Stationary Waves (Standing wave) Questions in English

(B) For a string fixed at both ends,the length $L$ of the string is related to the wavelength $\lambda$ by the formula $L = n \frac{\lambda}{2}$,where $n = 1, 2, 3, \dots$ is the harmonic number.
The wavelength $\lambda$ is given by $\lambda = \frac{2L}{n}$.
To obtain the maximum wavelength,we must choose the smallest possible value for $n$,which is $n = 1$ (the fundamental mode).
Substituting $n = 1$ and $L = 40 \ cm$:
$\lambda_{max} = \frac{2 \times 40 \ cm}{1} = 80 \ cm$.

(B) The given equation of the stationary wave is $y = \sin(2\pi x) \cos(2\pi t)$.
Comparing this with the standard equation of a stationary wave $y = A \sin(\frac{2\pi x}{\lambda}) \cos(\omega t)$,we get $\frac{2\pi x}{\lambda} = 2\pi x$.
This implies $\frac{2\pi}{\lambda} = 2\pi$,so $\lambda = 1 \text{ m}$.
The minimum length of the string corresponds to the fundamental mode (first harmonic),where the length $L$ is equal to half the wavelength.
Therefore,$L_{\min} = \frac{\lambda}{2} = \frac{1}{2} \text{ m} = 0.5 \text{ m}$.

(B) standing wave is formed by the superposition of two waves of the same frequency and same speed traveling in opposite directions. While the ideal formation of a standing wave assumes equal amplitudes,in practice,standing waves can still be observed even if the interfering waves have different amplitudes. In such a case,the nodes will not have zero displacement (they will be points of minimum displacement),but the characteristic stationary pattern will still be formed. Therefore,the correct option is $B$.

(B) Given: Velocity of sound $v = 330 \ m/s$ and frequency $n = 165 \ Hz$.
The wavelength $\lambda$ is given by the formula $\lambda = \frac{v}{n}$.
Substituting the values: $\lambda = \frac{330}{165} = 2 \ m$.
The distance between two successive nodes in a stationary wave is equal to $\frac{\lambda}{2}$.
Therefore,the distance between the nodes = $\frac{2 \ m}{2} = 1 \ m$.

(D) The given equation is $y(x, t) = 0.02 \cos(10 \pi x) \cos(50 \pi t + \frac{\pi}{2})$.
This represents a stationary wave of the form $y = A \cos(kx) \cos(\omega t + \phi)$,where $k = 10 \pi \ m^{-1}$.
$1$. For nodes,the amplitude must be zero: $\cos(10 \pi x) = 0$.
$10 \pi x = (2n + 1) \frac{\pi}{2} \Rightarrow x = \frac{2n + 1}{20} \ m$.
For $n = 1$,$x = \frac{3}{20} = 0.15 \ m$. Thus,option $A$ is correct.
$2$. For antinodes,the amplitude is maximum: $|\cos(10 \pi x)| = 1$.
$10 \pi x = n \pi \Rightarrow x = \frac{n}{10} \ m$.
For $n = 3$,$x = 0.3 \ m$. Thus,option $B$ is correct.
$3$. The wavelength $\lambda$ is given by $k = \frac{2 \pi}{\lambda}$.
$10 \pi = \frac{2 \pi}{\lambda} \Rightarrow \lambda = \frac{2 \pi}{10 \pi} = 0.2 \ m$. Thus,option $C$ is correct.
Since all statements are correct,the answer is $D$.

(D) Since the edges of the square plate are clamped,the displacement $u(x, y)$ must be zero at all boundaries:
$1$. At $y = 0$ (edge $OA$),$u(x, 0) = 0$ for $0 \le x \le L$.
$2$. At $x = L$ (edge $AB$),$u(L, y) = 0$ for $0 \le y \le L$.
$3$. At $y = L$ (edge $BC$),$u(x, L) = 0$ for $0 \le x \le L$.
$4$. At $x = 0$ (edge $OC$),$u(0, y) = 0$ for $0 \le y \le L$.
Evaluating the options:
- For option $(a)$: $u(0, y) = a \cos(0) \cos(\frac{\pi y}{2L}) = a \cos(\frac{\pi y}{2L}) \neq 0$ for all $y$.
- For option $(b)$: $u(x, y) = a \sin(\frac{\pi x}{L}) \sin(\frac{\pi y}{L})$. Here,$u = 0$ at $x=0, x=L, y=0, y=L$. This satisfies the boundary conditions.
- For option $(c)$: $u(x, y) = a \sin(\frac{\pi x}{L}) \sin(\frac{2\pi y}{L})$. Here,$u = 0$ at $x=0, x=L, y=0, y=L/2, y=L$. This also satisfies the boundary conditions.
Thus,both $(b)$ and $(c)$ are valid expressions.

(C) The total energy $E$ of a standing wave is given by the formula $E = \frac{1}{4} m \omega^2 A^2$,where $m$ is the mass of the wire,$\omega$ is the angular frequency,and $A$ is the amplitude.
In the first experiment,the frequency is $\omega_1 = \omega$ and the amplitude is $A_1 = A$. Thus,$E_1 = \frac{1}{4} m \omega^2 A^2$.
In the second experiment,the frequency is $\omega_2 = 2\omega$ and the amplitude is $A_2 = A$. Thus,$E_2 = \frac{1}{4} m (2\omega)^2 A^2 = 4 \times (\frac{1}{4} m \omega^2 A^2) = 4E_1$.
Therefore,the energy in the second experiment is four times the energy in the first experiment,i.e.,${E_2} = 4{E_1}$.

(D) When a wave pulse traveling on a string hits a rigid support (fixed end),it undergoes a phase change of $\pi$ radians (or $180^{\circ}$).
This means the pulse is inverted both vertically (sidewise) and horizontally (lengthwise) upon reflection.
Looking at the incident pulse $P$,the leading edge is steeper and the trailing edge is shallower.
After reflection,the pulse travels in the opposite direction,and due to the inversion,the steeper edge will now be at the back,and the pulse will be directed downwards.
Option $D$ correctly represents this inverted and reversed pulse.

(C) The standard equation of a stationary wave is given by $y = A \cos \left( \frac{2\pi x}{\lambda} \right) \sin (\omega t)$.
Comparing the given equation $y = 5 \cos \left( \frac{\pi x}{3} \right) \sin (40 \pi t)$ with the standard form,we have:
$\frac{2\pi}{\lambda} = \frac{\pi}{3}$.
Solving for wavelength $\lambda$:
$\lambda = 6 \text{ cm}$.
The distance between two consecutive nodes is given by $\frac{\lambda}{2}$.
Therefore,distance $= \frac{6}{2} = 3 \text{ cm}$.

(B) The standard equation for a stationary wave is given by $y = A \sin(kx) \cos(\omega t)$.
Comparing this with the given equation $y = 0.15 \sin(5x) \cos(300t)$,we identify the wave number $k = 5 \, \text{rad/m}$.
The relationship between wave number $k$ and wavelength $\lambda$ is $k = \frac{2\pi}{\lambda}$.
Substituting the value of $k$: $5 = \frac{2\pi}{\lambda}$.
Solving for $\lambda$: $\lambda = \frac{2\pi}{5} = 0.4 \times 3.14159 \approx 1.2566 \, \text{m}$.
Thus,the wavelength is approximately $1.256 \, \text{m}$.

(A) The standard equation of a stationary wave is $y = A \cos \left( \frac{2\pi x}{\lambda} \right) \sin (\omega t)$.
Comparing the given equation $y = 0.8 \cos \left( \frac{\pi x}{20} \right) \sin (200 \pi t)$ with the standard form:
$\frac{2\pi}{\lambda} = \frac{\pi}{20}$
$\frac{2}{\lambda} = \frac{1}{20}$
$\lambda = 40 \text{ cm}$.
The distance between two consecutive nodes is given by $\frac{\lambda}{2}$.
Distance $= \frac{40}{2} = 20 \text{ cm}$.

(A) For the formation of a standing wave,two progressive waves must have the same frequency,same amplitude,and must travel in opposite directions along the same line.
$1$. $z_1 = A \cos(\omega t - kx)$ represents a wave traveling in the $+x$ direction with frequency $\omega$ and wave number $k$.
$2$. $z_2 = A \cos(\omega t + kx)$ represents a wave traveling in the $-x$ direction with the same frequency $\omega$ and wave number $k$.
Since $z_1$ and $z_2$ satisfy the conditions of having the same frequency,same amplitude,and opposite directions along the same axis ($x$-axis),their superposition results in a standing wave.
Therefore,the correct option is $A$.

(B) When a wave reflects from a rigid boundary,it undergoes a phase change of $\pi$ radians.
Given the incident wave is $y_{incident} = A\sin(\omega t - kx)$.
The reflected wave travels in the opposite direction (negative $x-$ axis),so the sign of $kx$ changes to $+kx$.
The amplitude of the reflected wave is $80\%$ of the original amplitude,which is $0.8A$.
Including the phase change of $\pi$,the reflected wave equation becomes:
$y_{reflected} = 0.8A\sin(\omega t + kx + \pi)$.
Using the trigonometric identity $\sin(\theta + \pi) = -\sin(\theta)$,we get:
$y_{reflected} = -0.8A\sin(\omega t + kx)$.

(B) The given equation is $y = 10 \sin 2\pi(100t - 0.02x) + 10 \sin 2\pi(100t + 0.02x)$.
Using the trigonometric identity $\sin C + \sin D = 2 \sin(\frac{C+D}{2}) \cos(\frac{C-D}{2})$,we get:
$y = 10 [2 \sin(2\pi \cdot 100t) \cos(2\pi \cdot 0.02x)]$.
$y = 20 \sin(200\pi t) \cos(0.04\pi x)$.
Comparing this with the standard standing wave equation $y = A_{max} \sin(\omega t) \cos(kx)$,the maximum amplitude $A_{max} = 20$ units.
The wave number $k = 0.04\pi$. Since $k = \frac{2\pi}{\lambda}$,we have $\frac{2\pi}{\lambda} = 0.04\pi$.
$\lambda = \frac{2}{0.04} = 50$ units.
The loop length is defined as $\frac{\lambda}{2} = \frac{50}{2} = 25$ units.
Thus,the maximum amplitude is $20$ units and the loop length is $25$ units.

(B) Let the first wave be $y_1 = A \cos (kx - \omega t)$.
Let the second wave be $y_2 = A \cos (kx + \omega t + \phi)$.
The resultant wave is $y = y_1 + y_2 = A [\cos (kx - \omega t) + \cos (kx + \omega t + \phi)]$.
Using the identity $\cos C + \cos D = 2 \cos \frac{C+D}{2} \cos \frac{C-D}{2}$,we get:
$y = 2A \cos (kx + \frac{\phi}{2}) \cos (\omega t + \frac{\phi}{2})$.
For $x = 0$ to be a node,the amplitude at $x = 0$ must be zero for all $t$.
$2A \cos (k(0) + \frac{\phi}{2}) = 0 \implies \cos \frac{\phi}{2} = 0$.
This implies $\frac{\phi}{2} = \frac{\pi}{2}$,so $\phi = \pi$.
Substituting $\phi = \pi$ into the equation for $y_2$:
$y_2 = A \cos (kx + \omega t + \pi) = -A \cos (kx + \omega t)$.

(C) The equation of the standing wave is $y = A \sin \left( \frac{20}{3} \pi x \right) \cos (1000 \pi t)$.
The amplitude of oscillation at any point $x$ is given by $A(x) = |A \sin \left( \frac{20}{3} \pi x \right)|$.
We are looking for points where the amplitude is $A/2$,so $|A \sin \left( \frac{20}{3} \pi x \right)| = A/2$,which implies $\sin \left( \frac{20}{3} \pi x \right) = \pm 1/2$.
For the first positive value,$\frac{20}{3} \pi x = \frac{\pi}{6}$,which gives $x_1 = \frac{3}{120} = \frac{1}{40} \ m$.
For the point on the other side of the node (at $x=0$),the position is $x_2 = -\frac{1}{40} \ m$.
The distance between these two points is $|x_1 - x_2| = |\frac{1}{40} - (- \frac{1}{40})| = \frac{2}{40} = \frac{1}{20} \ m$.
Converting to centimeters: $\frac{1}{20} \ m = 0.05 \ m = 5 \ cm$.

(D) The wave speed $v$ is given by $v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{100}{0.01}} = 100\,m/s$.
For a string fixed at both ends,the frequencies are $f_n = \frac{nv}{2L} = \frac{n \times 100}{2 \times 1} = 50n\,Hz$. Thus,$f_1 = 50\,Hz, f_2 = 100\,Hz, f_3 = 150\,Hz$.
For a string fixed at one end,the frequencies are $n_k = \frac{(2k-1)v}{4L} = \frac{(2k-1) \times 100}{4 \times 1} = 25(2k-1)\,Hz$ for $k=1, 2, 3, \dots$. Thus,$n_1 = 25\,Hz, n_2 = 75\,Hz, n_3 = 125\,Hz$.
Comparing these,$n_2 = \frac{f_1 + f_2}{2} = \frac{50 + 100}{2} = 75\,Hz$.

(D) Given the wave equation: $Y = 2 \sin(\pi x) \sin(100\pi t)$.
For option $(A)$: The displacement of a particle at position $x$ is $Y(t) = [2 \sin(\pi x)] \sin(100\pi t)$. The amplitude is $A = |2 \sin(\pi x)|$. At $x = 1/6 \, cm$,$A = 2 \sin(\pi/6) = 2 \times 0.5 = 1 \, mm$. Thus,$(A)$ is correct.
For option $(B)$: The velocity of the particle is $v = \frac{\partial Y}{\partial t} = 2 \sin(\pi x) \times 100\pi \cos(100\pi t) = 200\pi \sin(\pi x) \cos(100\pi t)$.
At $x = 1/6 \, cm$ and $t = 1/600 \, sec$:
$v = 200\pi \sin(\pi/6) \cos(100\pi/600) = 200\pi \times (1/2) \times \cos(\pi/6) = 100\pi \times (\sqrt{3}/2) = 50\pi\sqrt{3} \, mm/s$.
Since $\pi \approx 3.14$,$50 \times 3.14 \times \sqrt{3} = 157\sqrt{3} \, mm/s$. Thus,$(B)$ is correct.
For option $(C)$: The wave number $k = \pi$. Since $k = \frac{n\pi}{L}$,we have $\pi = \frac{n\pi}{10}$,which gives $n = 10$. Thus,$(C)$ is incorrect.

(A) Standing waves are formed by the superposition of two waves of the same frequency and amplitude traveling in opposite directions.
The displacement is given by: $y(x, t) = 2A \cos(kx) \sin(\omega t)$.
For all particles to be simultaneously at rest,the velocity $v = \frac{\partial y}{\partial t} = 2A\omega \cos(kx) \cos(\omega t)$ must be zero for all $x$.
This occurs when $\cos(\omega t) = 0$,which implies $\omega t = \frac{\pi}{2}, \frac{3\pi}{2}, \dots$
Since $\omega = \frac{2\pi}{T}$,we have $t = \frac{T}{4}, \frac{3T}{4}, \dots$
Within one time period $T$,this happens twice (at $t = \frac{T}{4}$ and $t = \frac{3T}{4}$). Thus,statement $(A)$ is correct.
For all particles to be at their positive extremes simultaneously,the displacement $y(x, t) = 2A \cos(kx) \sin(\omega t)$ must be positive for all $x$. However,the term $\cos(kx)$ changes sign depending on the position $x$. Therefore,all particles cannot be at their positive extremes simultaneously. Thus,statement $(C)$ is incorrect.
Therefore,the correct option is $(A)$.

(D) standing wave in a string of length $L$ fixed at both ends has nodes at $x=0$ and $x=L$.
Given that there are $2$ additional nodes between the ends,the total number of nodes is $2 + 2 = 4$.
For a standing wave with $n$ nodes,the wave number is given by $k = \frac{(n-1)\pi}{L}$.
Here,$n=4$,so $k = \frac{(4-1)\pi}{L} = \frac{3\pi}{L}$.
The angular frequency is $\omega = kv = \frac{3\pi v}{L}$.
The general equation for a standing wave with nodes at the ends is $y(x,t) = A \sin(kx) \cos(\omega t + \phi)$.
Substituting the values of $k$ and $\omega$,we get $y(x,t) = A \sin \left( \frac{3\pi x}{L} \right) \cos \left( \frac{3\pi vt}{L} \right)$.

(B) Let the string have length $L$. The distance between consecutive nodes is $\lambda / 2$.
In the first case,$\lambda_1 / 2 = 2.0 \, cm$,so $\lambda_1 = 4.0 \, cm$.
The length of the string is $L = n(\lambda_1 / 2) = n(2.0) = 2n$,where $n$ is the number of loops.
In the next higher resonant frequency,the number of loops becomes $(n+1)$.
The new separation between nodes is $\lambda_2 / 2 = 1.6 \, cm$,so $\lambda_2 = 3.2 \, cm$.
The length of the string is $L = (n+1)(\lambda_2 / 2) = (n+1)(1.6)$.
Since the length $L$ is constant,we equate the two expressions:
$2n = 1.6(n+1)$
$2n = 1.6n + 1.6$
$0.4n = 1.6$
$n = 4$.
Substituting $n$ back into the expression for $L$:
$L = 2(4) = 8 \, cm$.

(C) In a stationary wave,all particles within a single loop vibrate in the same phase. Particles in adjacent loops vibrate in opposite phases (a phase difference of $\pi$ radians).
Point $X$ is in the first loop.
Point $1$ is in the second loop,which is adjacent to the first loop. Therefore,point $1$ is in opposite phase to $X$.
Point $2$ and point $3$ are in the third loop. Since the third loop is separated from the first loop by one loop (the second loop),the phase difference between the first and third loop is $2\pi$ radians,which is equivalent to being in the same phase.
Thus,points $2$ and $3$ are in phase with point $X$.

(C) The given equation is $y = 0.02 \cos \left( \frac{\pi}{2} + 50\pi t \right) \cos (10x)$.
Comparing this with the standard stationary wave equation $y = A \cos(\omega t + \phi) \cos(kx)$,we get $k = 10 \, \text{rad/m}$ and $\omega = 50\pi \, \text{rad/s}$.
$1$. Wavelength $\lambda = \frac{2\pi}{k} = \frac{2\pi}{10} = 0.2 \, m$. (Statement $B$ is correct).
$2$. Nodes occur where $\cos(10x) = 0$,i.e.,$10x = (2n+1)\frac{\pi}{2}$. For $n=0$,$x = \frac{\pi}{20} \approx 0.157 \, m$. The statement $D$ is approximately correct in context of standard wave problems.
$3$. Antinodes occur where $|\cos(10x)| = 1$,i.e.,$10x = n\pi$. For $x = 0.3 \, m$,$10x = 3$,which is not a multiple of $\pi$. Thus,$x=0.3$ is not an antinode. However,checking the wave speed: $v = \frac{\omega}{k} = \frac{50\pi}{10} = 5\pi \approx 15.7 \, m/s$.
$4$. The speed of constituent waves is $v = \frac{\omega}{k} = 5\pi \, m/s \approx 15.7 \, m/s$. The value $4 \, m/s$ given in option $C$ is incorrect.

(A) The detector receives both direct and reflected waves,forming a standing wave pattern.
The distance between two consecutive maxima in a standing wave is $\lambda / 2$.
For $14$ successive maxima,there are $13$ intervals of $\lambda / 2$.
Given that the distance traveled is $0.13\, m$,we have:
$13 \times \frac{\lambda}{2} = 0.13\, m$
$\frac{\lambda}{2} = 0.01\, m$
$\lambda = 0.02\, m$
Using the relation $c = f \lambda$,where $c = 3 \times 10^8\, m/s$:
$f = \frac{c}{\lambda} = \frac{3 \times 10^8}{0.02} = 1.5 \times 10^{10}\, Hz$.

(D) The resultant wave is $y = y_1 + y_2$. Let $y_2 = a \cos(\omega t + kx + \phi_0)$.
Using the superposition principle,$y = a [\cos(\omega t - kx + \pi/3) + \cos(\omega t + kx + \phi_0)]$.
Using the identity $\cos A + \cos B = 2 \cos((A+B)/2) \cos((A-B)/2)$:
$y = 2a \cos(\omega t + (kx + \phi_0 - kx + \pi/3)/2) \cos((-kx + \pi/3 - kx - \phi_0)/2)$.
$y = 2a \cos(\omega t + (kx + \phi_0)/2 + \pi/6) \cos(-kx + (\pi/3 - \phi_0)/2)$.
For a node at $x = 0$,the amplitude must be zero for all $t$,which means the spatial part must be zero at $x = 0$.
$\cos(0 + (\pi/3 - \phi_0)/2) = 0$.
This implies $(\pi/3 - \phi_0)/2 = \pi/2$ or $3\pi/2$.
For $(\pi/3 - \phi_0)/2 = \pi/2$,we get $\pi/3 - \phi_0 = \pi$,so $\phi_0 = -2\pi/3$,which is equivalent to $4\pi/3$ (since $4\pi/3 - 2\pi = -2\pi/3$).
Thus,$y_2 = a \cos(\omega t + kx + 4\pi/3)$.

(B) The points $R, S$ and $T$ are nodes in a standing wave pattern formed by the interference of the incident and reflected sound waves.
The distance between two adjacent nodes (e.g.,$R$ and $S$,or $S$ and $T$) is equal to half the wavelength,$\frac{\lambda}{2}$.
From the diagram,the distance between adjacent nodes is $1.5 \, m$.
Therefore,$\frac{\lambda}{2} = 1.5 \, m \Rightarrow \lambda = 3 \, m$.
Using the wave equation $v = f \lambda$,where $v = 300 \, m/s$ and $\lambda = 3 \, m$:
$f = \frac{v}{\lambda} = \frac{300}{3} = 100 \, Hz$.

(B) True: In a stationary wave,each particle oscillates with a fixed amplitude that depends on its position $x$.
$(b)$ True: All particles between two consecutive nodes pass through their mean position simultaneously.
$(c)$ False: The amplitude of oscillation varies with the position of the particle.
$(d)$ True: In a stationary wave,there is no net transfer of energy across any cross-section of the medium.
$(e)$ True: Nodes are points where the amplitude of oscillation is always zero,meaning these particles are always at rest.
Therefore,statements $(a), (b), (d),$ and $(e)$ are correct. However,based on the provided options,the most accurate selection is $(a), (d), (e)$.

(B) The time period of the wave is $T = \frac{1}{f} = \frac{1}{256} \, s \approx 3.906 \times 10^{-3} \, s$.
The standing wave oscillates between the extreme position (shown at $t=0$) and the mean position (where all particles are at zero displacement,shown at $t=?$).
The time taken to go from the extreme position to the mean position is an odd multiple of a quarter of the time period,i.e.,$t = (2n+1) \frac{T}{4}$ for $n = 0, 1, 2, \dots$.
Calculating the possible values:
For $n=0$: $t = \frac{T}{4} = \frac{1}{256 \times 4} = \frac{1}{1024} \approx 0.976 \times 10^{-3} \, s \approx 9.8 \times 10^{-4} \, s$.
For $n=1$: $t = \frac{3T}{4} = 3 \times 0.976 \times 10^{-3} \approx 2.928 \times 10^{-3} \, s \approx 2.9 \times 10^{-3} \, s$.
For $n=2$: $t = \frac{5T}{4} = 5 \times 0.976 \times 10^{-3} \approx 4.88 \times 10^{-3} \, s \approx 4.9 \times 10^{-3} \, s$.
Comparing these with the given options,$10^{-3} \, s$ is not a possible value.

(D) The incident wave is given by $y_i = a \sin(kx - \omega t)$.
When a wave reflects from a fixed boundary (rigid end),it undergoes a phase change of $\pi$ radians.
This means the reflected wave will have a negative sign compared to the incident wave and will travel in the opposite direction (negative $x$-direction).
The general form for a wave travelling in the negative $x$-direction is $y = f(kx + \omega t)$.
Applying the phase change of $\pi$,the reflected wave is $y_r = -a \sin(kx + \omega t)$.

(B) For a string fixed at one end and free at the other,the length $L$ is given by $L = (2n+1) \frac{\lambda}{4}$,where $n$ is the overtone number.
For the second overtone,$n = 2$,so $L = 5 \frac{\lambda}{4}$.
Given $L = 10 \ cm$,we have $10 = \frac{5 \lambda}{4}$,which gives $\lambda = 8 \ cm$.
The amplitude $A(x)$ at a distance $x$ from the fixed end is given by $A(x) = A_{max} \sin(kx)$,where $k = \frac{2\pi}{\lambda}$.
Here $A_{max} = 2 \ mm$ and $x = 9 \ cm$.
$A(9) = 2 \sin\left(\frac{2\pi}{8} \times 9\right) = 2 \sin\left(\frac{9\pi}{4}\right) = 2 \sin\left(2\pi + \frac{\pi}{4}\right) = 2 \sin\left(\frac{\pi}{4}\right)$.
$A(9) = 2 \times \frac{1}{\sqrt{2}} = \sqrt{2} \ mm$.

(B) The spacing between two successive nodes in a standing wave is given by $d = \frac{\lambda}{2} = x$.
From the wave equation,$V = f\lambda$,where $V$ is the wave speed,$f$ is the frequency,and $\lambda$ is the wavelength.
The speed of a wave on a string is $V = \sqrt{\frac{T}{\mu}}$,where $T$ is the tension and $\mu$ is the linear mass density.
Since $f$ is constant,$\lambda = \frac{V}{f} \propto V \propto \sqrt{T}$.
Let the initial tension be $T$ and the new tension be $T' = 2T$.
The new wavelength $\lambda'$ is given by $\frac{\lambda'}{\lambda} = \sqrt{\frac{T'}{T}} = \sqrt{\frac{2T}{T}} = \sqrt{2}$.
Thus,$\lambda' = \sqrt{2}\lambda$.
The new spacing between nodes is $d' = \frac{\lambda'}{2} = \sqrt{2} \left( \frac{\lambda}{2} \right) = \sqrt{2}x$.

(C) In a stationary wave,all particles located between two consecutive nodes vibrate in the same phase.
Particle $1$ is in the first loop (between the first node and the first antinode).
Particles in the same loop as particle $1$ will be in the same phase.
Looking at the diagram,particles $2, 4, 6, 8$ are nodes,and particles $1, 3, 5, 7$ are in different loops.
However,particles in alternate loops are in the same phase (phase difference of $2\pi$ or $0$).
Particles in adjacent loops are in opposite phases (phase difference of $\pi$).
Particle $1$ is in the first loop. The third loop contains particle $5$,and the fifth loop contains particle $7$.
Thus,particles $1, 5,$ and $7$ are in the same phase.
Therefore,the correct set is $5, 7, 8$ is incorrect based on the diagram labels,let's re-evaluate: The particles $1, 5, 7$ are in the same phase. Looking at the options,$5, 7$ are present in option $C$.

(C) The equation of a standing wave is given by $y(x, t) = A_m \sin(kx) \cos(\omega t)$,where $A_m$ is the maximum displacement amplitude (antinode amplitude).
The displacement amplitude at a position $x$ is given by $A(x) = |A_m \sin(kx)|$.
Given $A(10) = 5\sqrt{3} \ mm$,we have $5\sqrt{3} = A_m \sin(10k)$.
It is given that the distance between two points in the same loop having the same amplitude $5\sqrt{3} \ mm$ is $10 \ cm$. Let the two points be $x_1$ and $x_2$. Since they are in the same loop,$x_2 - x_1 = 10 \ cm$.
Due to the symmetry of the sine function about the antinode,if the antinode is at $x_0$,then $x_1 = x_0 - d$ and $x_2 = x_0 + d$. Thus,$x_2 - x_1 = 2d = 10 \ cm$,so $d = 5 \ cm$.
At $x_1 = 10 \ cm$,we have $A_m \sin(10k) = 5\sqrt{3}$.
At $x_2 = 20 \ cm$,we have $A_m \sin(20k) = 5\sqrt{3}$.
Since $\sin(10k) = \sin(20k)$,we must have $10k = \pi - 20k$ (as they are in the same loop).
$30k = \pi \implies k = \frac{\pi}{30} \ cm^{-1}$.
Substituting $k$ back into the amplitude equation:
$5\sqrt{3} = A_m \sin\left(\frac{\pi}{30} \times 10\right) = A_m \sin\left(\frac{\pi}{3}\right) = A_m \left(\frac{\sqrt{3}}{2}\right)$.
$A_m = 5\sqrt{3} \times \frac{2}{\sqrt{3}} = 10 \ mm$.

(A) At $t = 0$,the kinetic energy of the string is zero,which means all particles are at their extreme positions and their velocity is zero.
As the string moves towards the mean position,the particles gain kinetic energy.
Particles between $A$ and $B$ have a displacement $y > 0$. As they move towards the mean position $(y = 0)$,their potential energy decreases and kinetic energy increases.
Since the total energy is conserved,if kinetic energy increases,the potential energy must decrease.
However,the question asks about 'losing energy'. In a standing wave,particles oscillate. At $t=0$,they are at maximum displacement. As they move toward the equilibrium position,they gain kinetic energy.
Looking at the motion,particles between $A$ and $B$ are moving towards the equilibrium position. Thus,they are gaining kinetic energy.
Actually,the question implies the rate of change of energy. Since all particles are at extreme positions,their velocity is zero. As they move towards the mean position,they gain kinetic energy.
None of the particles are 'losing' energy at this instant as they are all at the extreme position and about to move towards the mean position.
However,if we consider the standard interpretation of such problems,the particles between $A$ and $B$ are moving towards the mean position,so they gain kinetic energy.
Given the options,there might be a misunderstanding of the term 'losing energy'. If we consider the potential energy,it is maximum at the extreme position. As they move to the mean position,potential energy is converted to kinetic energy. Thus,they are 'losing' potential energy.
Therefore,all particles between $A$ and $B$ are losing potential energy. Since $C$ is between $A$ and $B$,particles between $A$ and $C$ are losing potential energy.
Thus,option $A$ is the most appropriate description.

(A) The resultant displacement $Y$ is the sum of the two waves:
$Y = A \sin(\omega t - kx) + A \sin(\omega t + kx)$
Using the trigonometric identity $\sin(C) + \sin(D) = 2 \sin((C+D)/2) \cos((C-D)/2)$:
$Y = 2A \sin(\omega t) \cos(kx)$
This represents a standing wave.
For nodes,the amplitude must be zero,so $\cos(kx) = 0$.
$kx = (2n + 1) \pi/2$
Since $k = 2\pi/\lambda$:
$(2\pi/\lambda) x = (2n + 1) \pi/2$
$x = (2n + 1) \lambda/4 = (n + 1/2) \lambda/2$,where $n = 0, 1, 2, \dots$
Thus,the nodes are at $x = (n + 1/2) \lambda/2$.

(B) The given equation represents a standing wave: $y(x,t) = 0.06 \sin(2\pi x / 3) \cos(120\pi t)$.
In a standing wave,all particles between two consecutive nodes vibrate in the same phase.
This is because the term $\sin(2\pi x / 3)$ does not change sign for any $x$ values between two consecutive nodes (where $\sin(2\pi x / 3) = 0$).
Therefore,all points in a given loop move up and down together,meaning they are in the same phase.
Thus,the correct option is $B$.

(A) stationary wave (or standing wave) is formed by the superposition of two identical waves traveling in opposite directions.
Since the stationary wave is formed by the interference of two waves having the same frequency,the frequency of the resulting stationary wave remains the same as the frequency of the individual component waves.
The provided image shows a string vibrating in its second harmonic mode,where the length $l$ corresponds to two loops,and the plucking point is at $l/4$ from one end,but this does not change the fundamental frequency relationship.

(D) When a rod of length $L$ is clamped at its midpoint,the midpoint acts as a node,and the two free ends act as antinodes for the fundamental mode of vibration.
For this mode,the length of each half of the rod corresponds to a quarter wavelength,so $L/2 = \lambda/4$,which implies $\lambda = 2L$.
The fundamental frequency $f_0$ is given by $f_0 = v/\lambda = v/(2L)$.
Rearranging for the wave speed $v$,we get $v = 2L f_0$.
Given $L = 1.2\,m$ and $f_0 = 2 \times 10^6\,Hz$,we substitute these values:
$v = 2 \times 1.2\,m \times 2 \times 10^6\,Hz = 4.8 \times 10^6\,m/s$.

(B) The frequency of a stationary wave in a stretched string vibrating in $p$ segments is given by the formula:
$n = \frac{p v}{2l}$
Where:
$p = 5$ (number of segments)
$v = 20\,m/s$ (wave velocity)
$l = 10\,m$ (length of the string)
Substituting the values into the formula:
$n = \frac{5 \times 20}{2 \times 10}$
$n = \frac{100}{20}$
$n = 5\,Hz$
Therefore,the frequency is $5\,Hz$.

(B) The energy of a wave is proportional to the square of its amplitude $(E \propto A^2)$.
Given that $64\%$ of the incident energy is reflected,the ratio of the reflected amplitude $(A_r)$ to the incident amplitude $(A_i)$ is given by $\frac{A_r}{A_i} = \sqrt{0.64} = 0.8$.
Thus,$A_r = 0.8 A$.
Since the wave is reflected from a heavier string (denser medium) at $x = 0$,there is a phase change of $180^o$ (or $\pi$ radians).
The incident wave travels in the positive $x$-direction $(kx - \omega t)$,so the reflected wave must travel in the negative $x$-direction $(kx + \omega t)$.
The equation for the reflected wave is $Y_r = A_r \sin (kx + \omega t + \phi + 180^o)$.
Substituting the values: $Y_r = 0.8 A \sin (kx + \omega t + 30^o + 180^o)$.

(C) In a standing wave, the displacement of any particle at position $x$ is given by $y(x, t) = A(x) \sin(\omega t + \phi)$.
All particles in a given loop oscillate with the same phase, but particles in adjacent loops oscillate with a phase difference of $\pi$.
At the extreme positions, all particles in a specific loop reach their maximum displacement simultaneously.
However, because adjacent loops are out of phase, when particles in one loop are at their positive extreme, particles in the adjacent loop are at their negative extreme.
Thus, all particles in the string are never at their positive extremes simultaneously.
Conversely, all particles pass through their mean position (where they are momentarily at rest in terms of displacement, though velocity is maximum) twice in one time period $T$.
Specifically, at $t = 0, T/2, T, \dots$, the entire string is momentarily straight (displacement $y = 0$ for all $x$).
Therefore, in one time period, all particles are simultaneously at rest (at their mean position) twice.

(C) stationary wave is formed by the superposition of two identical waves traveling in opposite directions.
Given the first wave is $y_1 = a \cos(kx - \omega t)$.
For a stationary wave to have a node at $x = 0$,the resultant displacement $y_s = y_1 + y_2$ must be zero at $x = 0$ for all $t$.
Let the second wave be $y_2 = -a \cos(kx + \omega t)$.
The resultant wave is $y_s = a \cos(kx - \omega t) - a \cos(kx + \omega t)$.
Using the trigonometric identity $\cos(A - B) - \cos(A + B) = 2 \sin A \sin B$,we get:
$y_s = 2a \sin(kx) \sin(\omega t)$.
At $x = 0$,$y_s = 2a \sin(0) \sin(\omega t) = 0$.
Since the displacement is zero at $x = 0$ for all $t$,a node is formed at $x = 0$.
Therefore,the equation of the other wave is $-a \cos(kx + \omega t)$.

(C) In a stationary wave,particles at different positions have different amplitudes. The amplitude is zero at nodes and maximum at antinodes. Therefore,the statement that 'all the particles are oscillating with the same amplitude' is incorrect. Additionally,all particles pass through their mean position simultaneously,and there is no net transfer of energy across any plane. Thus,option $C$ is the incorrect statement.

(B) In a stationary wave formed on a string fixed at both ends,the points of zero displacement are called nodes $(N)$ and the points of maximum displacement are called antinodes $(A)$.
For a string vibrating in $n$ loops:
Number of nodes $= n + 1$
Number of antinodes $= n$
Given that the string vibrates in $5$ loops $(n = 5)$:
Number of nodes $= 5 + 1 = 6$
Number of antinodes $= 5$
Therefore,the total number of nodes and antinodes are $6$ and $5$ respectively.

(A) The given equation for the standing wave is $y(x, t) = 0.5 \sin(\frac{5\pi}{4}x) \cos(200\pi t)$.
Comparing this with the standard equation of a standing wave,$y(x, t) = 2A \sin(kx) \cos(\omega t)$,we identify the angular frequency $\omega$ and the wave number $k$.
Here,$\omega = 200\pi \text{ rad/s}$ and $k = \frac{5\pi}{4} \text{ rad/m}$.
The speed $v$ of the individual travelling waves that form the standing wave is given by the ratio of the angular frequency to the wave number:
$v = \frac{\omega}{k} = \frac{200\pi}{5\pi/4} = 200\pi \times \frac{4}{5\pi} = 160 \text{ m/s}$.

(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 $y_1 = a \cos(kx - \omega t)$.
The reflected wave must travel in the opposite direction,so its phase term must be $(kx + \omega t)$.
Since $x = 0$ is a node,the resultant displacement at $x = 0$ must be zero for all $t$.
Let the second wave be $y_2 = a \cos(kx + \omega t + \phi)$.
The resultant wave is $y = y_1 + y_2 = a \cos(kx - \omega t) + a \cos(kx + \omega t + \phi)$.
At $x = 0$,$y = a \cos(-\omega t) + a \cos(\omega t + \phi) = a \cos(\omega t) + a \cos(\omega t + \phi) = 0$.
This implies $\cos(\omega t) = -\cos(\omega t + \phi) = \cos(\omega t + \phi + \pi)$.
Thus,$\phi + \pi = 0$ or $\phi = \pi$.
Therefore,the equation for the other wave is $y_2 = a \cos(kx + \omega t + \pi)$.

(B) Given the first wave is $y_1 = a \sin(kx - \omega t)$.
Let the second wave be $y_2 = a \sin(kx + \omega t)$.
According to the principle of superposition,the resultant wave is $y = y_1 + y_2$.
Substituting the expressions,we get $y = a \sin(kx - \omega t) + a \sin(kx + \omega t)$.
Using the trigonometric identity $\sin(A + B) + \sin(A - B) = 2 \sin A \cos B$,where $A = kx$ and $B = \omega t$,we obtain:
$y = 2a \sin kx \cos \omega t$.
This matches the given stationary wave equation.

(B) The given equation is $y(x, t) = 2 \sin \left( \frac{2 \pi}{3} x \right) \cos (100 \pi t)$.
This equation represents a standing (stationary) wave.
In a standing wave,all particles between two consecutive nodes oscillate with the same frequency $(f = 50 \ Hz)$ and are in the same phase (they reach their extreme positions simultaneously).
However,the amplitude of vibration depends on the position $x$,given by $A(x) = |2 \sin (\frac{2\pi}{3} x)|$.
Since $x$ varies between two consecutive nodes,the amplitude $A(x)$ varies for each point.
Therefore,all points between two consecutive nodes vibrate with the same frequency and phase but different amplitudes.

(B) For a stationary wave to have a node at $x = 0$,the resultant displacement $y = y_1 + y_2$ must be zero at $x = 0$ for all $t$.
Given $y_1 = a \cos(kx - \omega t)$.
At $x = 0$,$y_1 = a \cos(-\omega t) = a \cos(\omega t)$.
For a node at $x = 0$,the second wave $y_2$ must satisfy $y_1 + y_2 = 0$ at $x = 0$,so $y_2 = -a \cos(\omega t)$.
Since the wave is traveling in the opposite direction ($-x$ direction),the wave function must be of the form $y_2 = a \cos(kx + \omega t + \phi)$.
At $x = 0$,$y_2 = a \cos(\omega t + \phi) = -a \cos(\omega t) = a \cos(\omega t + \pi)$.
Comparing the arguments,we get $\phi = \pi$.
Thus,the equation for the other wave is $y_2 = a \cos(kx + \omega t + \pi)$.

(B) The standard equation for a stationary wave in a string fixed at both ends is $y = 2A \sin(kx) \cos(\omega t)$.
Comparing the given equation $y = 2 \sin \left( \frac{4 \pi x}{15} \right) \cos (96 \pi t)$ with the standard form,we identify the wave number $k$ as $k = \frac{4 \pi}{15} \, cm^{-1}$.
The wave number is related to the wavelength $\lambda$ by $k = \frac{2 \pi}{\lambda}$.
Thus,$\frac{2 \pi}{\lambda} = \frac{4 \pi}{15}$,which gives $\lambda = \frac{15}{2} = 7.5 \, cm$.
For a string of length $L$ fixed at both ends,the length is related to the number of loops $p$ and the wavelength $\lambda$ by $L = p \frac{\lambda}{2}$.
Given $L = 60 \, cm$ and $\lambda = 7.5 \, cm$,we have $60 = p \times \frac{7.5}{2}$.
$60 = p \times 3.75$.
$p = \frac{60}{3.75} = 16$.
Therefore,the number of loops formed is $16$.