Principle of superposition of waves Questions in English

(C) The given wave equations are $y_1 = A_1 \sin(\omega t - kx)$ and $y_2 = A_2 \sin(\omega t - kx - \theta)$,where $A_1 = 2a$ and $A_2 = 2a$.
Using the principle of superposition,the resultant amplitude $A_R$ for two waves with amplitudes $A_1$ and $A_2$ and phase difference $\phi$ is given by $A_R = \sqrt{A_1^2 + A_2^2 + 2A_1 A_2 \cos \phi}$.
Here,$A_1 = A_2 = 2a$ and $\phi = \theta$.
$A_R = \sqrt{(2a)^2 + (2a)^2 + 2(2a)(2a) \cos \theta}$
$A_R = \sqrt{4a^2 + 4a^2 + 8a^2 \cos \theta} = \sqrt{8a^2(1 + \cos \theta)}$
Using the identity $1 + \cos \theta = 2 \cos^2(\theta / 2)$:
$A_R = \sqrt{8a^2 \cdot 2 \cos^2(\theta / 2)} = \sqrt{16a^2 \cos^2(\theta / 2)}$
$A_R = 4a \cos(\theta / 2)$.

(B) For destructive interference,the path difference between two waves is given by $\Delta x = (2n + 1) \frac{\lambda}{2}$,where $n = 0, 1, 2, ...$. For the first case of destructive interference $(n=0)$,the path difference is $\Delta x = \frac{\lambda}{2}$.
To achieve constructive interference,the path difference must be an integer multiple of the wavelength,i.e.,$\Delta x' = n\lambda$. The condition for constructive interference is $\Delta x' = m\lambda$ (where $m$ is an integer).
If we increase the path of one wave by $\frac{\lambda}{2}$,the new path difference becomes $\Delta x_{new} = \Delta x + \frac{\lambda}{2} = \frac{\lambda}{2} + \frac{\lambda}{2} = \lambda$. Since $\lambda$ is an integer multiple of the wavelength $(m=1)$,constructive interference occurs.

(D) The resultant amplitude $A_R$ of two waves with amplitudes $A_1$ and $A_2$ and phase difference $\phi$ is given by $A_R = \sqrt{A_1^2 + A_2^2 + 2 A_1 A_2 \cos \phi}$.
Given $A_1 = A_2 = A$ and $\phi = \pi / 2$.
Substituting these values: $A_R = \sqrt{A^2 + A^2 + 2 A^2 \cos(\pi / 2)}$.
Since $\cos(\pi / 2) = 0$,we get $A_R = \sqrt{2 A^2} = \sqrt{2} A$.
The frequency of the resultant wave remains the same as the frequency of the individual waves,which is $\omega$.

(A) The resultant amplitude $R$ of two waves with amplitudes $A_1$ and $A_2$ and phase difference $\phi$ is given by $R = \sqrt{A_1^2 + A_2^2 + 2A_1A_2 \cos(\phi)}$.
Given the phase difference $\phi = 2\pi$.
Since $\cos(2\pi) = 1$,the expression becomes $R = \sqrt{A_1^2 + A_2^2 + 2A_1A_2} = \sqrt{(A_1 + A_2)^2} = A_1 + A_2$.
This represents constructive interference,where the resultant amplitude is maximum.

(D) When two waves of the same frequency $f$ and amplitude $a$ superpose,the resultant amplitude $A$ is given by the formula:
$A = \sqrt{a^2 + a^2 + 2a^2 \cos \phi} = \sqrt{2a^2(1 + \cos \phi)} = \sqrt{4a^2 \cos^2(\phi/2)} = 2a \cos(\phi/2)$
Since the intensity $I$ of a wave is directly proportional to the square of its amplitude $(I \propto A^2)$,
$I \propto (2a \cos(\phi/2))^2 = 4a^2 \cos^2(\phi/2)$.
For constructive interference,the maximum intensity is proportional to $4a^2$.

(B) Let the amplitude of each wave be $a$. The resultant amplitude $A$ is given by the formula:
$A^2 = a_1^2 + a_2^2 + 2a_1a_2 \cos \phi$
Given $a_1 = a_2 = a$ and the resultant amplitude $A = a$,we substitute these values:
$a^2 = a^2 + a^2 + 2a^2 \cos \phi$
$a^2 = 2a^2 + 2a^2 \cos \phi$
$-a^2 = 2a^2 \cos \phi$
$\cos \phi = -1/2$
Since $\cos \phi = -1/2$,the phase difference $\phi$ is $2\pi / 3$ radians.

(D) The wavelength $\lambda$ is given by $\lambda = \frac{v}{f} = \frac{350}{350} = 1 m = 100 cm$.
The path difference $(\Delta x)$ between the waves at point $P$ is $\Delta x = AP - BP = 25 cm$.
The phase difference $(\Delta \phi)$ is calculated as $\Delta \phi = \frac{2\pi}{\lambda} \Delta x = \frac{2\pi}{100} \times 25 = \frac{\pi}{2} rad$.
Since the phase difference is $\frac{\pi}{2}$,the two waves are perpendicular to each other in terms of phase. The resultant amplitude $A$ is given by $A = \sqrt{a_1^2 + a_2^2 + 2a_1 a_2 \cos(\Delta \phi)}$.
Substituting the values: $A = \sqrt{(0.3)^2 + (0.4)^2 + 2(0.3)(0.4) \cos(\frac{\pi}{2})}$.
Since $\cos(\frac{\pi}{2}) = 0$,we get $A = \sqrt{0.09 + 0.16} = \sqrt{0.25} = 0.5 mm$.

(D) Given: Amplitude of each wave = $a$. Wavelength $\lambda = 1 \ m$. Path difference $\Delta x = AP - BP = 50 \ cm = 0.5 \ m$.
The phase difference due to path difference is $\phi_{path} = \frac{2\pi}{\lambda} \times \Delta x = \frac{2\pi}{1} \times 0.5 = \pi$.
Since source $A$ is ahead of $B$ by $\frac{\pi}{3}$,the total phase difference $\Delta \phi$ between the two waves at point $P$ is $\Delta \phi = \phi_{path} - \frac{\pi}{3} = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$.
The resultant amplitude $R$ is given by $R = \sqrt{a^2 + a^2 + 2a^2 \cos(\Delta \phi)}$.
Substituting $\Delta \phi = \frac{2\pi}{3}$:
$R = \sqrt{2a^2 + 2a^2 \cos(120^\circ)} = \sqrt{2a^2 + 2a^2(-0.5)} = \sqrt{2a^2 - a^2} = \sqrt{a^2} = a$.

(C) The resultant amplitude $A$ of two interfering waves with amplitudes $a_1$ and $a_2$ and phase difference $\phi$ is given by the formula:
$A = \sqrt{a_1^2 + a_2^2 + 2a_1a_2 \cos \phi}$
Given values are $a_1 = 0.3 \ cm$,$a_2 = 0.4 \ cm$,and $\phi = \pi/2$.
Substituting these values into the formula:
$A = \sqrt{(0.3)^2 + (0.4)^2 + 2(0.3)(0.4) \cos(\pi/2)}$
Since $\cos(\pi/2) = 0$,the expression simplifies to:
$A = \sqrt{0.09 + 0.16 + 0} = \sqrt{0.25} = 0.5 \ cm$.

(A) When two waves propagate in the same direction with the same frequency and velocity,the resultant amplitude $R$ is given by the formula $R = \sqrt{a_1^2 + a_2^2 + 2a_1a_2 \cos \phi}$.
Given that the waves are in the same phase,the phase difference $\phi = 0$.
Since $\cos(0) = 1$,the formula simplifies to $R = \sqrt{a_1^2 + a_2^2 + 2a_1a_2} = \sqrt{(a_1 + a_2)^2} = a_1 + a_2$.
Substituting the given amplitudes $a_1 = 2A$ and $a_2 = A$,we get $R = 2A + A = 3A$.

(C) For interference to occur,two waves must have a constant phase relationship.
Wave $(2)$ is $y = a \sin(\omega t - kx)$ and wave $(4)$ is $y = a \cos(\omega t - kx) = a \sin(\omega t - kx + \pi/2)$.
Since both waves $(2)$ and $(4)$ have the same frequency and travel in the same direction with a constant phase difference of $\pi/2$,they can produce a stable interference pattern.
Similarly,waves $(1)$ and $(3)$ have a constant phase relationship,but among the given options,only the pair $(2)$ and $(4)$ is listed correctly as a valid combination for interference.

(D) When two waves of the same frequency and intensity superimpose in opposite phases (phase difference of $\pi$ radians),destructive interference occurs.
In destructive interference,the resultant amplitude $A_{res} = A_1 - A_2$.
Since the intensities are equal,the amplitudes are also equal $(A_1 = A_2 = A)$.
Therefore,the resultant amplitude $A_{res} = A - A = 0$.
Since intensity $I \propto A^2$,the resultant intensity $I_{res} = 0$.
Thus,the intensity becomes zero,which is not mentioned in the options $A$,$B$,or $C$.

(A) The phase difference between the two waves is given by $\phi = (\omega t - \beta_2) - (\omega t - \beta_1) = \beta_1 - \beta_2$.
According to the principle of superposition,the resultant amplitude $A$ of two waves with amplitudes $A_1$ and $A_2$ and phase difference $\phi$ is given by the formula $A = \sqrt{A_1^2 + A_2^2 + 2A_1A_2 \cos \phi}$.
Substituting the value of $\phi$,we get $A = \sqrt{A_1^2 + A_2^2 + 2A_1A_2 \cos(\beta_1 - \beta_2)}$.

(A) The resultant wave is given by the superposition principle: $y = y_1 + y_2 = A\sin(\omega t - kx) + A\sin(\omega t - kx - \theta )$.
Using the trigonometric identity $\sin C + \sin D = 2\sin(\frac{C+D}{2})\cos(\frac{C-D}{2})$,we get:
$y = 2A \sin(\omega t - kx - \frac{\theta}{2}) \cos(\frac{\theta}{2})$.
The resultant amplitude $A_R$ is the coefficient of the sine term:
$A_R = 2A \cos(\frac{\theta}{2})$.
Alternatively,using the vector addition formula for amplitudes:
$A_R = \sqrt{A_1^2 + A_2^2 + 2A_1A_2\cos\theta} = \sqrt{A^2 + A^2 + 2A^2\cos\theta} = \sqrt{2A^2(1 + \cos\theta)}$.
Since $1 + \cos\theta = 2\cos^2(\frac{\theta}{2})$,we have $A_R = \sqrt{2A^2 \cdot 2\cos^2(\frac{\theta}{2})} = 2A\cos(\frac{\theta}{2})$.

(C) The first wave is $y_1 = a \sin(\omega t + \frac{\pi}{6})$.
The second wave is $y_2 = a \cos(\omega t) = a \sin(\omega t + \frac{\pi}{2})$.
The phase difference between the two waves is $\phi = (\omega t + \frac{\pi}{2}) - (\omega t + \frac{\pi}{6}) = \frac{\pi}{2} - \frac{\pi}{6} = \frac{\pi}{3}$.
The resultant amplitude $A$ of two waves with equal amplitude $a$ is given by $A = \sqrt{a_1^2 + a_2^2 + 2a_1 a_2 \cos \phi}$.
Substituting $a_1 = a$,$a_2 = a$,and $\phi = \frac{\pi}{3}$:
$A = \sqrt{a^2 + a^2 + 2(a)(a) \cos(\frac{\pi}{3})}$
$A = \sqrt{2a^2 + 2a^2(\frac{1}{2})}$
$A = \sqrt{2a^2 + a^2} = \sqrt{3a^2} = \sqrt{3} a$.

(B) The resultant amplitude $A$ of two superimposing waves with equal amplitude $a$ is given by the formula: $A^2 = a^2 + a^2 + 2a^2 \cos \phi$,where $\phi$ is the phase difference.
Given that the resultant amplitude $A$ is equal to the amplitude of the individual waves,$A = a$.
Substituting this into the equation: $a^2 = a^2 + a^2 + 2a^2 \cos \phi$.
$a^2 = 2a^2 + 2a^2 \cos \phi$.
$-a^2 = 2a^2 \cos \phi$.
$\cos \phi = -\frac{1}{2}$.
Since $\cos(120^\circ) = -\frac{1}{2}$,the phase difference $\phi = \frac{2\pi}{3}$ radians.

(B) The path difference between the waves reaching the detector $D$ from $L_1$ and $L_2$ is given by:
$\Delta x = L_2D - L_1D$
From the geometry,$L_1D = 40 \, m$ and the distance between $L_1$ and $L_2$ is $9 \, m$. Thus,$L_2D = \sqrt{40^2 + 9^2} = \sqrt{1600 + 81} = \sqrt{1681} = 41 \, m$.
$\Delta x = 41 \, m - 40 \, m = 1 \, m$.
For constructive interference (maxima),the condition is $\Delta x = n\lambda$,where $n = 1, 2, 3, ...$.
For the first maximum,we take $n = 1$:
$1 \, m = 1 \cdot \lambda \implies \lambda = 1 \, m$.
The frequency $f$ is given by $f = \frac{v}{\lambda}$,where $v = 330 \, m/s$.
$f = \frac{330 \, m/s}{1 \, m} = 330 \, Hz$.

(C) Let the amplitudes be $A_1 = 10 \,\mu m$,$A_2 = 4 \,\mu m$,and $A_3 = 7 \,\mu m$.
The waves arrive with successive phase differences of $\frac{\pi}{2}$.
Let the phase of the first wave be $0$,the second be $\frac{\pi}{2}$,and the third be $\pi$.
Using the phasor method,the resultant amplitude $A_R$ is given by the vector sum of the amplitudes:
$A_R = \sqrt{(\sum A_i \cos \phi_i)^2 + (\sum A_i \sin \phi_i)^2}$
$A_x = A_1 \cos(0) + A_2 \cos(\frac{\pi}{2}) + A_3 \cos(\pi) = 10(1) + 4(0) + 7(-1) = 10 - 7 = 3 \,\mu m$
$A_y = A_1 \sin(0) + A_2 \sin(\frac{\pi}{2}) + A_3 \sin(\pi) = 10(0) + 4(1) + 7(0) = 4 \,\mu m$
Resultant amplitude $A_R = \sqrt{A_x^2 + A_y^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \,\mu m$.

(B) The initial distance between the centres of the two pulses is $d = 8 \ cm$.
Each pulse moves towards the other with a speed of $v = 2 \ cm/s$.
The relative speed of the two pulses is $v_{rel} = v + v = 2 + 2 = 4 \ cm/s$.
The time taken for the pulses to meet is $t = d / v_{rel} = 8 \ cm / 4 \ cm/s = 2 \ s$.
After $2 \ s$,the two pulses overlap completely.
Since the pulses have opposite phases (one is a crest and the other is a trough of equal magnitude),their displacements cancel each other out at every point,making the string momentarily straight.
Because the string is straight,there is no deformation,and thus the potential energy is zero.
However,the particles of the string still possess velocity at this instant,so the total energy of the system is purely kinetic.

(C) The speed of each pulse is $v = 2.5 \ cm/s$.
The time elapsed is $t = 2 \ s$.
The distance traveled by each pulse is $d = v \times t = 2.5 \ cm/s \times 2 \ s = 5 \ cm$.
Initially,the pulses are $10 \ cm$ apart. Since they are moving towards each other,after $2 \ s$,each pulse covers $5 \ cm$,effectively closing the $10 \ cm$ gap.
When the two pulses meet,they are in mutually opposite phases (one is a crest and the other is a trough). According to the principle of superposition,the resultant displacement of the string will be the algebraic sum of the individual displacements.
Since the pulses have equal and opposite amplitudes,their sum will be zero at the point of intersection.
Therefore,the string will appear as a straight line.

(D) Both sound waves and light waves are wave phenomena that exhibit the property of interference. Interference occurs when two waves of the same frequency and constant phase difference superpose,resulting in a new wave pattern with a different amplitude. Since both sound and light satisfy the conditions for wave superposition,they both produce interference patterns. Other options are incorrect because sound waves are mechanical and longitudinal,while light waves are electromagnetic and transverse,and they travel at vastly different speeds.

(B) The given wave equations are $y_1 = 4\sin \omega t$ and $y_2 = 3\sin (\omega t + \pi/3)$.
Comparing these with the standard form $y = a\sin(\omega t + \phi)$,we get amplitudes $a_1 = 4$ and $a_2 = 3$,and the phase difference $\phi = \pi/3$.
The amplitude $A$ of the resulting wave is given by the formula $A = \sqrt{a_1^2 + a_2^2 + 2a_1a_2 \cos \phi}$.
Substituting the values: $A = \sqrt{4^2 + 3^2 + 2(4)(3) \cos(\pi/3)}$.
Since $\cos(\pi/3) = 0.5$,we have $A = \sqrt{16 + 9 + 24(0.5)} = \sqrt{25 + 12} = \sqrt{37}$.
Calculating the square root,$A \approx 6.08$,which is approximately $6$.

(D) The resultant amplitude $R$ of two waves with amplitudes $a_1$ and $a_2$ and phase difference $\delta$ is given by $R^2 = a_1^2 + a_2^2 + 2a_1 a_2 \cos \delta$.
Here,$a_1 = a$ and $a_2 = a$.
Substituting these values,we get $R^2 = a^2 + a^2 + 2(a)(a) \cos \delta$.
$R^2 = 2a^2 + 2a^2 \cos \delta = 2a^2(1 + \cos \delta)$.
Using the trigonometric identity $1 + \cos \delta = 2 \cos^2 \left( \frac{\delta}{2} \right)$,we get:
$R^2 = 2a^2 \times 2 \cos^2 \left( \frac{\delta}{2} \right) = 4a^2 \cos^2 \left( \frac{\delta}{2} \right)$.
Since intensity $I \propto R^2$,the resultant intensity is proportional to $4a^2 \cos^2 \left( \frac{\delta}{2} \right)$.

(D) Interference is a phenomenon that occurs when two or more coherent waves superpose.
Coherent waves are waves that have the same frequency and a constant phase difference over time.
In this problem,the waves are stated to be 'independent'.
Independent sources of light emit waves with random phase changes,meaning they are incoherent.
Therefore,interference cannot be observed between independent sources.
Thus,none of the given pairs represent the phenomenon of interference.

(D) The resultant amplitude $A$ of two waves with amplitudes $a_1$ and $a_2$ and phase difference $\phi$ is given by the formula: $A = \sqrt{a_1^2 + a_2^2 + 2 a_1 a_2 \cos \phi}$.
Here,$a_1 = a$,$a_2 = a$,and the phase difference $\phi = \frac{\pi}{3} - 0 = \frac{\pi}{3}$.
Substituting these values into the formula:
$A = \sqrt{a^2 + a^2 + 2(a)(a) \cos(\frac{\pi}{3})}$
Since $\cos(\frac{\pi}{3}) = \frac{1}{2}$,we get:
$A = \sqrt{a^2 + a^2 + 2a^2(\frac{1}{2})}$
$A = \sqrt{2a^2 + a^2} = \sqrt{3a^2} = \sqrt{3} a$.

(A) The resultant wave is $y = y_1 + y_2$.
$y = A_1 \sin(wt - \beta_1) + A_2 \sin(wt - \beta_2)$.
Using the identity $\sin(x - y) = \sin x \cos y - \cos x \sin y$:
$y = A_1(\sin wt \cos \beta_1 - \cos wt \sin \beta_1) + A_2(\sin wt \cos \beta_2 - \cos wt \sin \beta_2)$.
Grouping the terms:
$y = (A_1 \cos \beta_1 + A_2 \cos \beta_2) \sin wt - (A_1 \sin \beta_1 + A_2 \sin \beta_2) \cos wt$.
Let $X = A_1 \cos \beta_1 + A_2 \cos \beta_2$ and $Y = A_1 \sin \beta_1 + A_2 \sin \beta_2$.
The resultant amplitude $A$ is $\sqrt{X^2 + Y^2}$.
$A = \sqrt{(A_1 \cos \beta_1 + A_2 \cos \beta_2)^2 + (A_1 \sin \beta_1 + A_2 \sin \beta_2)^2}$.
Expanding the squares:
$A = \sqrt{A_1^2(\cos^2 \beta_1 + \sin^2 \beta_1) + A_2^2(\cos^2 \beta_2 + \sin^2 \beta_2) + 2A_1A_2(\cos \beta_1 \cos \beta_2 + \sin \beta_1 \sin \beta_2)}$.
Using trigonometric identities $\sin^2 \theta + \cos^2 \theta = 1$ and $\cos(A - B) = \cos A \cos B + \sin A \sin B$:
$A = \sqrt{A_1^2 + A_2^2 + 2A_1A_2 \cos(\beta_1 - \beta_2)}$.

(A) The resultant intensity $I_R$ of two interfering waves with intensities $I_1$ and $I_2$ and phase difference $\phi$ is given by the formula: $I_R = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi$.
Given $I_1 = I$,$I_2 = I$,and $\phi = 120^o$.
Substituting these values into the formula:
$I_R = I + I + 2\sqrt{I \cdot I} \cos(120^o)$.
Since $\cos(120^o) = -1/2$,we get:
$I_R = 2I + 2I(-1/2) = 2I - I = I$.
Therefore,the resultant intensity is $I$.

(C) The resultant amplitude $A$ of two interfering waves is given by the formula:
$A = \sqrt{a_1^2 + a_2^2 + 2a_1 a_2 \cos \phi}$
Here,$a_1 = 4$,$a_2 = 3$,and the phase difference $\phi = \pi / 3$.
Substituting these values into the formula:
$A = \sqrt{4^2 + 3^2 + 2(4)(3) \cos(\pi / 3)}$
Since $\cos(\pi / 3) = 0.5$:
$A = \sqrt{16 + 9 + 24(0.5)}$
$A = \sqrt{25 + 12}$
$A = \sqrt{37}$
$A \approx 6.08$
Rounding to the nearest integer,the amplitude is $6$.

(B) The resultant amplitude $A$ of two waves with amplitudes $A_1$ and $A_2$ and phase difference $\phi$ is given by the formula: $A = \sqrt{A_1^2 + A_2^2 + 2A_1A_2 \cos \phi}$.
Given $A_1 = A_2 = a$ and the resultant amplitude $A = a$.
Substituting these values into the formula: $a = \sqrt{a^2 + a^2 + 2a^2 \cos \phi}$.
Squaring both sides: $a^2 = 2a^2 + 2a^2 \cos \phi$.
Dividing by $a^2$: $1 = 2 + 2 \cos \phi$.
Rearranging gives: $2 \cos \phi = -1$,so $\cos \phi = -1/2$.
Therefore,the phase difference $\phi = 120^\circ$ or $\phi = 2\pi / 3$ radians.

(D) The intensity $I$ of a wave is proportional to the square of its amplitude,$I \propto A^2$.
When two waves with amplitudes $a_1$ and $a_2$ superimpose,the resultant amplitude $A_R$ can range from $|a_1 - a_2|$ to $(a_1 + a_2)$.
For maximum intensity (constructive interference),the resultant amplitude is $A_R = a_1 + a_2$.
Given $a_1 = a_2 = a$,the resultant amplitude is $A_R = a + a = 2a$.
The resultant intensity $I$ is proportional to the square of the resultant amplitude:
$I \propto (A_R)^2$
$I \propto (2a)^2$
$I \propto 4a^2$.

(D) The resultant amplitude $A_R$ of two superimposing waves with amplitudes $a_1$ and $a_2$ and phase difference $\phi$ is given by:
$A_R = \sqrt{a_1^2 + a_2^2 + 2a_1a_2 \cos \phi}$
Given $a_1 = a_2 = A$ and $\phi = \pi / 2$:
$A_R = \sqrt{A^2 + A^2 + 2A^2 \cos(\pi / 2)}$
Since $\cos(\pi / 2) = 0$,we get:
$A_R = \sqrt{2A^2} = \sqrt{2} A$
When two waves of the same frequency superimpose,the resultant wave has the same frequency as the individual waves.
Therefore,the resultant frequency is $\omega$.

(A) The relationship between time difference $(\Delta t)$ and phase difference $(\phi)$ is given by $\phi = \frac{2\pi}{T} \times \Delta t$.
Given $\Delta t = \frac{3T}{2}$,we substitute this into the formula:
$\phi = \frac{2\pi}{T} \times \frac{3T}{2} = 3\pi$.
The resultant intensity $I_R$ of two waves with intensities $I_1 = x$ and $I_2 = y$ is given by $I_R = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos(\phi)$.
Substituting $\phi = 3\pi$ and $\cos(3\pi) = -1$:
$I_R = x + y + 2\sqrt{xy}(-1) = x + y - 2\sqrt{xy} = (\sqrt{x} - \sqrt{y})^2$.
Therefore,the correct resultant intensity is $(\sqrt{x} - \sqrt{y})^2$.

(B) The sound wave splits into two paths at the junction. One path is straight (length $L_1 = 2r$) and the other path is semicircular (length $L_2 = \pi r$).
For a minimum of sound to be heard at the detector $D$,the path difference between the two waves must be an odd multiple of half the wavelength,i.e.,$\Delta x = (2n + 1) \frac{\lambda}{2}$.
For the smallest radius $r$,we take $n = 0$,so $\Delta x = \frac{\lambda}{2}$.
The path difference is $\Delta x = L_2 - L_1 = \pi r - 2r = r(\pi - 2)$.
Given $\lambda = 32 \ cm$,we have $r(\pi - 2) = \frac{32}{2} = 16$.
Therefore,$r = \frac{16}{\pi - 2} \approx \frac{16}{3.14 - 2} = \frac{16}{1.14} \approx 14 \ cm$.

(C) The resultant intensity $I$ of two interfering waves of equal intensity $I_0$ is given by $I = 4I_0 \cos^2(\phi/2)$,where $\phi$ is the phase difference between the waves.
For waves arriving in phase,the phase difference $\phi_1 = 0^{\circ}$.
Thus,$I_1 = 4I_0 \cos^2(0/2) = 4I_0 \cos^2(0) = 4I_0$.
For waves arriving $90^{\circ}$ out of phase,the phase difference $\phi_2 = 90^{\circ}$.
Thus,$I_2 = 4I_0 \cos^2(90^{\circ}/2) = 4I_0 \cos^2(45^{\circ}) = 4I_0 \times (1/\sqrt{2})^2 = 4I_0 \times (1/2) = 2I_0$.
The ratio of the intensities is $\frac{I_1}{I_2} = \frac{4I_0}{2I_0} = \frac{2}{1}$.

(A) Given waves are $y_1 = 5 \sin (\omega t - kx)$ and $y_2 = -5 \cos (\omega t - kx - 150^{\circ})$.
We know that $-\cos(\theta) = \sin(\theta - 90^{\circ})$.
So,$y_2 = 5 \sin (\omega t - kx - 150^{\circ} - 90^{\circ}) = 5 \sin (\omega t - kx - 240^{\circ})$.
Alternatively,$y_2 = 5 \sin (\omega t - kx - 240^{\circ} + 360^{\circ}) = 5 \sin (\omega t - kx + 120^{\circ})$.
The resultant amplitude $A$ of two waves $A_1 \sin(\phi_1)$ and $A_2 \sin(\phi_2)$ is given by $A = \sqrt{A_1^2 + A_2^2 + 2A_1A_2 \cos(\Delta\phi)}$.
Here $A_1 = 5$,$A_2 = 5$,and the phase difference $\Delta\phi = 120^{\circ} - 0^{\circ} = 120^{\circ}$.
$A = \sqrt{5^2 + 5^2 + 2(5)(5) \cos(120^{\circ})}$.
Since $\cos(120^{\circ}) = -0.5$,we have $A = \sqrt{25 + 25 + 50(-0.5)} = \sqrt{50 - 25} = \sqrt{25} = 5$.

(A) The amplitudes of the waves are $a_{1} = 10 \, \mu m$,$a_{2} = 4 \, \mu m$,and $a_{3} = 7 \, \mu m$.
The phase difference between the $1^{st}$ and $2^{nd}$ wave is $\phi_{12} = \pi / 2$,and the phase difference between the $2^{nd}$ and $3^{rd}$ wave is $\phi_{23} = \pi / 2$. Thus,the phase difference between the $1^{st}$ and $3^{rd}$ wave is $\phi_{13} = \pi$.
First,we combine the $1^{st}$ and $3^{rd}$ waves. Since they are in opposite phase $(\pi)$,their resultant amplitude $A_{13}$ is:
$A_{13} = |a_{1} - a_{3}| = |10 - 7| = 3 \, \mu m$.
Now,we combine this resultant $A_{13}$ with the $2^{nd}$ wave. The phase difference between the $2^{nd}$ wave and the resultant of the $1^{st}$ and $3^{rd}$ waves is $\pi / 2$ (since the resultant $A_{13}$ is in the direction of the $1^{st}$ wave).
The final resultant amplitude $A$ is given by:
$A = \sqrt{A_{13}^{2} + a_{2}^{2} + 2 A_{13} a_{2} \cos(\pi / 2)}$
$A = \sqrt{3^{2} + 4^{2} + 0} = \sqrt{9 + 16} = \sqrt{25} = 5 \, \mu m$.

(A) The person receives sound via two paths: the direct path of length $d_1 = 6\, m$ and the reflected path of length $d_2 = 5\, m + 5\, m = 10\, m$.
The path difference between the two waves is $\Delta x = d_2 - d_1 = 10\, m - 6\, m = 4\, m$.
For constructive interference (maximum sound intensity),the path difference must be an integral multiple of the wavelength $\lambda$:
$\Delta x = n\lambda$,where $n = 1, 2, 3, \dots$
Substituting the values,we get $4 = n\lambda$,which implies $\lambda = \frac{4}{n}$.
To find the maximum wavelength,we must choose the smallest positive integer value for $n$,which is $n = 1$.
Therefore,$\lambda_{\max} = \frac{4}{1} = 4\, m$.

(B) The resultant intensity $I_R$ of two interfering waves is given by the formula: $I_R = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi$,where $\phi$ is the phase difference.
Given $I_1 = I$ and $I_2 = 4I$.
At point $A$,the phase difference $\phi_A = \pi / 2$. Substituting this into the formula:
$I_A = I + 4I + 2\sqrt{I \cdot 4I} \cos(\pi / 2) = 5I + 2(2I)(0) = 5I$.
At point $B$,the phase difference $\phi_B = \pi$. Substituting this into the formula:
$I_B = I + 4I + 2\sqrt{I \cdot 4I} \cos(\pi) = 5I + 2(2I)(-1) = 5I - 4I = I$.
The difference between the resultant intensities at $A$ and $B$ is $I_A - I_B = 5I - I = 4I$.

(C) The resultant wave function at $x = 0$ is $y = y_1 + y_2$.
Substituting $x = 0$ into the wave functions:
$y_1 = A \cos(-kvt) = A \cos(kvt)$
$y_2 = A \cos(kvt + \phi)$
Using the superposition principle,$y = A [\cos(kvt) + \cos(kvt + \phi)]$.
Using the trigonometric identity $\cos C + \cos D = 2 \cos(\frac{C+D}{2}) \cos(\frac{C-D}{2})$:
$y = 2A \cos(kvt + \frac{\phi}{2}) \cos(\frac{\phi}{2})$.
For constructive interference,the amplitude must be maximum,which occurs when $|\cos(\frac{\phi}{2})| = 1$,implying $\frac{\phi}{2} = n\pi$,or $\phi = 2n\pi$ (where $n = 0, 1, 2, ...$).
For destructive interference,the amplitude must be zero,which occurs when $\cos(\frac{\phi}{2}) = 0$,implying $\frac{\phi}{2} = (2n+1)\frac{\pi}{2}$,or $\phi = (2n+1)\pi$ (where $n = 0, 1, 2, ...$).
Checking the options,for $n=0$ in the destructive interference condition,$\phi = \pi$. Thus,option $C$ is correct.

(C) Let the equations of the two waves be $y_1 = A \sin(kx - \omega t)$ and $y_2 = A \sin(kx - \omega t + \frac{\pi}{2})$.
Since $\sin(kx - \omega t + \frac{\pi}{2}) = \cos(kx - \omega t)$,the resultant wave is $y = y_1 + y_2 = A \sin(kx - \omega t) + A \cos(kx - \omega t)$.
Using the identity $a \sin \theta + b \cos \theta = \sqrt{a^2 + b^2} \sin(\theta + \phi)$,where $\phi = \tan^{-1}(b/a)$,we get:
$y = \sqrt{A^2 + A^2} \sin(kx - \omega t + \frac{\pi}{4}) = \sqrt{2}A \sin(kx - \omega t + \frac{\pi}{4})$.
The resultant wave has an amplitude of $\sqrt{2}A$,which is different from the individual amplitude $A$.
However,the wavelength $\lambda = \frac{2\pi}{k}$ and the frequency $f = \frac{\omega}{2\pi}$ (and thus the velocity $v = f\lambda = \frac{\omega}{k}$) remain the same as those of the individual waves.
Therefore,the resultant wave has the same wavelength and velocity as the individual waves,but a different amplitude.

(B) The circumference of the circular tube is $2\pi R$.
The two paths from $S$ to $D$ are along the arcs of the circle. Since $S$ and $D$ are at right angles,the shorter path length is $L_1 = \frac{1}{4}(2\pi R) = \frac{\pi R}{2}$.
The longer path length is $L_2 = \frac{3}{4}(2\pi R) = \frac{3\pi R}{2}$.
The path difference between the two waves is $\Delta x = L_2 - L_1 = \frac{3\pi R}{2} - \frac{\pi R}{2} = \pi R$.
For destructive interference (minima) to occur at $D$,the path difference must be an odd multiple of half-wavelengths:
$\Delta x = (2n + 1) \frac{\lambda}{2}$,where $n = 0, 1, 2, \dots$
Equating the path difference:
$\pi R = (2n + 1) \frac{\lambda}{2}$
$\lambda = \frac{2\pi R}{2n + 1}$
To find the maximum value of $\lambda$,we choose the smallest non-negative integer $n = 0$:
$\lambda_{\text{max}} = \frac{2\pi R}{2(0) + 1} = 2\pi R$.

(A) The path difference is given by $\Delta x = S_1P - S_2P = \frac{\lambda}{2}$.
The phase difference $\Delta \phi$ is calculated as $\Delta \phi = k \Delta x = \frac{2 \pi}{\lambda} \times \Delta x$.
Substituting the value of $\Delta x$,we get $\Delta \phi = \frac{2 \pi}{\lambda} \times \frac{\lambda}{2} = \pi$.
Since the phase difference is $\pi$,the two waves interfere destructively at point $P$.
The resultant amplitude $A_R$ is given by $A_R = |A_1 - A_2|$,where $A_1 = 2a$ and $A_2 = a$.
Thus,$A_R = |2a - a| = a$.
The resultant oscillation at $P$ will be a sinusoidal wave with an amplitude of $a$.

(C) The displacement of the wave from $S_1$ at point $A$ at time $t$ is $y_{A1} = 5 \sin(\pi(t - 1))$.
The displacement of the wave from $S_2$ at point $A$ at time $t$ is $y_{A2} = 5 \sin(\pi(t - 0.5) + \pi/6)$.
Comparing the phases at point $A$:
Phase of wave $1$: $\phi_1 = \pi t - \pi$.
Phase of wave $2$: $\phi_2 = \pi t - 0.5\pi + \pi/6 = \pi t - \pi/2 + \pi/6 = \pi t - \pi/3$.
The phase difference $\Delta\phi = \phi_2 - \phi_1 = (\pi t - \pi/3) - (\pi t - \pi) = \pi - \pi/3 = 2\pi/3$.
The resulting amplitude $A_{res} = \sqrt{A_1^2 + A_2^2 + 2A_1 A_2 \cos(\Delta\phi)}$.
Given $A_1 = A_2 = 5 \text{ mm}$ and $\Delta\phi = 2\pi/3$:
$A_{res} = \sqrt{5^2 + 5^2 + 2(5)(5) \cos(2\pi/3)} = \sqrt{25 + 25 + 50(-0.5)} = \sqrt{50 - 25} = \sqrt{25} = 5 \text{ mm}$.

(C) At $t = 0 \ s$,the triangular pulse is between $x = 0 \ m$ and $x = 2 \ m$ moving to the right at $1 \ m/s$. The rectangular pulse is between $x = 4 \ m$ and $x = 7 \ m$ moving to the left at $1 \ m/s$.
At $t = 3 \ s$,the triangular pulse has moved $3 \ m$ to the right,occupying the region between $x = 3 \ m$ and $x = 5 \ m$.
The rectangular pulse has moved $3 \ m$ to the left,occupying the region between $x = 1 \ m$ and $x = 4 \ m$.
Therefore,the combined shape of the string at $t = 3 \ s$ is the superposition of these two pulses,which spans from $x = 1 \ m$ to $x = 5 \ m$.

(B) Let the source be at $S$ and the observer at $O$. The direct distance $SO = 120\,m$.
The ceiling is at a height $h = 25\,m$. The reflected wave travels from $S$ to the midpoint of the ceiling and then to $O$.
Since the ceiling is at the midpoint,the horizontal distance from $S$ to the reflection point is $60\,m$ and from the reflection point to $O$ is $60\,m$.
The path length of the reflected wave is $d = 2 \times \sqrt{60^2 + 25^2} = 2 \times \sqrt{3600 + 625} = 2 \times \sqrt{4225} = 2 \times 65 = 130\,m$.
The path difference between the reflected wave and the direct wave is $\Delta x = 130\,m - 120\,m = 10\,m$.
For constructive interference,the path difference must be an integer multiple of the wavelength: $\Delta x = n\lambda$,where $n = 1, 2, 3, \dots$.
Thus,$10 = n\lambda \Rightarrow \lambda = \frac{10}{n}$.
For $n = 1, 2, 3, \dots$,the possible wavelengths are $\lambda = 10\,m, 5\,m, \frac{10}{3}\,m, \dots$.

(D) The first wave is $y_1 = a \sin \frac{2\pi}{\lambda} (vt - x)$.
The second wave is $y_2 = a \cos \frac{2\pi}{\lambda} (vt - x) = a \sin \left( \frac{2\pi}{\lambda} (vt - x) + \frac{\pi}{2} \right)$.
Comparing these with the standard form $y = A \sin(\omega t - kx + \phi)$,the phase difference between the two waves is $\Delta \phi = \frac{\pi}{2}$.
The resultant amplitude $A_{res}$ of two waves with equal amplitude $a$ and phase difference $\Delta \phi$ is given by $A_{res} = \sqrt{a^2 + a^2 + 2a^2 \cos(\Delta \phi)}$.
Substituting $\Delta \phi = 90^{\circ}$,we get $A_{res} = \sqrt{a^2 + a^2 + 2a^2 \cos(90^{\circ})} = \sqrt{2a^2 + 0} = a\sqrt{2}$.

(B) The resultant amplitude $A$ of two interfering waves with amplitudes $a_1$ and $a_2$ and phase difference $\phi$ is given by the formula:
$A = \sqrt{a_1^2 + a_2^2 + 2 a_1 a_2 \cos \phi}$
Here,$a_1 = 4$,$a_2 = 3$,and $\phi = \frac{\pi}{3} = 60^{\circ}$.
Substituting these values into the formula:
$A = \sqrt{4^2 + 3^2 + 2(4)(3) \cos 60^{\circ}}$
$A = \sqrt{16 + 9 + 24 \times 0.5}$
$A = \sqrt{25 + 12}$
$A = \sqrt{37} \approx 6.08$
Thus,the amplitude of the resulting wave is approximately $6$.

(D) Interference is a fundamental property of all wave phenomena. Both light waves (electromagnetic) and sound waves (mechanical/longitudinal) exhibit the phenomenon of interference when two or more waves overlap in space.

(A) The two waves are given by $Y_1 = A_1 \sin(\omega t - \beta_1)$ and $Y_2 = A_2 \sin(\omega t - \beta_2)$.
The phase difference between the two waves is $\phi = (\omega t - \beta_1) - (\omega t - \beta_2) = \beta_2 - \beta_1$.
The resultant amplitude $A$ of two superimposing waves with amplitudes $A_1$ and $A_2$ and phase difference $\phi$ is given by the formula:
$A = \sqrt{A_1^2 + A_2^2 + 2A_1A_2 \cos \phi}$.
Substituting $\phi = \beta_2 - \beta_1$ into the formula,and noting that $\cos(\beta_2 - \beta_1) = \cos(\beta_1 - \beta_2)$:
$A = \sqrt{A_1^2 + A_2^2 + 2A_1A_2 \cos(\beta_1 - \beta_2)}$.

(B) Given the displacement equation: $y = 4 \cos^2(t/2) \sin(1000t)$
Using the trigonometric identity $2 \cos^2 \theta = 1 + \cos(2\theta)$,we can rewrite the expression as:
$y = 2 [2 \cos^2(t/2)] \sin(1000t)$
$y = 2(1 + \cos t) \sin(1000t)$
$y = 2 \sin(1000t) + 2 \sin(1000t) \cos t$
Using the product-to-sum identity $2 \sin A \cos B = \sin(A+B) + \sin(A-B)$,we get:
$y = 2 \sin(1000t) + \sin(1000t + t) + \sin(1000t - t)$
$y = 2 \sin(1000t) + \sin(1001t) + \sin(999t)$
Thus,the given expression represents the superposition of three waves.