fundamental frequency $f$ is given by the formula $f = \frac{v}{2L}$,where $v$ is the velocity of sound and $L$ is the length of the tube.
Given: $f = 350 \ Hz$ and $v = 350 \ m/s$.
Substituting the values into the formula: $350 = \frac{350}{2L}$.
Simplifying the equation: $1 = \frac{1}{2L}$,which gives $2L = 1$.
Therefore,$L = 0.5 \ m$.
Converting the length into centimeters: $L = 0.5 \times 100 \ cm = 50 \ cm$.

(B) The fundamental frequency of an open organ pipe is given by $n = \frac{v}{2l}$.
Given,the beat frequency is $4 \ Hz$,so $\Delta n = n_1 - n_2 = 4$.
Substituting the values: $\frac{v}{2l_1} - \frac{v}{2l_2} = 4$.
Given $l_1 = 1.00 \ m$ and $l_2 = 1.025 \ m$.
$\frac{v}{2} \left( \frac{1}{1.00} - \frac{1}{1.025} \right) = 4$.
$\frac{v}{2} \left( \frac{1.025 - 1.00}{1.00 \times 1.025} \right) = 4$.
$\frac{v}{2} \left( \frac{0.025}{1.025} \right) = 4$.
$v = \frac{4 \times 2 \times 1.025}{0.025} = \frac{8.2}{0.025} = 328 \ m/s$.

(A) In a pipe open at one end (closed pipe),the boundary conditions require a node at the closed end and an antinode at the open end.
For a pipe of length $L$,the allowed wavelengths are given by $\lambda_n = \frac{4L}{n}$,where $n = 1, 3, 5, ...$.
The corresponding frequencies are $f_n = \frac{nv}{4L}$,where $v$ is the speed of sound.
Since $n$ can only take odd integer values,only odd harmonics are present in a pipe open at one end.

(D) Let the length of the pipe be $l$ and the speed of sound be $v$.
The fundamental frequency of an open organ pipe is given by $f_{open} = \frac{v}{2l}$.
The frequencies of a closed organ pipe are given by $f_n = \frac{nv}{4l}$,where $n$ is an odd integer $(n = 1, 3, 5, ...)$.
The third harmonic of a closed pipe corresponds to $n = 3$,so $f_{closed, 3} = \frac{3v}{4l}$.
According to the problem,the frequency of the third harmonic of the closed pipe is $100 \ Hz$ higher than the fundamental frequency of the open pipe:
$f_{closed, 3} = f_{open} + 100$
Substituting the expressions:
$\frac{3v}{4l} = \frac{v}{2l} + 100$
Rearranging the terms:
$\frac{3v}{4l} - \frac{2v}{4l} = 100$
$\frac{v}{4l} = 100$
We need to find the fundamental frequency of the open pipe,which is $f_{open} = \frac{v}{2l}$.
Since $\frac{v}{4l} = 100$,then $\frac{v}{2l} = 2 \times 100 = 200 \ Hz$.

(C) For an open organ pipe (tube $A$) of length $l$,the fundamental frequency is given by $n_A = \frac{v}{2l}$,where $v$ is the speed of sound.
For a closed organ pipe (tube $B$) of the same length $l$,the fundamental frequency is given by $n_B = \frac{v}{4l}$.
Taking the ratio of the fundamental frequency of tube $A$ to tube $B$:
$\frac{n_A}{n_B} = \frac{v/2l}{v/4l} = \frac{4l}{2l} = \frac{2}{1}$.
Therefore,the ratio is $2:1$.

(B) stationary wave is formed by the superposition of two waves of the same frequency and amplitude traveling in opposite directions.
Given the incident wave is $y_1 = a \sin (\omega t - kx)$.
For a closed organ pipe,the reflection at the closed end introduces a phase change of $\pi$.
Thus,the reflected wave is $y_2 = a \sin (\omega t + kx + \pi)$.
Using the trigonometric identity $\sin (\theta + \pi) = -\sin \theta$,we get $y_2 = -a \sin (\omega t + kx)$.
The superposition of these two waves is $y = y_1 + y_2 = a \sin (\omega t - kx) - a \sin (\omega t + kx)$.
Using the identity $\sin C - \sin D = 2 \cos \frac{C+D}{2} \sin \frac{C-D}{2}$,we get $y = 2a \cos \omega t \sin (-kx) = -2a \sin kx \cos \omega t$.
Note: The sign depends on the coordinate system origin; the wave $y = -a \sin (\omega t + kx)$ is the correct reflected component.

(C) In an open pipe vibrating in its fundamental mode,the displacement nodes are formed at the ends of the pipe,and a displacement antinode is formed at the center of the pipe.
Pressure variation is inversely related to displacement variation. Where displacement is maximum (antinode),pressure variation is minimum (node). Where displacement is minimum (node),pressure variation is maximum (antinode).
Therefore,the pressure variation is maximum at the displacement nodes,which are located at the ends of the pipe.

(C) The given frequencies are $100 \ Hz$,$300 \ Hz$,and $500 \ Hz$.
These frequencies follow the ratio $1:3:5$.
For a pipe closed at one end,the allowed frequencies are given by $f_n = n \cdot f_1$,where $n = 1, 3, 5, \dots$ (odd harmonics only).
Since the observed frequencies are odd multiples of the fundamental frequency $(100 \ Hz)$,the pipe must be closed at one end and open at the other.

(B) The fundamental frequency of an open pipe of length $L_o = 0.5 \ m$ is given by $f_o = \frac{v}{2L_o}$.
The first overtone of a closed pipe of length $l_c$ is the third harmonic,given by $f_c = \frac{3v}{4l_c}$.
According to the problem,$f_o = f_c$,so:
$\frac{v}{2 \times 0.5} = \frac{3v}{4l_c}$.
Simplifying the equation:
$\frac{v}{1} = \frac{3v}{4l_c}$.
Canceling $v$ from both sides:
$1 = \frac{3}{4l_c}$.
Solving for $l_c$:
$4l_c = 3 \implies l_c = \frac{3}{4} = 0.75 \ m$.

(B) In a closed organ pipe,only odd harmonics are produced. The fundamental frequency is $f_1 = 50 \ Hz$. The possible frequencies are given by $f_n = n \times f_1$,where $n$ is an odd integer $(n = 1, 3, 5, 7, \dots)$.
Thus,the possible frequencies are $50 \ Hz, 150 \ Hz, 250 \ Hz, 350 \ Hz$,etc.
Since $100 \ Hz$ is an even multiple of the fundamental frequency,it cannot be produced by a closed organ pipe.
Therefore,the correct option is $B$.

(B) The distance between two successive nodes in a stationary wave is $\frac{\lambda}{2}$.
For $6$ successive nodes,there are $5$ such intervals.
Therefore,the total distance is $d = 5 \times \frac{\lambda}{2} = 85 \text{ cm}$.
Solving for wavelength $\lambda$: $\lambda = \frac{2 \times 85}{5} = 34 \text{ cm} = 0.34 \text{ m}$.
The speed of sound $v$ is given by $v = f \lambda$,where $f = 1000 \text{ Hz}$.
$v = 1000 \times 0.34 = 340 \text{ m/s}$.

(B) For a closed pipe,the allowed frequencies are odd harmonics of the fundamental frequency $(n)$,given by $n, 3n, 5n, 7n, \dots$
For an open pipe,the allowed frequencies are all harmonics of the fundamental frequency $(n)$,given by $n, 2n, 3n, 4n, \dots$
Given frequencies are $255, 425,$ and $595 \text{ Hz}$.
Let the fundamental frequency be $n$. We check for the greatest common divisor $(GCD)$ of these frequencies.
$255 = 3 \times 85$
$425 = 5 \times 85$
$595 = 7 \times 85$
Since all given frequencies are odd multiples of $85 \text{ Hz}$ $(3n, 5n, 7n)$,the pipe must be closed at one end.
The fundamental frequency (base frequency) is $n = 85 \text{ Hz}$.

(A) In a closed organ pipe,the fundamental frequency (first harmonic) corresponds to a length $l_1 = \frac{\lambda}{4}$.
The third harmonic in a closed organ pipe corresponds to a length $l_3 = \frac{3\lambda}{4}$.
Therefore,the relationship between the length of the third harmonic and the fundamental frequency is $l_3 = 3 \times l_1$.
Given $l_1 = 24.7 \, cm$,we calculate $l_3 = 3 \times 24.7 \, cm = 74.1 \, cm$.

(C) The fundamental frequency of an open pipe is given by $f_1 = \frac{v}{2l}$.
Given: $v = 330 \ m/s$,$l = 33 \ cm = 0.33 \ m$.
$f_1 = \frac{330}{2 \times 0.33} = \frac{330}{0.66} = 500 \ Hz$.
The given frequency is $f = 1000 \ Hz$.
We know that $f = p \times f_1$,where $p$ is the harmonic number.
$1000 = p \times 500 \implies p = 2$.
Thus,the frequency corresponds to the second harmonic of the pipe.

(C) The frequency of a closed organ pipe of length $l$ is given by $n = \frac{v}{4l}$.
Given lengths are $l_1 = 100 \,cm = 1 \,m$ and $l_2 = 101 \,cm = 1.01 \,m$.
The number of beats per second is $n_b = \frac{16}{20} = 0.8 \,Hz$.
The beat frequency is the difference between the fundamental frequencies: $n_b = n_1 - n_2 = \frac{v}{4l_1} - \frac{v}{4l_2}$.
Substituting the values: $0.8 = \frac{v}{4} \left( \frac{1}{1} - \frac{1}{1.01} \right)$.
$0.8 = \frac{v}{4} \left( \frac{1.01 - 1}{1.01} \right) = \frac{v}{4} \left( \frac{0.01}{1.01} \right)$.
$0.8 = \frac{0.01v}{4.04}$.
$v = \frac{0.8 \times 4.04}{0.01} = 80 \times 4.04 = 323.2 \,m/s$.

(A) For an open organ pipe of length $L$,the fundamental frequency is given by $n = \frac{v}{2L}$.
The frequencies of the harmonics produced in an open organ pipe are given by $f_k = k \cdot n$,where $k = 1, 2, 3, 4, ...$.
This means that both even and odd harmonics are present in an open organ pipe.
Therefore,the frequencies are $n, 2n, 3n, 4n, ...$.

(D) The velocity of sound $v$ in a resonance tube is given by the formula $v = 2n(l_2 - l_1)$,where $n$ is the frequency of the tuning fork,$l_1$ is the first resonance length,and $l_2$ is the second resonance length.
Given: $n = 512 \ Hz$,$l_1 = 30.7 \ cm$,$l_2 = 63.2 \ cm$.
Substituting the values:
$v = 2 \times 512 \times (63.2 - 30.7) \ cm/s$
$v = 1024 \times 32.5 \ cm/s = 33280 \ cm/s$.
The actual speed of sound is given as $v_0 = 332 \ m/s = 33200 \ cm/s$.
The error in the velocity of sound is $\Delta v = |v - v_0| = |33280 - 33200| \ cm/s = 80 \ cm/s$.

(C) For an open pipe,the frequency of the $N^{th}$ harmonic is given by $n = \frac{N \cdot v}{2l}$,where $N$ is the order of the harmonic,$v$ is the speed of sound,and $l$ is the length of the pipe.
Rearranging the formula to solve for $N$:
$N = \frac{n \cdot 2l}{v}$
Given:
$n = 480 \ Hz$
$l = 1 \ m$
$v = 330 \ m/s$
Substituting the values:
$N = \frac{480 \times 2 \times 1}{330}$
$N = \frac{960}{330} \approx 2.91$
Since the harmonic order must be an integer,we round to the nearest integer,which is $3$. Therefore,it is the third harmonic.

(A) For an organ pipe open at one end,the frequency of the first overtone is given by $n_c = \frac{3v}{4l_c}$,where $v$ is the speed of sound and $l_c$ is the length of the closed pipe.
For an organ pipe open at both ends,the frequency of the third harmonic is given by $n_o = \frac{3v}{2l_o}$,where $l_o$ is the length of the open pipe.
Since the pipes are in resonance,$n_c = n_o$.
Equating the two expressions: $\frac{3v}{4l_c} = \frac{3v}{2l_o}$.
Simplifying this,we get $\frac{1}{4l_c} = \frac{1}{2l_o}$,which implies $\frac{l_c}{l_o} = \frac{2}{4} = \frac{1}{2}$.
Thus,the ratio of the length of the closed pipe to the open pipe is $1:2$.

(B) For an open tube of length $l$,the fundamental frequency is given by $n_0 = \frac{v}{2l}$.
When the tube is dipped in water such that $75\%$ of its length is inside the water,the effective length of the air column becomes $l' = l - 0.75l = 0.25l = \frac{l}{4}$.
The tube now acts as a closed organ pipe (closed at one end by water).
The fundamental frequency of this closed tube is $n = \frac{v}{4l'}$.
Substituting $l' = \frac{l}{4}$,we get $n = \frac{v}{4(l/4)} = \frac{v}{l}$.
Since $n_0 = \frac{v}{2l}$,we have $v = 2l n_0$.
Substituting this into the expression for $n$,we get $n = \frac{2l n_0}{l} = 2n_0$.
Therefore,the ratio of the new frequency of the tube to the frequency of the string is $\frac{n}{n_0} = 2$.

(B) Let $x$ be the end correction.
For the fundamental mode (first resonance),the length of the air column is $l_1 = 0.1 \ m$. The resonance condition is given by $f = \frac{v}{4(l_1 + x)}$.
For the first overtone (second resonance),the length of the air column is $l_2 = 0.35 \ m$. The resonance condition is given by $f = \frac{3v}{4(l_2 + x)}$.
Since the same tuning fork is used,the frequency $f$ remains constant. Therefore,$\frac{v}{4(l_1 + x)} = \frac{3v}{4(l_2 + x)}$.
Simplifying the equation: $l_2 + x = 3(l_1 + x)$.
Substituting the values: $0.35 + x = 3(0.1 + x)$.
$0.35 + x = 0.3 + 3x$.
$0.05 = 2x$.
$x = 0.025 \ m$.

(C) The frequency of the first overtone of a closed organ pipe of length $L_1 = L$ is given by $f_1 = \frac{3v_1}{4L_1}$,where $v_1 = \sqrt{\frac{1}{\beta \rho_1}}$ and $\beta$ is the compressibility.
The frequency of the first overtone of an open organ pipe of length $L_2$ is given by $f_2 = \frac{2v_2}{2L_2} = \frac{v_2}{L_2}$,where $v_2 = \sqrt{\frac{1}{\beta \rho_2}}$.
Given that the frequencies are equal $(f_1 = f_2)$:
$\frac{3}{4L} \sqrt{\frac{1}{\beta \rho_1}} = \frac{1}{L_2} \sqrt{\frac{1}{\beta \rho_2}}$
Canceling the common term $\sqrt{\frac{1}{\beta}}$ from both sides:
$\frac{3}{4L\sqrt{\rho_1}} = \frac{1}{L_2\sqrt{\rho_2}}$
Solving for $L_2$:
$L_2 = \frac{4L}{3} \sqrt{\frac{\rho_1}{\rho_2}}$

(C) For an open pipe,the frequency of the $2^{nd}$ harmonic is given by $f_1 = \frac{2v}{2L} = \frac{v}{L}$.
When one end is closed,the pipe becomes a closed organ pipe. The resonant frequencies for a closed pipe are given by $f_n = \frac{nv}{4L}$,where $n$ is an odd integer $(n = 1, 3, 5, \dots)$.
We are given that $f_2 > f_1$. Substituting $v = f_1 L$ into the closed pipe formula,we get $f_2 = \frac{n(f_1 L)}{4L} = \frac{n}{4}f_1$.
Since $f_2 > f_1$,we must have $\frac{n}{4} > 1$,which implies $n > 4$.
The smallest odd integer greater than $4$ is $n = 5$.
Therefore,$f_2 = \frac{5}{4}f_1$ and $n = 5$.