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

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

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

(NONE) For an open-ended pipe (both ends open), the $k^{th}$ harmonic has $k+1$ nodes.
For the $5^{th}$ harmonic $(k=5)$, the number of nodes $n = 5 + 1 = 6$.
For a pipe with one end closed, the $p^{th}$ harmonic (where $p$ must be odd) has $\frac{p+1}{2}$ nodes.
For the $9^{th}$ harmonic $(p=9)$, the number of nodes $m = \frac{9+1}{2} = 5$.
Therefore, the ratio $\frac{n}{m} = \frac{6}{5}$.

(C) Let the length of both pipes be $L$.
For an open pipe, the frequency of the $n^{th}$ harmonic is $f_n = n \cdot \frac{v}{2L}$. The first overtone is the second harmonic $(n=2)$, so $f_{open} = 2 \cdot \frac{v}{2L} = \frac{v}{L}$.
For a closed pipe, the frequency of the $n^{th}$ harmonic is $f_n = (2n-1) \cdot \frac{v}{4L}$. The first overtone is the third harmonic $(n=2)$, so $f_{closed} = 3 \cdot \frac{v}{4L}$.
The beat frequency is $|f_{open} - f_{closed}| = 3$.
$|\frac{v}{L} - \frac{3v}{4L}| = 3$ implies $\frac{v}{4L} = 3$ implies $\frac{v}{L} = 12$.
Now, the new length of the open pipe is $L' = \frac{L}{3}$ and the new length of the closed pipe is $L'' = 3L$.
The new frequency of the open pipe in the first overtone is $f'_{open} = 2 \cdot \frac{v}{2L'} = \frac{v}{L/3} = 3 \cdot \frac{v}{L} = 3 \cdot 12 = 36 \text{ Hz}$.
The new frequency of the closed pipe in the first overtone is $f''_{closed} = 3 \cdot \frac{v}{4L''} = 3 \cdot \frac{v}{4(3L)} = \frac{v}{4L} = 3 \text{ Hz}$.
The number of beats produced is $|36 - 3| = 33$.
Wait, re-evaluating the question: if the first overtone of the closed pipe is $3 \cdot \frac{v}{4L} = 3 \text{ Hz}$, then $f_{open} = 12 \text{ Hz}$.
New $f'_{open} = \frac{v}{L/3} = 3 \cdot 12 = 36$.
New $f''_{closed} = \frac{3v}{4(3L)} = \frac{v}{4L} = 3$.
Beats = $36 - 3 = 33$.
Since $33$ is not in the options, let's re-check the overtone definition. First overtone of closed pipe is $3^{rd}$ harmonic $(3v/4L)$. First overtone of open pipe is $2^{nd}$ harmonic $(2v/2L = v/L)$.
$|v/L - 3v/4L| = v/4L = 3$.
If $v/L = 12$, then $v/4L = 3$.
New $f_{open} = 2 \cdot \frac{v}{2(L/3)} = 3 \cdot \frac{v}{L} = 36$.
New $f_{closed} = 3 \cdot \frac{v}{4(3L)} = \frac{1}{4} \cdot \frac{v}{4L} = \frac{3}{4} = 0.75$.
Perhaps the question implies fundamental frequency? If $f_{open} = v/2L$ and $f_{closed} = v/4L$, then $|v/2L - v/4L| = v/4L = 3$.
Then $v/2L = 6$.
New $f_{open} = \frac{v}{2(L/3)} = 3 \cdot \frac{v}{2L} = 18$.
New $f_{closed} = \frac{v}{4(3L)} = \frac{1}{3} \cdot \frac{v}{4L} = 1$.
$|18 - 1| = 17$.

(D) In a closed pipe of length $L$, the fundamental mode (first harmonic) has a node at the closed end and an antinode at the open end.
For the fundamental mode, the length of the pipe is $L = \frac{\lambda}{4}$, where $\lambda$ is the wavelength.
The time $t$ taken for the sound wave to travel from the open end to the closed end (a distance $L$) is given by $t = \frac{L}{v}$, where $v$ is the speed of sound.
Since $v = f\lambda$, we have $t = \frac{L}{f\lambda}$.
Substituting $L = \frac{\lambda}{4}$, we get $t = \frac{\lambda/4}{f\lambda} = \frac{1}{4f}$.
Rearranging for frequency $f$, we get $f = \frac{1}{4t}$.

(A) Let the fundamental frequency of the open pipe be $f_o = \frac{v}{2L}$.
The first overtone of the open pipe is $f_{o1} = 2f_o$.
For a closed pipe of the same length $L$, the fundamental frequency is $f_c = \frac{v}{4L} = \frac{f_o}{2}$.
The second overtone of the closed pipe is $f_{c2} = 5f_c = 5 \left( \frac{f_o}{2} \right) = 2.5f_o$.
According to the problem, $f_{c2} - f_{o1} = 100 \text{ Hz}$.
Substituting the values: $2.5f_o - 2f_o = 100 \text{ Hz}$.
$0.5f_o = 100 \text{ Hz}$.
$f_o = 200 \text{ Hz}$.

(B) Let the total length of the pipe be $L$. For a pipe open at both ends, the fundamental frequency is given by $f = \frac{v}{2L}$, where $v$ is the speed of sound in air.
When the pipe is dipped in water such that $\frac{2}{3}$ of its length is submerged, the length of the air column remaining above the water is $L' = L - \frac{2}{3}L = \frac{1}{3}L$.
This air column now acts as a pipe closed at one end (the water surface) and open at the other end.
The fundamental frequency $f'$ of a pipe closed at one end is given by $f' = \frac{v}{4L'}$.
Substituting $L' = \frac{1}{3}L$ into the formula, we get $f' = \frac{v}{4(\frac{1}{3}L)} = \frac{3v}{4L}$.
Since $f = \frac{v}{2L}$, we can write $v = 2Lf$.
Substituting this into the expression for $f'$, we get $f' = \frac{3(2Lf)}{4L} = \frac{6f}{4} = \frac{3}{2}f$.

(D) Let the fundamental frequency of the open pipe be $f_o = \frac{v}{2L}$.
The second overtone of an open pipe is given by $f_{o,2} = 3f_o = 3 \times \frac{v}{2L}$.
Let the fundamental frequency of the closed pipe be $f_c = \frac{v}{4L}$.
The third overtone of a closed pipe is the $7^{th}$ harmonic, given by $f_{c,3} = 7f_c = 7 \times \frac{v}{4L}$.
According to the problem, $f_{c,3} - f_{o,2} = 150 \text{ Hz}$.
Substituting the expressions: $\frac{7v}{4L} - \frac{3v}{2L} = 150$.
$\frac{7v - 6v}{4L} = 150 \implies \frac{v}{4L} = 150 \text{ Hz}$.
Since $f_o = \frac{v}{2L} = 2 \times \frac{v}{4L}$, we have $f_o = 2 \times 150 = 300 \text{ Hz}$.

(B) For a pipe closed at one end, the resonance occurs when the length of the air column $L$ is an odd multiple of a quarter of the wavelength $\lambda$.
The condition for resonance is $L = (2n - 1) \frac{\lambda}{4}$, where $n = 1, 2, 3, ...$
First, calculate the wavelength $\lambda$ using the formula $v = f \lambda$, where $v = 332$ m/s and $f = 415$ Hz.
$\lambda = \frac{v}{f} = \frac{332}{415} = 0.8$ m = $80$ cm.
Now, substitute $\lambda = 80$ cm into the resonance condition:
$L = (2n - 1) \frac{80}{4} = (2n - 1) \times 20$ cm.
For $n = 1, L = 20$ cm.
For $n = 2, L = 60$ cm.
For $n = 3, L = 100$ cm.
Comparing these values with the given options, $40$ cm is not an odd multiple of $20$ cm, therefore it is not in resonance.

(D) For a pipe closed at one end, the frequencies of the harmonics are given by $f_n = (2n + 1)f_0$, where $n = 0, 1, 2, ...$ represents the overtone number.
For the fourth overtone, $n = 4$.
The frequency is $f_4 = (2(4) + 1)f_0 = 9f_0$.
The general formula for the number of nodes in a closed pipe vibrating in the $n$-th overtone is $x = n + 1$.
Substituting $n = 4$, we get $x = 4 + 1 = 5$.
The general formula for the number of antinodes in a closed pipe vibrating in the $n$-th overtone is $y = n + 1$.
Substituting $n = 4$, we get $y = 4 + 1 = 5$.
Therefore, the values of $x$ and $y$ are $5$ and $5$ respectively.

(D) For an open pipe of length $L_0$, the frequency of the $n^{th}$ overtone is given by $f_n = (n+1) \frac{v}{2L_0}$.
For the fifth overtone $(n=5)$, the frequency is $f_{5, \text{open}} = (5+1) \frac{v}{2L_0} = \frac{6v}{2L_0} = \frac{3v}{L_0}$.
For a pipe closed at one end of length $L_c$, the frequency of the $n^{th}$ overtone is given by $f_n = (2n+1) \frac{v}{4L_c}$.
For the fifth overtone $(n=5)$, the frequency is $f_{5, \text{closed}} = (2 \times 5 + 1) \frac{v}{4L_c} = \frac{11v}{4L_c}$.
Since the frequencies are in unison, $f_{5, \text{open}} = f_{5, \text{closed}}$.
Therefore, $\frac{3v}{L_0} = \frac{11v}{4L_c}$.
Rearranging for the ratio $\frac{L_0}{L_c}$, we get $\frac{L_0}{L_c} = \frac{3 \times 4}{11} = \frac{12}{11}$.

(C) For an organ pipe of length $L$ closed at one end, the fundamental frequency is given by $\nu = \frac{v}{4L}$, where $v$ is the speed of sound.
When the pipe is cut into two equal parts, each part has a length of $L' = \frac{L}{2}$.
One part remains closed at one end, so its fundamental frequency is $\nu_1 = \frac{v}{4L'} = \frac{v}{4(L/2)} = \frac{2v}{4L} = 2\nu$.
The other part is now open at both ends, so its fundamental frequency is $\nu_2 = \frac{v}{2L'} = \frac{v}{2(L/2)} = \frac{v}{L} = 4 \times (\frac{v}{4L}) = 4\nu$.
Thus, the frequencies are $2\nu$ and $4\nu$.

(B) For a closed pipe of length $L_c$, the frequency of the $n^{th}$ overtone is given by $f_c = (2n + 1) \frac{v}{4L_c}$, where $n$ is the overtone number. For the third overtone $(n = 3)$, $f_c = (2 \times 3 + 1) \frac{v}{4L_c} = \frac{7v}{4L_c}$.
For an open pipe of length $L_0$, the frequency of the $n^{th}$ overtone is given by $f_0 = (n + 1) \frac{v}{2L_0}$. For the third overtone $(n = 3)$, $f_0 = (3 + 1) \frac{v}{2L_0} = \frac{4v}{2L_0} = \frac{2v}{L_0}$.
Given that the frequencies are equal, $\frac{7v}{4L_c} = \frac{2v}{L_0}$.
Rearranging the terms to find the ratio $L_c : L_0$, we get $\frac{L_c}{L_0} = \frac{7}{4 \times 2} = \frac{7}{8}$.
Thus, the ratio $L_c : L_0$ is $7:8$.

(D) For a resonance tube open at one end, the resonance condition is given by $L + e = (2n - 1) \frac{\lambda}{4}$, where $L$ is the length of the air column, $e$ is the end correction, and $n = 1, 2, 3, \dots$
For the first resonance $(n=1)$: $L_1 + e = \frac{\lambda}{4}$.
Given $L_1 = 18$ cm and $e = 1.1$ cm, we have $18 + 1.1 = \frac{\lambda}{4} \implies 19.1 = \frac{\lambda}{4} \implies \lambda = 76.4$ cm.
For the second resonance $(n=2)$: $L_2 + e = \frac{3\lambda}{4}$.
Substituting the values: $L_2 + 1.1 = 3 \times 19.1$.
$L_2 + 1.1 = 57.3$.
$L_2 = 57.3 - 1.1 = 56.2$ cm.

(A) For a tube closed at one end, the resonance occurs when the length of the air column is an odd multiple of $\frac{\lambda}{4}$.
Let the two shortest lengths be $l_1$ and $l_2$.
$l_1 = \frac{\lambda}{4}$
$l_2 = \frac{3\lambda}{4}$
Subtracting the two equations:
$l_2 - l_1 = \frac{3\lambda}{4} - \frac{\lambda}{4} = \frac{2\lambda}{4} = \frac{\lambda}{2}$
Therefore, $\lambda = 2(l_2 - l_1)$.
The speed of sound $v$ is given by $v = n\lambda$.
Substituting the value of $\lambda$:
$v = n \times 2(l_2 - l_1) = 2n(l_2 - l_1)$.

(B) The frequency of the $n^{th}$ harmonic for a string fixed at both ends is given by $f_n = n \cdot f_1$, where $f_1$ is the fundamental frequency.
Given that the wire resonates at $320 \text{ Hz}$ in the $P^{th}$ harmonic, we have $f_P = P \cdot f_1 = 320 \text{ Hz}$.
The next higher frequency in a successive mode is the $(P+1)^{th}$ harmonic, given as $f_{P+1} = (P+1) \cdot f_1 = 400 \text{ Hz}$.
Subtracting the first equation from the second: $(P+1)f_1 - P f_1 = 400 - 320$.
This gives $f_1 = 80 \text{ Hz}$.
Substituting $f_1$ back into the first equation: $P \cdot 80 = 320$.
Therefore, $P = 320 / 80 = 4$.