Principle of superposition of waves Questions in English

(C) When two identical pulses with opposite phases (one crest and one trough) overlap,they undergo destructive interference.
At the instant of complete overlap,the net displacement of every point on the string becomes zero.
Since the potential energy of a string is proportional to the square of the displacement (or the square of the slope),the potential energy at this instant is zero.
However,the particles of the string are still moving with maximum transverse velocity at this instant due to the superposition of the velocities of the two pulses.
Therefore,the total energy of the string at this instant is entirely in the form of kinetic energy.

(C) The resultant amplitude $R$ of two waves with amplitudes $A_1$ and $A_2$ and a phase difference $\phi$ is given by the formula: $R = \sqrt{A_1^2 + A_2^2 + 2A_1A_2 \cos \phi}$.
Given $A_1 = 10$,$A_2 = 20$,and the phase difference $\phi = \pi/2$.
Substituting these values into the formula:
$R = \sqrt{10^2 + 20^2 + 2(10)(20) \cos(\pi/2)}$.
Since $\cos(\pi/2) = 0$,the expression simplifies to:
$R = \sqrt{100 + 400 + 0} = \sqrt{500}$.
$R = \sqrt{100 \times 5} = 10\sqrt{5}$.

(C) The given equation is $y = 3\, \sin\, \omega t + 4\, \cos\, \omega t$.
This is of the form $y = A_1\, \sin\, \omega t + A_2\, \cos\, \omega t$,where $A_1 = 3\, cm$ and $A_2 = 4\, cm$.
The resultant amplitude $A$ is given by the formula $A = \sqrt{A_1^2 + A_2^2}$.
Substituting the values,we get $A = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5\, cm$.

(C) The resultant amplitude $A$ of two waves with amplitudes $a_1$ and $a_2$ and a phase difference $\phi$ is given by the formula:
$A = \sqrt{a_1^2 + a_2^2 + 2 a_1 a_2 \cos \phi}$
Given the equations of the waves:
$y_1 = 10 \sin(2000 \pi t + 2x)$
$y_2 = 10 \sin(2000 \pi t + 2x + \frac{\pi}{2})$
Here,$a_1 = 10$,$a_2 = 10$,and the phase difference $\phi = \frac{\pi}{2} = 90^\circ$.
Substituting these values into the formula:
$A = \sqrt{10^2 + 10^2 + 2(10)(10) \cos(90^\circ)}$
Since $\cos(90^\circ) = 0$:
$A = \sqrt{100 + 100 + 0} = \sqrt{200} = 10\sqrt{2} \approx 14.14 \ unit$.
Thus,the resultant amplitude is approximately $14.1 \ unit$.

(D) The phase difference $\Delta \phi$ is related to the path difference $\Delta x$ by the formula:
$\Delta \phi = \frac{2 \pi}{\lambda} \Delta x$
From the graph,the path difference between the two waves is $\Delta x = \frac{\lambda}{4}$.
Substituting this value:
$\Delta \phi = \frac{2 \pi}{\lambda} \times \frac{\lambda}{4} = \frac{\pi}{2}$
The resultant intensity $I$ of two waves with equal intensity $I_0$ and phase difference $\phi$ is given by:
$I = I_0 + I_0 + 2\sqrt{I_0 I_0} \cos \phi = 2I_0(1 + \cos \phi)$
Substituting $\phi = \frac{\pi}{2}$:
$I = 2I_0(1 + \cos \frac{\pi}{2})$
Since $\cos \frac{\pi}{2} = 0$,we get:
$I = 2I_0(1 + 0) = 2I_0$

(A) Let the amplitude of each wave be $A$. Since intensity $I \propto A^2$,we have $I_0 = kA^2$,where $k$ is a constant.
The resultant wave equation is given by the sum of the three waves:
$y = A \sin(\omega t) + A \sin(\omega t - \frac{\pi}{4}) + A \sin(\omega t + \frac{\pi}{4})$
Using the trigonometric identity $\sin(x - y) + \sin(x + y) = 2 \sin x \cos y$:
$y = A \sin(\omega t) + A [2 \sin(\omega t) \cos(\frac{\pi}{4})]$
Since $\cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$:
$y = A \sin(\omega t) + A [2 \sin(\omega t) \cdot \frac{1}{\sqrt{2}}]$
$y = A \sin(\omega t) + \sqrt{2} A \sin(\omega t)$
$y = (1 + \sqrt{2}) A \sin(\omega t)$
The resultant amplitude $A_R = (1 + \sqrt{2}) A$.
The resultant intensity $I = k A_R^2 = k [(1 + \sqrt{2}) A]^2 = k A^2 (1 + \sqrt{2})^2$.
Since $I_0 = k A^2$:
$I = I_0 (1 + 2 + 2\sqrt{2}) = I_0 (3 + 2 \times 1.414) = I_0 (3 + 2.828) = 5.828 I_0$.
Thus,the resultant intensity is approximately $5.8 I_0$.

(N/A) Principle of Superposition: When a particle of a medium is simultaneously influenced by two or more waves,its net displacement is the vector sum of the displacements that would occur under the influence of each individual wave.
Explanation:
$1$. When two waves travel in opposite directions,they maintain their individual characteristics even after they pass through each other.
$2$. During the superposition,the medium particles exhibit a displacement pattern that is the algebraic sum of the individual wave displacements.
$3$. The provided figure illustrates the superposition of two equal and opposite wave pulses. At $t = 2 \ s$ (part $c$),the two pulses overlap such that their displacements cancel each other out,resulting in a net displacement of zero at that instant. After passing through each other,the pulses continue to move in their original directions,unchanged in shape or velocity.

(N/A) Let $y_{1}(x, t)$ and $y_{2}(x, t)$ be the displacements that any element of the medium would experience if each wave travelled alone.
The displacement $y(x, t)$ of an element of the medium when the waves overlap is given by the superposition principle as:
$y(x, t) = y_{1}(x, t) + y_{2}(x, t) \quad \dots (1)$
If we have $n$ waves moving in the medium,the resultant waveform is the algebraic sum of the wave functions of the individual waves.
Let the individual wave functions be:
$y_{1} = f_{1}(x - vt)$
$y_{2} = f_{2}(x - vt)$
$y_{n} = f_{n}(x - vt)$
Then the resultant wave function $y$ is the sum of these individual functions:
$y = f_{1}(x - vt) + f_{2}(x - vt) + \dots + f_{n}(x - vt)$
Therefore,the resultant wave function can be expressed as:
$y = \sum_{i=1}^{n} f_{i}(x - vt)$
where $i = 1, 2, 3, \dots, n$.

(N/A) Consider two progressive harmonic waves traveling along a stretched string with the same angular frequency $(\omega)$,angular wave number $(k)$,and amplitude $(a)$.
Let the two waves be represented as:
$y_{1}(x, t) = a \sin(kx - \omega t)$
$y_{2}(x, t) = a \sin(kx - \omega t + \phi)$
where $\phi$ is the constant phase difference between the two waves.
According to the principle of superposition,the resultant displacement $y(x, t)$ is the algebraic sum of the individual displacements:
$y(x, t) = y_{1}(x, t) + y_{2}(x, t)$
$y(x, t) = a \sin(kx - \omega t) + a \sin(kx - \omega t + \phi)$
Using the trigonometric identity $\sin C + \sin D = 2 \sin \left( \frac{C+D}{2} \right) \cos \left( \frac{C-D}{2} \right)$,where $C = kx - \omega t$ and $D = kx - \omega t + \phi$:
$y(x, t) = 2a \sin \left( \frac{kx - \omega t + kx - \omega t + \phi}{2} \right) \cos \left( \frac{kx - \omega t - (kx - \omega t + \phi)}{2} \right)$
$y(x, t) = 2a \sin \left( kx - \omega t + \frac{\phi}{2} \right) \cos \left( -\frac{\phi}{2} \right)$
Since $\cos(-\theta) = \cos(\theta)$:
$y(x, t) = [2a \cos(\frac{\phi}{2})] \sin(kx - \omega t + \frac{\phi}{2})$
This is the equation of the resultant progressive wave with an amplitude of $2a \cos(\frac{\phi}{2})$ and an initial phase of $\frac{\phi}{2}$.

(N/A) The amplitude of the resultant wave is a function of the phase difference $\phi$ between the two component waves.
The resultant amplitude $A$ is given by the formula: $A(\phi) = 2a \cos(\frac{\phi}{2})$,where $a$ is the amplitude of each individual wave.
$1$. When the two waves are in phase,the phase difference is $\phi = 0$. Substituting this into the formula,we get $A = 2a \cos(0) = 2a$. This is the maximum possible amplitude,known as constructive interference.
$2$. When the two waves are completely out of phase,the phase difference is $\phi = \pi$. Substituting this into the formula,we get $A = 2a \cos(\frac{\pi}{2}) = 0$. This results in zero amplitude,known as destructive interference.
$3$. For any other phase difference,the resultant amplitude $A$ will lie in the range $0 \leq A \leq 2a$.

(N/A) The phenomenon of the superposition of two waves is called interference.
It has two types:
$(1)$ Constructive interference
$(2)$ Destructive interference
$(1)$ Constructive interference: When the crest of one wave is superposed on the crest of another wave,or the trough of one wave is superposed on the trough of another wave,the amplitudes add up for the resultant wave. This is called constructive interference.
$(2)$ Destructive interference: When the crest of one wave is superposed on the trough of another wave,or the trough of one wave is superposed on the crest of another wave,the amplitudes subtract for the resultant wave. This is called destructive interference.

(N/A) The principle of superposition states that when two or more waves travel through a medium simultaneously,each wave acts independently of the others. The resultant displacement $y$ at any point and at any instant is the algebraic sum of the individual displacements $y_1, y_2, y_3, \dots$ produced by each wave at that point.
Mathematically,it is expressed as: $y = y_1 + y_2 + y_3 + \dots = \sum_{i=1}^{n} y_i$.

Let the two waves be represented by the equations:
$y_1 = A_1 \sin(\omega t)$
$y_2 = A_2 \sin(\omega t + \phi)$
where $A_1$ and $A_2$ are the amplitudes,$\omega$ is the angular frequency,$t$ is time,and $\phi$ is the initial phase difference.
According to the principle of superposition,the resultant displacement $y$ is the vector sum of the individual displacements:
$y = y_1 + y_2 = A_1 \sin(\omega t) + A_2 \sin(\omega t + \phi)$
Using the trigonometric identity $\sin(a + b) = \sin a \cos b + \cos a \sin b$:
$y = A_1 \sin(\omega t) + A_2 (\sin(\omega t) \cos \phi + \cos(\omega t) \sin \phi)$
$y = (A_1 + A_2 \cos \phi) \sin(\omega t) + (A_2 \sin \phi) \cos(\omega t)$
Let $A_1 + A_2 \cos \phi = R \cos \theta$ and $A_2 \sin \phi = R \sin \theta$,where $R$ is the resultant amplitude and $\theta$ is the phase constant.
Then,$y = R \cos \theta \sin(\omega t) + R \sin \theta \cos(\omega t) = R \sin(\omega t + \theta)$
where $R = \sqrt{A_1^2 + A_2^2 + 2A_1 A_2 \cos \phi}$ and $\tan \theta = \frac{A_2 \sin \phi}{A_1 + A_2 \cos \phi}$.

(N/A) The figure $(a)$ shows two needles oscillating with equal phase,representing two coherent sources. Consider two needles moving periodically up and down in an identical fashion in a trough of water as shown in the figure; they produce two water waves.
At any particular point,the phase difference between the displacements produced by each of the waves does not change with time,so both sources are called coherent sources.
Figure $(b)$ shows the position of crests (solid circles) and troughs (dashed circles) at a given instant of time.
Now consider a point $P$ as shown in figure $(a)$ for which $S_{1}P = S_{2}P$.
Since waves from $S_{1}$ and $S_{2}$ will take the same time to travel to point $P$,they arrive with the same phase.
The displacement produced by the source $S_{1}$ at point $P$ is given by $y_{1} = a \cos \omega t$,and the displacement produced by the source $S_{2}$ at point $P$ is given by $y_{2} = a \cos \omega t$,where $a$ is the amplitude.
According to the superposition principle,the resultant displacement at $P$ is:
$y = y_{1} + y_{2} = a \cos \omega t + a \cos \omega t$
$\therefore y = 2a \cos \omega t$

(B) Let the position of the listener be $L$. Initially,the listener is at a distance $x = 2\, m$ from $S_{2}$.
The distance from $S_{1}$ is $S_{1}L = \sqrt{x^2 + (1.5)^2} = \sqrt{2^2 + 1.5^2} = \sqrt{4 + 2.25} = \sqrt{6.25} = 2.5\, m$.
The path difference is $\Delta x = S_{1}L - S_{2}L = 2.5 - 2 = 0.5\, m$.
Since $\lambda = 1\, m$,we have $\Delta x = \frac{\lambda}{2}$,which corresponds to a minimum (destructive interference).
When the listener moves away from $S_{1}$ while keeping the distance from $S_{2}$ fixed at $2\, m$,the path difference $\Delta x = S_{1}L - S_{2}L$ increases.
The next intensity maximum occurs when the path difference $\Delta x = n\lambda$. For the adjacent maximum,$n = 1$.
Thus,$\Delta x = 1\, m$.
Let $d$ be the distance from $S_{1}$ to the listener at this new position.
Then,$d - 2 = 1$,which gives $d = 3\, m$.

(A) The given wave equations are ${y}_{1} = {A}_{1} \sin {k}({x} - {vt})$ and ${y}_{2} = {A}_{2} \sin {k}({x} - {vt} + {x}_{0})$.
The phase difference $\Delta \phi$ between the two waves is given by $\Delta \phi = {k} \cdot {x}_{0}$.
Given ${k} = 6.28 \, {cm}^{-1}$ and ${x}_{0} = 3.5 \, {cm}$.
Since $6.28 \approx 2\pi$,we have $\Delta \phi = 6.28 \times 3.5 = 2\pi \times 3.5 = 7\pi$.
The resultant amplitude ${A}_{R}$ is given by ${A}_{R} = \sqrt{{A}_{1}^{2} + {A}_{2}^{2} + 2{A}_{1}{A}_{2} \cos(\Delta \phi)}$.
Substituting the values: ${A}_{R} = \sqrt{12^{2} + 5^{2} + 2(12)(5) \cos(7\pi)}$.
Since $\cos(7\pi) = -1$,we get ${A}_{R} = \sqrt{144 + 25 - 120} = \sqrt{49} = 7 \, {mm}$.

(A) Given amplitudes are $A_{1} = 5 \, cm$ and $A_{2} = 3 \, cm$.
The phase difference $\Delta \phi$ between the two waves is determined by the difference in their arguments:
$\Delta \phi = 2 \pi (x - vt + 1.5) - 2 \pi (x - vt) = 2 \pi (1.5) = 3 \pi$.
The resultant amplitude $A_{net}$ is given by the formula:
$A_{net} = \sqrt{A_{1}^{2} + A_{2}^{2} + 2 A_{1} A_{2} \cos(\Delta \phi)}$.
Substituting the values:
$A_{net} = \sqrt{5^{2} + 3^{2} + 2(5)(3) \cos(3 \pi)}$.
Since $\cos(3 \pi) = -1$:
$A_{net} = \sqrt{25 + 9 + 30(-1)} = \sqrt{34 - 30} = \sqrt{4} = 2 \, cm$.

(C) The resultant amplitude $A_R$ of two superimposing waves with individual amplitudes $A_1 = A_2 = A$ and phase difference $\phi$ is given by the formula:
$A_R = \sqrt{A_1^2 + A_2^2 + 2A_1A_2 \cos \phi}$
Given that $A_R = \sqrt{3}A$,we substitute the values:
$\sqrt{3}A = \sqrt{A^2 + A^2 + 2A^2 \cos \phi}$
Squaring both sides:
$3A^2 = 2A^2 + 2A^2 \cos \phi$
$3A^2 - 2A^2 = 2A^2 \cos \phi$
$A^2 = 2A^2 \cos \phi$
$\cos \phi = \frac{1}{2}$
Therefore,$\phi = 60^{\circ}$.

(A) For a minimum intensity at point $P$ (directly in front of source $R$),the path difference between the sound waves reaching $P$ from $S$ and $R$ must be an odd multiple of half the wavelength.
The path difference is given by $\Delta x = SP - RP$.
Given the geometry,$RP = 12 \,m$ and $RS = 5 \,m$. Thus,$SP = \sqrt{RP^2 + RS^2} = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13 \,m$.
The path difference is $\Delta x = 13 \,m - 12 \,m = 1 \,m$.
For destructive interference (minima),$\Delta x = (2n + 1) \frac{\lambda}{2}$,where $n = 0, 1, 2, \dots$.
Substituting $\lambda = \frac{v}{f}$,we get $1 = (2n + 1) \frac{v}{2f}$,which implies $f = \frac{(2n + 1)v}{2}$.
Given $v = 330 \,m/s$,we have $f = \frac{(2n + 1) \times 330}{2} = (2n + 1) \times 165$.
For $n = 0$,$f = 165 \,Hz$.
For $n = 1$,$f = 3 \times 165 = 495 \,Hz$.
For $n = 2$,$f = 5 \times 165 = 825 \,Hz$.
Comparing with the given options,$495 \,Hz$ is a possible value.

(C) The distance from the first speaker to point $P$ is $d_1 = 4 \,m$. The distance from the second speaker to point $P$ is $d_2 = \sqrt{3^2 + 4^2} = 5 \,m$.
The path difference between the two sound waves at point $P$ is $\Delta L = d_2 - d_1 = 5 \,m - 4 \,m = 1 \,m$.
For the sound to be least intense (destructive interference),the path difference must be an odd multiple of half the wavelength:
$\Delta L = (2n + 1) \frac{\lambda_1}{2}$,where $n = 0, 1, 2, \dots$
For $n = 0$,$1 \,m = \frac{\lambda_1}{2} \Rightarrow \lambda_1 = 2 \,m$.
For the sound to be most intense (constructive interference),the path difference must be an integer multiple of the wavelength:
$\Delta L = n \lambda_2$,where $n = 1, 2, 3, \dots$
For $n = 1$,$1 \,m = \lambda_2 \Rightarrow \lambda_2 = 1 \,m$.
Thus,the possible values are $\lambda_1 = 2 \,m$ and $\lambda_2 = 1 \,m$. The correct option is $(c)$.

(D) The amplitudes of the two waves are $A_1 = A_0$ and $A_2 = x A_0$.
The maximum resultant amplitude $(A_{max})$ occurs when the waves are in phase:
$A_{max} = A_1 + A_2 = A_0 + x A_0 = (x+1) A_0$.
The minimum resultant amplitude $(A_{min})$ occurs when the waves are out of phase. Given $x > 1$,$A_2 > A_1$,so:
$A_{min} = A_2 - A_1 = x A_0 - A_0 = (x-1) A_0$.
The difference between the maximum and minimum resultant amplitude is:
Difference $= A_{max} - A_{min} = (x+1) A_0 - (x-1) A_0$.
Difference $= (x + 1 - x + 1) A_0 = 2 A_0$.

(B) The resultant wave is $x' = x_1 + x_2 = a[\sin(\omega t + \phi_1) + \sin(\omega t + \phi_2)]$.
Using the trigonometric identity $\sin A + \sin B = 2 \sin(\frac{A+B}{2}) \cos(\frac{A-B}{2})$,we get:
$x' = 2a \sin(\omega t + \frac{\phi_1 + \phi_2}{2}) \cos(\frac{\phi_1 - \phi_2}{2})$.
The amplitude of the resultant wave is $A_{res} = |2a \cos(\frac{\phi_1 - \phi_2}{2})|$.
Given that the resultant amplitude is equal to the amplitude of the individual waves $(a)$:
$|2a \cos(\frac{\phi_1 - \phi_2}{2})| = a$.
$\cos(\frac{\phi_1 - \phi_2}{2}) = \frac{1}{2}$.
Let $\Delta\phi = \phi_1 - \phi_2$. Then $\cos(\frac{\Delta\phi}{2}) = \frac{1}{2}$.
$\frac{\Delta\phi}{2} = \frac{\pi}{3}$.
$\Delta\phi = \frac{2\pi}{3}$.

(B) The resultant wave is given by the principle of superposition: $y = y_1 + y_2$.
Substituting the given equations:
$y = A \sin \left(kx - \omega t + \frac{\pi}{6}\right) + A \sin \left(kx - \omega t - \frac{\pi}{6}\right)$.
Using the trigonometric identity $\sin C + \sin D = 2 \sin \left(\frac{C+D}{2}\right) \cos \left(\frac{C-D}{2}\right)$:
Let $C = kx - \omega t + \frac{\pi}{6}$ and $D = kx - \omega t - \frac{\pi}{6}$.
Then $\frac{C+D}{2} = kx - \omega t$ and $\frac{C-D}{2} = \frac{\pi}{6}$.
Therefore,$y = 2A \sin(kx - \omega t) \cos\left(\frac{\pi}{6}\right)$.
Since $\cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}$,we get:
$y = 2A \sin(kx - \omega t) \cdot \frac{\sqrt{3}}{2} = A \sqrt{3} \sin(kx - \omega t)$.

(C) The resultant intensity $I$ of two interfering waves is given by the formula:
$I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi$
Given $I_1 = 2 \, W/m^2$,$I_2 = 3 \, W/m^2$,and $I = 5 \, W/m^2$.
Substituting these values into the equation:
$5 = 2 + 3 + 2\sqrt{2 \times 3} \cos \phi$
$5 = 5 + 2\sqrt{6} \cos \phi$
$0 = 2\sqrt{6} \cos \phi$
Since $2\sqrt{6} \neq 0$,we must have $\cos \phi = 0$.
Therefore,the phase difference $\phi = \frac{\pi}{2}$.

(A) The resultant amplitude $R$ of two waves with equal amplitude $A$ and phase difference $\Delta \phi$ is given by the formula: $R = 2A \cos \left(\frac{\Delta \phi}{2}\right)$.
Given $A = 8\,cm$ and $R = 8\,cm$.
Substituting the values: $8 = 2(8) \cos \left(\frac{\Delta \phi}{2}\right)$.
This simplifies to: $1 = 2 \cos \left(\frac{\Delta \phi}{2}\right)$,which means $\cos \left(\frac{\Delta \phi}{2}\right) = \frac{1}{2}$.
Since $\cos(60^{\circ}) = \frac{1}{2}$,we have $\frac{\Delta \phi}{2} = 60^{\circ}$.
Therefore,the phase difference $\Delta \phi = 120^{\circ}$.

(C) Given the first wave equation: $y_1 = 10 \sin \left(\omega t + \frac{\pi}{3}\right)$.
For the second wave equation: $y_2 = 5[\sin (\omega t) + \sqrt{3} \cos \omega t]$.
Multiply and divide by $2$ inside the bracket: $y_2 = 5 \times 2 \left[\frac{1}{2} \sin (\omega t) + \frac{\sqrt{3}}{2} \cos (\omega t)\right]$.
Using the identity $\sin(A+B) = \sin A \cos B + \cos A \sin B$,we have $\sin(\omega t + \frac{\pi}{3}) = \sin(\omega t) \cos(\frac{\pi}{3}) + \cos(\omega t) \sin(\frac{\pi}{3}) = \frac{1}{2} \sin(\omega t) + \frac{\sqrt{3}}{2} \cos(\omega t)$.
Thus,$y_2 = 10 \sin \left(\omega t + \frac{\pi}{3}\right)$.
Since both waves have the same amplitude $A_1 = A_2 = 10 \text{ cm}$ and the same phase $\phi_1 = \phi_2 = \frac{\pi}{3}$,they are in phase.
The resultant amplitude $A_R$ for two waves in phase is $A_R = A_1 + A_2$.
$A_R = 10 + 10 = 20 \text{ cm}$.

(A) The given equations of the two progressive waves are $y_1 = 4 \sin(2x - 6t)$ and $y_2 = 3 \sin(2x - 6t - \pi/2)$.
Comparing these with the standard form $y = A \sin(\omega t - kx + \phi)$,we get amplitudes $A_1 = 4$ and $A_2 = 3$.
The phase difference between the two waves is $\Delta\phi = \pi/2$.
The resultant amplitude $A$ of two superimposed waves is given by the formula $A = \sqrt{A_1^2 + A_2^2 + 2A_1A_2 \cos(\Delta\phi)}$.
Substituting the values,$A = \sqrt{4^2 + 3^2 + 2(4)(3) \cos(\pi/2)}$.
Since $\cos(\pi/2) = 0$,we have $A = \sqrt{16 + 9 + 0} = \sqrt{25} = 5$.

(C) The resultant amplitude $A_R$ of the superposition of waves with phase angles $\phi_1, \phi_2, \phi_3, \phi_4$ is given by the vector sum of individual amplitudes $A_0$ (where $I_0 = k A_0^2$).
Let the waves be represented as complex numbers: $z_1 = A_0 e^{i0} = A_0$,$z_2 = A_0 e^{i\pi/3}$,$z_3 = A_0 e^{i2\pi/3}$,and $z_4 = A_0 e^{i\pi} = -A_0$.
The sum is $S = A_0(1 + e^{i\pi/3} + e^{i2\pi/3} - 1) = A_0(e^{i\pi/3} + e^{i2\pi/3})$.
Using the identity $e^{i\theta} = \cos \theta + i \sin \theta$,we have $S = A_0 [(\cos \pi/3 + i \sin \pi/3) + (\cos 2\pi/3 + i \sin 2\pi/3)]$.
$S = A_0 [(1/2 + i\sqrt{3}/2) + (-1/2 + i\sqrt{3}/2)] = A_0 (i\sqrt{3})$.
The resultant intensity $I = |S|^2 = A_0^2 |i\sqrt{3}|^2 = I_0 (3) = 3I_0$.
Thus,$n = 3$.

(D) The resultant wave is given by $y = y_1 + y_2 = A \sin(kx - \omega t + \phi)$.
Using the phasor addition method for two waves with amplitudes $A_1 = 4$ and $A_2 = 2$ and a phase difference $\Delta\phi = \frac{2\pi}{3} = 120^{\circ}$,the resultant amplitude $A$ is:
$A = \sqrt{A_1^2 + A_2^2 + 2A_1A_2 \cos(\Delta\phi)}$
$A = \sqrt{4^2 + 2^2 + 2(4)(2) \cos(120^{\circ})}$
$A = \sqrt{16 + 4 + 16(-0.5)} = \sqrt{20 - 8} = \sqrt{12} = 2\sqrt{3}$.
The phase angle $\phi$ of the resultant wave is given by:
$\tan \phi = \frac{A_2 \sin(\Delta\phi)}{A_1 + A_2 \cos(\Delta\phi)}$
$\tan \phi = \frac{2 \sin(120^{\circ})}{4 + 2 \cos(120^{\circ})} = \frac{2(\frac{\sqrt{3}}{2})}{4 + 2(-0.5)} = \frac{\sqrt{3}}{4 - 1} = \frac{\sqrt{3}}{3} = \frac{1}{\sqrt{3}}$.
Thus,$\phi = \frac{\pi}{6}$.

(D) For completely destructive interference (minima),the path difference $\Delta x$ between the two waves must be an odd multiple of half the wavelength.
Mathematically,$\Delta x = (2n - 1) \frac{\lambda}{2}$,where $n = 1, 2, 3, \dots$
This can also be written as $\Delta x = 0.5 \lambda, 1.5 \lambda, 2.5 \lambda, 3.5 \lambda, 4.5 \lambda, 5.5 \lambda, \dots$
Comparing this with the given options,$5.5 \lambda$ corresponds to the case where $n = 6$ (i.e.,$\Delta x = (2 \times 6 - 1) \frac{\lambda}{2} = 11 \frac{\lambda}{2} = 5.5 \lambda$).
Therefore,the correct option is $5.5 \lambda$.

(D) The given equation is $y = \frac{1}{\sqrt{a}} \sin \omega t \pm \frac{1}{\sqrt{b}} \cos \omega t$.
We know that $\cos \omega t = \sin(\omega t + \frac{\pi}{2})$.
Substituting this,we get $y = \frac{1}{\sqrt{a}} \sin \omega t \pm \frac{1}{\sqrt{b}} \sin(\omega t + \frac{\pi}{2})$.
This represents the superposition of two simple harmonic motions with amplitudes $A_1 = \frac{1}{\sqrt{a}}$ and $A_2 = \frac{1}{\sqrt{b}}$,and a phase difference of $\phi = \frac{\pi}{2}$.
The resultant amplitude $A$ is given by $A = \sqrt{A_1^2 + A_2^2 + 2A_1 A_2 \cos \phi}$.
Since $\cos \frac{\pi}{2} = 0$,the formula simplifies to $A = \sqrt{A_1^2 + A_2^2}$.
Substituting the values,$A = \sqrt{(\frac{1}{\sqrt{a}})^2 + (\frac{1}{\sqrt{b}})^2} = \sqrt{\frac{1}{a} + \frac{1}{b}} = \sqrt{\frac{a+b}{ab}}$.

(C) The formula for the resultant amplitude $R$ of two superimposed waves is given by:
$R = \sqrt{A_1^2 + A_2^2 + 2 A_1 A_2 \cos \phi}$
Given that the amplitude of each wave is $A_1 = A_2 = A$ and the phase difference is $\phi = \frac{\pi}{2}$ (which is $90^{\circ}$).
Substituting these values into the formula:
$R = \sqrt{A^2 + A^2 + 2 A^2 \cos 90^{\circ}}$
Since $\cos 90^{\circ} = 0$,the expression simplifies to:
$R = \sqrt{A^2 + A^2 + 0} = \sqrt{2 A^2} = \sqrt{2} A$

(D) The intensity $I$ of a sound wave is proportional to the square of its amplitude $A$,i.e.,$I \propto A^2$.
Given amplitudes are $A_1 = 3$ and $A_2 = 5$.
The maximum amplitude is $A_{max} = A_1 + A_2 = 3 + 5 = 8$.
The minimum amplitude is $A_{min} = |A_1 - A_2| = |3 - 5| = 2$.
The ratio of maximum intensity to minimum intensity is given by:
$\frac{I_{max}}{I_{min}} = \left( \frac{A_{max}}{A_{min}} \right)^2$
$\frac{I_{max}}{I_{min}} = \left( \frac{8}{2} \right)^2 = (4)^2 = 16$.
Thus,the ratio is $16:1$.

(C) The resultant amplitude '$R$' of two waves with amplitudes '$A_1$' and '$A_2$' and a phase difference '$\phi$' is given by the formula: $R = \sqrt{A_1^2 + A_2^2 + 2A_1 A_2 \cos(\phi)}$.
Given: $A_1 = A$,$A_2 = A$,and $\phi = \frac{\pi}{2}$.
Substituting these values into the formula:
$R = \sqrt{A^2 + A^2 + 2(A)(A) \cos(\frac{\pi}{2})}$
Since $\cos(\frac{\pi}{2}) = 0$,the expression simplifies to:
$R = \sqrt{A^2 + A^2 + 0} = \sqrt{2A^2} = \sqrt{2} A$.
Therefore,the maximum amplitude of the resultant wave is $\sqrt{2} A$.

(A) The resultant amplitude $A$ of two waves with amplitudes $a_1$ and $a_2$ and phase difference $\phi = \phi_1 - \phi_2$ is given by the formula:
$A^2 = a_1^2 + a_2^2 + 2a_1 a_2 \cos \phi$
Given that $a_1 = a_2 = a$ and the resultant amplitude $A = a$,we substitute these into the equation:
$a^2 = a^2 + a^2 + 2(a)(a) \cos \phi$
$a^2 = 2a^2 + 2a^2 \cos \phi$
Subtracting $2a^2$ from both sides:
$-a^2 = 2a^2 \cos \phi$
Dividing by $2a^2$:
$\cos \phi = -\frac{1}{2}$
Therefore,the phase difference is:
$\phi = \phi_1 - \phi_2 = \cos ^{-1}\left(-\frac{1}{2}\right)$

(A) The path difference is given as $\Delta x = \lambda/3$.
The phase difference $\phi$ is related to path difference by the formula $\phi = \frac{2\pi}{\lambda} \times \Delta x$.
Substituting the value,$\phi = \frac{2\pi}{\lambda} \times \frac{\lambda}{3} = \frac{2\pi}{3} = 120^{\circ}$.
The resultant amplitude $R$ of two waves with equal amplitude $A$ is given by $R = \sqrt{A^2 + A^2 + 2A^2 \cos \phi}$.
Substituting $\phi = 120^{\circ}$ and $\cos(120^{\circ}) = -0.5$:
$R = \sqrt{A^2 + A^2 + 2A^2(-0.5)}$
$R = \sqrt{2A^2 - A^2}$
$R = \sqrt{A^2} = A$.

(B) Given,the ratio of intensities of two waves is $\frac{I_1}{I_2} = \frac{9}{1}$.
Since intensity $I \propto a^2$,where $a$ is the amplitude,we have:
$\frac{I_1}{I_2} = \left(\frac{a_1}{a_2}\right)^2 = \frac{9}{1}$
Taking the square root on both sides:
$\frac{a_1}{a_2} = \frac{3}{1} \Rightarrow a_1 = 3a_2$.
The ratio of maximum to minimum intensity is given by:
$\frac{I_{\max}}{I_{\min}} = \frac{(a_1 + a_2)^2}{(a_1 - a_2)^2}$
Substituting $a_1 = 3a_2$:
$\frac{I_{\max}}{I_{\min}} = \frac{(3a_2 + a_2)^2}{(3a_2 - a_2)^2} = \frac{(4a_2)^2}{(2a_2)^2} = \frac{16a_2^2}{4a_2^2} = \frac{4}{1}$.

(B) Let the amplitude of each wave be $A$. The intensity $I$ is proportional to the square of the amplitude,so $I \propto A^2$. Let $I_0 = kA^2$ be the intensity of each wave.
Case $1$: When the waves arrive in phase,the phase difference $\phi = 0$. The resultant amplitude is $A_R = A + A = 2A$. The resultant intensity is $I_1 = k(2A)^2 = 4kA^2 = 4I_0$.
Case $2$: When the waves arrive $90^{\circ}$ ($\pi/2$ radians) out of phase,the resultant amplitude is $A_R = \sqrt{A^2 + A^2 + 2AA \cos(90^{\circ})} = \sqrt{2A^2} = A\sqrt{2}$. The resultant intensity is $I_2 = k(A\sqrt{2})^2 = 2kA^2 = 2I_0$.
The ratio of the resultant intensities is $\frac{I_1}{I_2} = \frac{4I_0}{2I_0} = \frac{2}{1}$.

(D) Let the equations of the two waves be $y_1 = a \sin(\omega t - kx)$ and $y_2 = a \sin(\omega t - kx + \phi)$.
When they superpose,the resultant wave is $y = y_1 + y_2$.
Using the trigonometric identity $\sin A + \sin B = 2 \sin(\frac{A+B}{2}) \cos(\frac{A-B}{2})$,we get:
$y = 2a \cos(\frac{\phi}{2}) \sin(\omega t - kx + \frac{\phi}{2})$.
The resultant amplitude is $A_R = |2a \cos(\frac{\phi}{2})|$.
Given that the resultant amplitude is equal to the individual amplitude $a$,we have:
$a = |2a \cos(\frac{\phi}{2})| \implies \cos(\frac{\phi}{2}) = \pm \frac{1}{2}$.
Considering the magnitude,$\cos(\frac{\phi}{2}) = \frac{1}{2}$.
Since $\cos(\frac{\pi}{3}) = \frac{1}{2}$,we have $\frac{\phi}{2} = \frac{\pi}{3}$.
Therefore,the phase difference is $\phi = \frac{2\pi}{3}$.

(D) The resultant amplitude of waves at the point $P$ is given by the formula: $A = \sqrt{A_1^2 + A_2^2 + 2 A_1 A_2 \cos \phi}$,where $\phi$ is the phase difference,and $A_1$ and $A_2$ are the amplitudes of the sound waves.
First,we calculate the wavelength $\lambda$ of the sound wave: $\lambda = \frac{v}{f} = \frac{350 m s^{-1}}{350 Hz} = 1 m = 100 cm$.
The path difference is given as $\Delta x = AP - BP = 25 cm$.
The phase difference $\phi$ is calculated as: $\phi = \frac{2 \pi}{\lambda} \Delta x = \frac{2 \pi}{100 cm} \times 25 cm = \frac{\pi}{2}$.
Now,substitute the values into the resultant amplitude formula:
$A = \sqrt{0.3^2 + 0.4^2 + 2 \times 0.3 \times 0.4 \cos(\frac{\pi}{2})}$
Since $\cos(\frac{\pi}{2}) = 0$,the expression simplifies to:
$A = \sqrt{0.3^2 + 0.4^2} = \sqrt{0.09 + 0.16} = \sqrt{0.25} = 0.5 mm$.

(B) Given that, the amplitudes of the two waves are $a_1 = a_2 = 10 \,mm$.
The phase difference between the waves is $\phi = 90^{\circ}$.
The resultant amplitude $A$ of two interfering waves is given by the formula:
$A = \sqrt{a_1^2 + a_2^2 + 2 a_1 a_2 \cos \phi}$
Substituting the given values into the formula:
$A = \sqrt{(10)^2 + (10)^2 + 2(10)(10) \cos 90^{\circ}}$
Since $\cos 90^{\circ} = 0$, the expression simplifies to:
$A = \sqrt{100 + 100 + 0}$
$A = \sqrt{200}$
$A = 10 \sqrt{2} \,mm$

(D) The given displacement equation is $y = \frac{1}{\sqrt{a}} \sin \omega t \pm \frac{1}{\sqrt{b}} \cos \omega t$.
We can rewrite $\cos \omega t$ as $\sin(\omega t + \frac{\pi}{2})$.
Thus,$y = \frac{1}{\sqrt{a}} \sin \omega t \pm \frac{1}{\sqrt{b}} \sin(\omega t + \frac{\pi}{2})$.
This is a superposition of two simple harmonic motions with amplitudes $A_1 = \frac{1}{\sqrt{a}}$ and $A_2 = \frac{1}{\sqrt{b}}$,and a phase difference of $\phi = \frac{\pi}{2}$.
The resultant amplitude $A$ is given by $A = \sqrt{A_1^2 + A_2^2 + 2A_1A_2 \cos \phi}$.
Since $\cos(\frac{\pi}{2}) = 0$,the formula simplifies to $A = \sqrt{A_1^2 + A_2^2}$.
Substituting the values,$A = \sqrt{(\frac{1}{\sqrt{a}})^2 + (\frac{1}{\sqrt{b}})^2} = \sqrt{\frac{1}{a} + \frac{1}{b}}$.
Taking the common denominator,$A = \sqrt{\frac{a+b}{ab}}$.

(D) When two waves of same frequency $f$ and same amplitude $a$ superimpose,the resultant amplitude $A$ depends on the phase difference. Assuming constructive interference (maximum intensity),the resultant amplitude is $A = a + a = 2 a$.
Since the intensity $I$ of a wave is directly proportional to the square of its amplitude,we have $I \propto A^2$.
Substituting the value of $A$,we get $I \propto (2 a)^2 = 4 a^2$.
Thus,the total intensity is directly proportional to $4 a^2$.

(B) The resultant displacement is given by $x = x_1 + x_2 + x_3$.
Substituting the given expressions:
$x = A \cos \omega t + 2 A \sin \omega t + \sqrt{2} A \cos (\omega t + \frac{\pi}{4})$.
Using the trigonometric identity $\cos (\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta$:
$x = A \cos \omega t + 2 A \sin \omega t + \sqrt{2} A (\cos \omega t \cos \frac{\pi}{4} - \sin \omega t \sin \frac{\pi}{4})$.
Since $\cos \frac{\pi}{4} = \sin \frac{\pi}{4} = \frac{1}{\sqrt{2}}$:
$x = A \cos \omega t + 2 A \sin \omega t + \sqrt{2} A (\frac{1}{\sqrt{2}} \cos \omega t - \frac{1}{\sqrt{2}} \sin \omega t)$.
$x = A \cos \omega t + 2 A \sin \omega t + A \cos \omega t - A \sin \omega t$.
$x = 2 A \cos \omega t + A \sin \omega t$.
The resultant amplitude $R$ for a displacement of the form $x = a \cos \omega t + b \sin \omega t$ is $R = \sqrt{a^2 + b^2}$.
Therefore,$R = \sqrt{(2 A)^2 + A^2} = \sqrt{4 A^2 + A^2} = \sqrt{5} A$.

(A) The resultant intensity $I_R$ of two waves with individual intensities $I_1$ and $I_2$ and a phase difference $\phi$ is given by the formula:
$I_R = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos(\phi)$
Given that $I_1 = I_2 = I$ and $\phi = \frac{\pi}{2}$,we substitute these values into the formula:
$I_R = I + I + 2\sqrt{I \cdot I} \cos(\frac{\pi}{2})$
Since $\cos(\frac{\pi}{2}) = 0$,the expression simplifies to:
$I_R = 2I + 2I(0) = 2I$
Therefore,the resultant intensity is $2I$.

(B) The intensity $I$ of a wave is proportional to the square of its amplitude $A$,i.e.,$I \propto A^2$.
For two waves with amplitudes $A_1$ and $A_2$,the maximum intensity $I_{max}$ occurs at constructive interference,where $I_{max} \propto (A_1 + A_2)^2$.
The minimum intensity $I_{min}$ occurs at destructive interference,where $I_{min} \propto (A_1 - A_2)^2$.
Given the ratio of intensities $\frac{I_{max}}{I_{min}} = \frac{9}{4}$,we have:
$\frac{(A_1 + A_2)^2}{(A_1 - A_2)^2} = \frac{9}{4}$
Taking the square root on both sides:
$\frac{A_1 + A_2}{A_1 - A_2} = \frac{3}{2}$
Cross-multiplying gives:
$2(A_1 + A_2) = 3(A_1 - A_2)$
$2A_1 + 2A_2 = 3A_1 - 3A_2$
$5A_2 = A_1$
Therefore,the ratio $\frac{A_2}{A_1} = \frac{1}{5} = 0.2$.

(B) Given that the amplitude of each sinusoidal wave is the same.
Let $a_1 = a_2 = a$.
The phase difference is $\phi = 120^{\circ}$.
The resultant amplitude is $A = 2 \,mm$.
The formula for the resultant amplitude of two interfering waves is given by:
$A = \sqrt{a_1^2 + a_2^2 + 2 a_1 a_2 \cos \phi}$
Substituting the given values:
$2 = \sqrt{a^2 + a^2 + 2 a^2 \cos 120^{\circ}}$
Since $\cos 120^{\circ} = -\frac{1}{2}$, we have:
$2 = \sqrt{a^2 + a^2 + 2 a^2 (-0.5)}$
$2 = \sqrt{2a^2 - a^2}$
$2 = \sqrt{a^2}$
$2 = a$
Therefore, the amplitude of each wave is $2 \,mm$.

(A) Given equations are:
$y_1 = a \sin (kx - \omega t)$
$y_2 = a \sin (-(kx - \omega t) + \phi)$
Using the identity $\sin(-\theta) = -\sin(\theta)$,we can rewrite $y_2$ as:
$y_2 = -a \sin (kx - \omega t - \phi)$
Using $\sin(\theta - \phi) = \sin \theta \cos \phi - \cos \theta \sin \phi$,the superposition $y = y_1 + y_2$ is:
$y = a \sin(kx - \omega t) + a \sin(kx - \omega t + \phi)$
Using the sum-to-product formula $\sin A + \sin B = 2 \sin(\frac{A+B}{2}) \cos(\frac{A-B}{2})$:
$y = 2a \sin(\frac{kx - \omega t + kx - \omega t + \phi}{2}) \cos(\frac{kx - \omega t - (kx - \omega t + \phi)}{2})$
$y = 2a \sin(kx - \omega t + \frac{\phi}{2}) \cos(-\frac{\phi}{2})$
Since $\cos(-\theta) = \cos(\theta)$:
$y = [2a \cos(\frac{\phi}{2})] \sin(kx - \omega t + \frac{\phi}{2})$
The amplitude of the resultant wave is the coefficient of the sine term,which is $2a \cos(\frac{\phi}{2})$.

(A) The path length of the straight route is $2r$ (the diameter of the semicircle).
The path length of the semicircular route is $\pi r$.
The path difference $\Delta x$ between the two routes is $\Delta x = \pi r - 2r = r(\pi - 2)$.
For maximum amplitude (constructive interference),the path difference must be an integral multiple of the wavelength $\lambda$,so $\Delta x = n\lambda$,where $n = 1, 2, 3, \dots$.
Substituting the path difference,we get $r(\pi - 2) = n\lambda$.
Since the velocity of sound is $v = f\lambda$,we have $\lambda = \frac{v}{f}$.
Substituting this into the equation: $r(\pi - 2) = n \frac{v}{f}$.
Rearranging for frequency $f$: $f = n \left[ \frac{v}{r(\pi - 2)} \right]$.
Thus,the frequencies of the resultant waves of maximum amplitude are integral multiples of $\frac{v}{r(\pi - 2)}$.

(B) Let the distance of point $A$ from the line joining the speakers be $D = 40 \ m$. The speakers $L_1$ and $L_2$ are separated by $10 \ m$,so their coordinates relative to the midpoint of the line joining them are $(0, 5)$ and $(0, -5)$. Point $A$ is at $(40, 0)$.
When the recorder moves to point $B$ at a distance of $25 \ m$ from $A$,its coordinates are $(40, 25)$.
The distances from the speakers to point $B$ are:
$L_1B = \sqrt{40^2 + (25-5)^2} = \sqrt{40^2 + 20^2} = \sqrt{1600 + 400} = \sqrt{2000} = 20\sqrt{5} \ m$.
Given $\sqrt{5} = 2.23$,$L_1B = 20 \times 2.23 = 44.6 \ m$.
$L_2B = \sqrt{40^2 + (25+5)^2} = \sqrt{40^2 + 30^2} = \sqrt{1600 + 900} = \sqrt{2500} = 50 \ m$.
The path difference at point $B$ is $\Delta x = L_2B - L_1B = 50 - 44.6 = 5.4 \ m$.
Since the recorder undergoes $10$ cycles of minima and maxima,point $B$ corresponds to the $10^{\text{th}}$ maxima,so $\Delta x = n\lambda$,where $n = 10$.
$5.4 = 10 \times \lambda \implies \lambda = 0.54 \ m$.
The frequency $f$ is given by $f = \frac{v}{\lambda} = \frac{324}{0.54} = 600 \ Hz$.