Wave Equation and Characteristics of Waves Questions in English

(C) The maximum particle velocity is given by $v_{\max} = \omega A$,where $\omega$ is the angular frequency and $A$ is the amplitude.
The wave propagation velocity is given by $v = \frac{\omega}{k}$,where $k$ is the wave number.
The ratio of maximum particle velocity to wave velocity is $\frac{v_{\max}}{v} = \frac{\omega A}{\omega / k} = kA$.
From the given wave equation $y = 60 \cos (1800t - 6x)$,we identify:
Amplitude $A = 60 \text{ microns} = 60 \times 10^{-6} \text{ m}$.
Wave number $k = 6 \text{ m}^{-1}$.
Therefore,the ratio is $\frac{v_{\max}}{v} = kA = 6 \times (60 \times 10^{-6}) = 360 \times 10^{-6} = 3.6 \times 10^{-4}$.

(B) The standard form of a wave equation is $y = A \sin(\omega t - kx)$.
Comparing the given equation $y = 0.30 \sin(314t - 1.57x)$ with the standard form,we get:
Angular frequency $\omega = 314 \, \text{rad/s}$
Wave number $k = 1.57 \, \text{rad/m}$
The speed of the wave $v$ is given by the formula $v = \frac{\omega}{k}$.
Substituting the values: $v = \frac{314}{1.57} = 200 \, \text{m/s}$.
Therefore,the speed of the wave is $200 \, \text{m/s}$.

(A) The standard equation of a progressive wave is given by $y = a \sin(\omega t - kx)$.
Comparing the given equation $y = a \sin(400\pi t - \pi x / 6.85)$ with the standard equation,we get the angular frequency $\omega = 400\pi \, rad/s$.
The relationship between frequency $f$ and angular frequency $\omega$ is given by $\omega = 2\pi f$.
Therefore,$f = \frac{\omega}{2\pi} = \frac{400\pi}{2\pi} = 200 \, Hz$.

(A) The phase of a wave is given by $\phi = 2\pi f t$.
For the first wave,$\phi_1 = 2\pi f_1 t = 2\pi \times 20 \times 0.6 = 24\pi$.
For the second wave,$\phi_2 = 2\pi f_2 t = 2\pi \times 30 \times 0.6 = 36\pi$.
The phase difference is $\Delta \phi = |\phi_2 - \phi_1| = |36\pi - 24\pi| = 12\pi$.
Since $12\pi$ is an even multiple of $\pi$,the phase difference is equivalent to $0$ (Zero).

(D) The relationship between path difference $\Delta x$ and phase difference $\phi$ is given by $\Delta x = \frac{\lambda}{2\pi} \times \phi$.
Given $\Delta x = 0.8 \ m$ and $\phi = 90^o = \frac{\pi}{2} \text{ radians}$.
Substituting the values: $0.8 = \frac{\lambda}{2\pi} \times \frac{\pi}{2} = \frac{\lambda}{4}$.
Therefore,$\lambda = 0.8 \times 4 = 3.2 \ m$.
The velocity of the wave $v$ is given by $v = f \lambda$,where $f = 120 \ Hz$.
$v = 120 \times 3.2 = 384 \ m/s$.

(B) The standard equation of a transverse wave is given by $y = a\sin 2\pi \left[ \frac{t}{T} - \frac{x}{\lambda} \right]$.
Comparing the given equation $y = 5\sin 2\pi \left[ \frac{t}{0.04} - \frac{x}{40} \right]$ with the standard equation,we identify the terms.
The term $\frac{x}{\lambda}$ corresponds to $\frac{x}{40}$.
Therefore,by direct comparison,the wavelength $\lambda = 40 \ cm$.

(D) The standard equation of a progressive wave is given by $y = a \sin(kx - \omega t)$.
Comparing the given equation $y = a \sin(0.01x - 2t)$ with the standard equation,we get:
Wave number $k = 0.01 \, cm^{-1}$
Angular frequency $\omega = 2 \, rad/s$
The velocity of propagation of the wave is given by $v = \frac{\omega}{k}$.
Substituting the values,we get $v = \frac{2}{0.01} = 200 \, cm/s$.

(D) The standard equation of a simple harmonic progressive wave is $y = A \sin(kx - \omega t)$.
Comparing this with the given equation $y = 8 \sin(2\pi(0.1x - 2t)) = 8 \sin(0.2\pi x - 4\pi t)$,we get the wave number $k = 0.2\pi \, rad/cm$.
The phase difference $\Delta \phi$ between two particles separated by a distance $\Delta x$ is given by $\Delta \phi = k \Delta x$.
Given $\Delta x = 2.0 \, cm$,we have $\Delta \phi = (0.2\pi) \times 2.0 = 0.4\pi \, radians$.
To convert radians to degrees,we multiply by $\frac{180^o}{\pi}$:
$\Delta \phi = 0.4\pi \times \frac{180^o}{\pi} = 0.4 \times 180^o = 72^o$.

(A) The standard equation of a progressive wave is given by $y = a \sin(\omega t - kx)$.
Comparing the given equation $y = a \sin(200t - x)$ with the standard equation,we get:
$\omega = 200 \text{ rad/s}$
$k = 1 \text{ rad/m}$
The velocity of the wave $v$ is given by the formula $v = \frac{\omega}{k}$.
Substituting the values,$v = \frac{200}{1} = 200 \text{ m/s}$.
Therefore,the correct option is $A$.

(A) The standard equation of a traveling wave is given by $y = A \sin(\omega t - kx)$.
Given equation is $y = 7 \sin \{ \pi (2t - 2x) \} = 7 \sin(2\pi t - 2\pi x)$.
Comparing this with the standard equation,we get angular frequency $\omega = 2\pi \, \text{rad/s}$ and wave number $k = 2\pi \, \text{rad/m}$.
The velocity of the wave $v$ is given by the formula $v = \frac{\omega}{k}$.
Substituting the values,$v = \frac{2\pi}{2\pi} = 1 \, m/s$.

(B) The standard equation of a wave is given by $y = a \cos (\omega t - kx)$.
Given equation: $y = 20 \cos (50\pi t - \pi x)$.
Comparing the given equation with the standard equation,we get the wave number $k = \pi$.
We know that the wave number $k$ is related to the wavelength $\lambda$ by the formula $k = \frac{2\pi}{\lambda}$.
Substituting the value of $k$: $\pi = \frac{2\pi}{\lambda}$.
Solving for $\lambda$: $\lambda = \frac{2\pi}{\pi} = 2 \, cm$.

(D) The standard wave equation is given by $y = a \sin(\omega t + kx)$.
Comparing the given equation $y = 0.001 \sin(100t + x)$ with the standard form:
$1$. Angular frequency $\omega = 100 \ rad/s$. Since $\omega = 2\pi f$,the frequency $f = \frac{100}{2\pi} = \frac{50}{\pi} \ Hz$.
$2$. Wave number $k = 1 \ m^{-1}$. Since $k = \frac{2\pi}{\lambda}$,the wavelength $\lambda = 2\pi \ m$.
$3$. Wave velocity $v = \frac{\omega}{k} = \frac{100}{1} = 100 \ m/s$.
$4$. Since there is a positive sign between the $t$ and $x$ terms $(100t + x)$,the wave is travelling in the negative $x$-direction.
Therefore,the correct option is $(d)$.

(B) The given wave equation is $y = A \sin 2\pi \left( \frac{t}{T} - \frac{x}{\lambda} \right)$.
Comparing this with the standard wave equation $y = A \sin (\omega t - kx)$,we have $\omega = \frac{2\pi}{T}$ and $k = \frac{2\pi}{\lambda}$.
The particle velocity $v_p$ is given by $\frac{\partial y}{\partial t} = A \omega \cos 2\pi \left( \frac{t}{T} - \frac{x}{\lambda} \right)$.
The maximum particle velocity is $v_{p,max} = A \omega = A \left( \frac{2\pi}{T} \right)$.
The wave velocity $v$ is given by $v = \frac{\omega}{k} = \frac{2\pi / T}{2\pi / \lambda} = \frac{\lambda}{T}$.
According to the problem,$v_{p,max} = 4v$.
Substituting the expressions: $A \left( \frac{2\pi}{T} \right) = 4 \left( \frac{\lambda}{T} \right)$.
Canceling $T$ from both sides: $2\pi A = 4\lambda$.
Therefore,$\lambda = \frac{2\pi A}{4} = \frac{\pi A}{2}$.

(D) The standard wave equation is given by $y = A \sin(\omega t - kx)$.
Comparing this with the given equation $y = 10^{-4} \sin(100t - x/10)$,we get:
Angular frequency $\omega = 100 \, rad/s$.
Wave number $k = 1/10 \, m^{-1}$.
The velocity of the wave $v$ is given by the formula $v = \frac{\omega}{k}$.
Substituting the values,$v = \frac{100}{1/10} = 1000 \, m/s$.

(C) wave travelling in the positive $x-$direction is represented by the equation:
$y = A \sin \left( \frac{2\pi}{\lambda} (vt - x) \right)$
Given values are $A = 0.2\;m,$ $v = 360\;m/s,$ and $\lambda = 60\;m.$
Substituting these values into the equation:
$y = 0.2 \sin \left( \frac{2\pi}{60} (360t - x) \right)$
$y = 0.2 \sin \left( 2\pi \left( \frac{360}{60}t - \frac{x}{60} \right) \right)$
$y = 0.2 \sin \left[ 2\pi \left( 6t - \frac{x}{60} \right) \right]$
Thus,the correct expression is $y = 0.2 \sin \left[ 2\pi \left( 6t - \frac{x}{60} \right) \right].$

(A) The standard equation of a travelling wave is given by $y = A \sin(\omega t - kx + \phi)$.
Comparing the given equation $y = 7 \sin(7\pi t - 0.4\pi x + \frac{\pi}{3})$ with the standard equation,we get:
Angular frequency $\omega = 7\pi \ rad/s$.
Wave number $k = 0.4\pi \ rad/m$.
The velocity of the wave $v$ is given by the ratio of angular frequency to wave number:
$v = \frac{\omega}{k}$
$v = \frac{7\pi}{0.4\pi}$
$v = \frac{7}{0.4} = \frac{70}{4} = 17.5 \ m/s$.
Therefore,the velocity of the wave is $17.5 \ m/s$.

(A) The standard equation of a progressive wave is given by $y = A \sin (\omega t - kx + \phi)$.
Comparing this with the given equation $y = 8 \sin \left[ \pi \left( \frac{t}{10} - \frac{x}{4} \right) + \frac{\pi}{3} \right]$,we can rewrite it as $y = 8 \sin \left( \frac{\pi t}{10} - \frac{\pi x}{4} + \frac{\pi}{3} \right)$.
The wave number $k$ is the coefficient of $x$,which is $k = \frac{\pi}{4}$.
We know that the relationship between wave number $k$ and wavelength $\lambda$ is $k = \frac{2\pi}{\lambda}$.
Substituting the value of $k$: $\frac{\pi}{4} = \frac{2\pi}{\lambda}$.
Solving for $\lambda$: $\lambda = 2 \times 4 = 8 \, m$.

(A) The standard wave equation is $y = a \sin(kx - \omega t)$.
Comparing the given equation $y = 0.07 \sin(12\pi x - 3000\pi t)$ with the standard form:
Amplitude $a = 0.07 \ m$.
Angular wave number $k = 12\pi \ rad/m$.
Angular frequency $\omega = 3000\pi \ rad/s$.
We know that $k = \frac{2\pi}{\lambda}$,so $\lambda = \frac{2\pi}{k} = \frac{2\pi}{12\pi} = \frac{1}{6} \ m$.
We know that $\omega = 2\pi n$,so frequency $n = \frac{\omega}{2\pi} = \frac{3000\pi}{2\pi} = 1500 \ Hz$.
The wave velocity $v = \frac{\omega}{k} = \frac{3000\pi}{12\pi} = 250 \ m/s$.
Thus,the correct values are $\lambda = 1/6 \ m$ and $v = 250 \ m/s$.

(B) The standard equation of a progressive wave is $y = A \sin(\omega t + kx + \phi)$.
Comparing $y = 25 \sin(20t + 5x)$ with the standard equation:
$1$. Amplitude $A = 25$ units.
$2$. Angular frequency $\omega = 20 \text{ rad/s}$ and wave number $k = 5 \text{ m}^{-1}$.
$3$. Wave velocity $v = \frac{\omega}{k} = \frac{20}{5} = 4$ units.
$4$. Maximum particle velocity $v_{p, \text{max}} = A\omega = 25 \times 20 = 500$ units.
Comparing these with the options:
Option $(a)$ is true $(A = 25)$.
Option $(b)$ is false because the maximum particle velocity is $500$ units,not $100$ units.
Option $(c)$ is true $(v = 4)$.
Option $(d)$ is true $(v_{p, \text{max}} = 500)$.
Therefore,the statement that is not true is $(b)$.

(B) The standard equation of a plane progressive wave is given by $y = a \cos (\omega t - kx)$.
Comparing the given equation $y = 25 \cos (2\pi t - \pi x)$ with the standard equation,we get the amplitude $a = 25$.
The angular frequency $\omega$ is given by $\omega = 2\pi$.
Since $\omega = 2\pi f$,where $f$ is the frequency,we have $2\pi f = 2\pi$.
Therefore,the frequency $f = 1 \text{ Hz}$.
Thus,the amplitude is $25$ and the frequency is $1 \text{ Hz}$.

(B) The standard equation of a travelling wave is given by $y = A \sin(\omega t - kx + \phi)$.
Comparing the given equation $y = 10^{-4} \sin(600t - 2x + \frac{\pi}{3})$ with the standard equation,we get:
Angular frequency $\omega = 600 \ rad \ s^{-1}$
Wave number $k = 2 \ m^{-1}$
The speed of the wave $v$ is given by the formula $v = \frac{\omega}{k}$.
Substituting the values,$v = \frac{600}{2} = 300 \ m \ s^{-1}$.
Therefore,the speed of the wave motion is $300 \ m \ s^{-1}$.

(B) The standard wave equation is given by $y = A \sin(\omega t + kx + \phi)$.
Comparing the given equation $y = 10^{-6} \sin(100t + 20x + \pi/4)$ with the standard equation,we get:
Angular frequency $\omega = 100 \ rad/s$
Wave number $k = 20 \ rad/m$
The speed of the wave $v$ is given by the ratio of the coefficient of $t$ to the coefficient of $x$:
$v = \frac{\omega}{k} = \frac{100}{20} = 5 \ m/s$.

(D) The standard form of a traveling wave equation is given by $y = a \sin \frac{2\pi}{\lambda} (vt - x)$.
Comparing the given equation $y = 0.08 \sin \frac{2\pi}{\lambda} (200t - x)$ with the standard form,we can identify the coefficient of $t$ as the wave velocity $v$.
Thus,by direct comparison,$v = 200 \ m/s$.

(B) The relationship between phase difference $(\Delta \phi)$ and path difference $(\Delta x)$ is given by: $\Delta \phi = \frac{2\pi}{\lambda} \times \Delta x$.
Given: $\Delta \phi = \frac{\pi}{2}$,$\Delta x = 0.8 \ m$,and frequency $f = 120 \ Hz$.
Substituting the values: $\frac{\pi}{2} = \frac{2\pi}{\lambda} \times 0.8$.
Solving for wavelength $(\lambda)$: $\lambda = 4 \times 0.8 = 3.2 \ m$.
The velocity of the wave $(v)$ is given by $v = f \times \lambda$.
$v = 120 \times 3.2 = 384 \ m/s$.

(A) The standard equation of a plane progressive wave is $y = A \sin(\omega t - kx)$.
Comparing the given equation $y = 0.1 \sin \left( 200\pi t - \frac{20\pi x}{17} \right)$ with the standard equation:
$1$. Angular frequency $\omega = 200\pi \ rad/s$. Since $\omega = 2\pi n$,we have $2\pi n = 200\pi$,which gives frequency $n = 100 \ Hz$.
$2$. Wave number $k = \frac{20\pi}{17} \ rad/m$. Since $k = \frac{2\pi}{\lambda}$,we have $\lambda = \frac{2\pi}{k} = \frac{2\pi}{20\pi/17} = 1.7 \ m$.
$3$. Wave speed $v = \frac{\omega}{k} = \frac{200\pi}{20\pi/17} = 170 \ m/s$.
Thus,the frequency,wavelength,and speed are $100 \ Hz, 1.7 \ m, 170 \ m/s$ respectively.

(B) The standard equation of a travelling wave is given by $y = a \sin(kx - \omega t)$.
Comparing the given equation $y = 0.5 \sin(20x - 400t)$ with the standard form,we identify the wave number $k = 20 \text{ rad/m}$ and the angular frequency $\omega = 400 \text{ rad/s}$.
The velocity of the wave $v$ is given by the ratio of angular frequency to wave number:
$v = \frac{\omega}{k}$
Substituting the values:
$v = \frac{400}{20} = 20 \text{ m/s}$.
Thus,the velocity of the wave is $20 \text{ m/s}$.

(D) Given: Velocity $v = 10\,m/s$,Frequency $f = 100\,Hz$,Path difference $\Delta x = 2.5\,cm = 0.025\,m$.
First,calculate the wavelength $\lambda$ using the relation $v = f\lambda$:
$\lambda = \frac{v}{f} = \frac{10}{100} = 0.1\,m = 10\,cm$.
Now,use the formula for phase difference $\Delta \phi$:
$\Delta \phi = \frac{2\pi}{\lambda} \times \Delta x$.
Substituting the values:
$\Delta \phi = \frac{2\pi}{10\,cm} \times 2.5\,cm = \frac{2\pi}{4} = \frac{\pi}{2}$ radians.

(B) Given equations are:
$y_1 = 10^{-6} \sin [100t + (x/50) + 0.5]$
$y_2 = 10^{-6} \cos [100t + (x/50)]$
To compare the phases,convert the cosine function into a sine function using the identity $\cos(\theta) = \sin(\theta + \pi/2)$:
$y_2 = 10^{-6} \sin [100t + (x/50) + \pi/2]$
Since $\pi/2 \approx 1.57$,we have:
$y_2 = 10^{-6} \sin [100t + (x/50) + 1.57]$
The phase of the first wave is $\phi_1 = 100t + (x/50) + 0.5$.
The phase of the second wave is $\phi_2 = 100t + (x/50) + 1.57$.
The phase difference $\Delta\phi$ is given by:
$\Delta\phi = |\phi_2 - \phi_1|$
$\Delta\phi = |(100t + x/50 + 1.57) - (100t + x/50 + 0.5)|$
$\Delta\phi = 1.57 - 0.5 = 1.07 \, rad$.

(B) In a simple harmonic wave motion,a particle oscillates about its mean position.
When a particle is at the trough (the point of maximum negative displacement),it is at one of its extreme positions.
The time taken to travel from an extreme position to the mean position is exactly one-fourth of the total time period $(T)$.
Therefore,the particle will reach the mean position after a time of $\frac{T}{4}$.

(A) The standard equation of a transverse wave is given by $Y = a \sin(kx - \omega t)$.
Comparing the given equation $Y = 2 \sin(kx - 2t)$ with the standard form,we get the amplitude $a = 2$ and the angular frequency $\omega = 2$.
The particle velocity $v_p$ is the derivative of displacement $Y$ with respect to time $t$:
$v_p = \frac{dY}{dt} = \frac{d}{dt} [2 \sin(kx - 2t)] = 2 \cos(kx - 2t) \times (-2) = -4 \cos(kx - 2t)$.
The maximum particle velocity is given by the magnitude of the coefficient of the cosine term:
$v_{max} = a \omega = 2 \times 2 = 4 \, units$.

(C) The first wave equation is ${y_1} = 10\sin \left( {3\pi t + \frac{\pi }{3}} \right)$. The amplitude of this wave is $A_1 = 10$.
The second wave equation is ${y_2} = 5(\sin 3\pi t + \sqrt 3 \cos 3\pi t)$.
We can rewrite this by multiplying and dividing by $2$:
${y_2} = 5 \times 2 \left( \frac{1}{2} \sin 3\pi t + \frac{\sqrt 3}{2} \cos 3\pi t \right)$.
Using the trigonometric identity $\sin(A + B) = \sin A \cos B + \cos A \sin B$,where $\cos \frac{\pi}{3} = \frac{1}{2}$ and $\sin \frac{\pi}{3} = \frac{\sqrt 3}{2}$:
${y_2} = 10 \left( \sin 3\pi t \cos \frac{\pi}{3} + \cos 3\pi t \sin \frac{\pi}{3} \right) = 10 \sin \left( 3\pi t + \frac{\pi}{3} \right)$.
The amplitude of the second wave is $A_2 = 10$.
The ratio of their amplitudes is $\frac{A_1}{A_2} = \frac{10}{10} = 1:1$.

(A) The given equation is $y = A \cos^2 \left( 2\pi nt - 2\pi \frac{x}{\lambda} \right)$.
Using the trigonometric identity $\cos^2 \theta = \frac{1 + \cos(2\theta)}{2}$,we can rewrite the equation as:
$y = \frac{A}{2} \left[ 1 + \cos \left( 4\pi nt - \frac{4\pi x}{\lambda} \right) \right] = \frac{A}{2} + \frac{A}{2} \cos \left( 4\pi nt - \frac{4\pi x}{\lambda} \right)$.
The oscillating part of the wave is $\frac{A}{2} \cos \left( 4\pi nt - \frac{4\pi x}{\lambda} \right)$.
Comparing this with the standard wave equation $y = a \cos(\omega t - kx)$:
Amplitude $a = A/2$.
Angular frequency $\omega = 4\pi n$,so frequency $f = \frac{\omega}{2\pi} = \frac{4\pi n}{2\pi} = 2n$.
Wave number $k = \frac{4\pi}{\lambda}$,so wavelength $\lambda' = \frac{2\pi}{k} = \frac{2\pi}{4\pi/\lambda} = \frac{\lambda}{2}$.
Thus,the wave has amplitude $A/2$,frequency $2n$,and wavelength $\lambda/2$.

(D) The equation $y = a \sin (kx - \omega t)$ represents a general traveling wave equation.
In the case of mechanical waves (like sound waves),$y$ can represent the displacement of particles or pressure variations.
In the case of electromagnetic waves,$y$ represents the oscillating electric field $(E)$ or magnetic field $(B)$ components.
Therefore,$y$ is a general physical quantity that oscillates and propagates through space.
Thus,the correct option is $D$.

(A) The general equation for a wave traveling in the positive $x$-direction is given by $y = f(x - vt)$,where $v$ is the wave velocity.
At $t = 0$,the equation is $y = f(x) = \frac{1}{1 + x^2}$.
At $t = 2$ seconds,the equation is $y = f(x - v(2)) = \frac{1}{1 + (x - 2v)^2}$.
Comparing this with the given equation at $t = 2$,which is $y = \frac{1}{1 + (x - 1)^2}$,we can equate the terms inside the square:
$x - 2v = x - 1$
$2v = 1$
$v = 0.5 \ m/s$.
Therefore,the velocity of the wave is $0.5 \ m/s$.

(C) From the figure,the distance between two consecutive nodes or a node and an adjacent antinode is $\frac{\lambda}{4}$. Thus,$ab = \frac{\lambda}{4}$,which implies $\lambda = 4ab$.
Statement $I$: The speed of the wave $v = n\lambda = n(4ab) = 4n \times ab$. This statement is correct.
Statement $II$: The point $a$ is at a crest and $d$ is at a node. For $a$ to be in the same phase as $d$,the wave must travel a distance such that $d$ reaches the crest position. The distance between $a$ and $d$ is $\frac{3\lambda}{4}$. The time taken is $t = \frac{\text{distance}}{\text{speed}} = \frac{3\lambda/4}{n\lambda} = \frac{3}{4n} \text{ s}$. The statement says $\frac{4}{3n} \text{ s}$,which is incorrect.
Statement $III$: The path difference between $b$ (node) and $e$ (crest) is $\frac{3\lambda}{4}$. The phase difference $\Delta\phi = \frac{2\pi}{\lambda} \times \Delta x = \frac{2\pi}{\lambda} \times \frac{3\lambda}{4} = \frac{3\pi}{2}$. This statement is correct.
Therefore,statements $I$ and $III$ are correct.

(D) Two points are in the same phase if they have the same displacement from the equilibrium position and are moving in the same direction.
This occurs when the distance between the two points is an integer multiple of the wavelength,$\lambda$.
In the given wave diagram,points $B$ and $F$ are at the same vertical displacement from the equilibrium line and are both moving in the same direction (downwards).
The horizontal distance between $B$ and $F$ is exactly one wavelength,$\lambda$.
Therefore,points $B$ and $F$ are in the same phase.

(C) The equation of the wave is $y = A \sin (\omega t - kx)$.
To find the slope of the curve,we differentiate $y$ with respect to $x$ at a constant time $t$:
$\frac{dy}{dx} = A \cos (\omega t - kx) \cdot (-k) = -kA \cos (\omega t - kx)$.
At point $B$,the wave crosses the $x$-axis,meaning $y = 0$. Thus,$\sin (\omega t - kx) = 0$,which implies $\cos (\omega t - kx) = \pm 1$.
For a wave traveling in the $+ve$ $x$-direction,at the point where the wave crosses the axis moving downwards (as seen in the figure),the slope is negative. Specifically,the magnitude of the slope is $|\frac{dy}{dx}| = |-kA \cos (\omega t - kx)| = kA$.
Alternatively,using the relation between particle velocity $v_p$ and wave velocity $v$: $v_p = -v \cdot (\text{slope})$. At point $B$,the particle velocity is at its maximum magnitude,$|v_p| = \omega A$. Since $v = \frac{\omega}{k}$,the magnitude of the slope is $|\frac{dy}{dx}| = \frac{|v_p|}{v} = \frac{\omega A}{\omega / k} = kA$.

(C) From the figure,the amplitude $A = 0.05 \ m$.
The distance covered by $2.5$ wavelengths is $0.25 \ m$.
So,$2.5 \lambda = 0.25 \ m \implies \lambda = 0.1 \ m$.
The wave speed $v = 330 \ m/s$.
The frequency $f = \frac{v}{\lambda} = \frac{330}{0.1} = 3300 \ Hz$.
The wave equation for a wave moving in the positive $x$-direction is given by $y = A \sin 2\pi (ft - \frac{x}{\lambda})$.
Substituting the values: $y = 0.05 \sin 2\pi (3300 \, t - \frac{x}{0.1})$.
$y = 0.05 \sin 2\pi (3300 \, t - 10 \, x)$.

(B) Given the wave function $y = a_0 \sin(\omega t - kx)$.
At time $t = 0$ and position $x = \frac{\pi}{2k}$,the displacement is:
$y = a_0 \sin(\omega(0) - k(\frac{\pi}{2k}))$
$y = a_0 \sin(-\frac{\pi}{2})$
$y = -a_0 \sin(\frac{\pi}{2}) = -a_0$
From the provided graph,the point representing the maximum displacement in the negative direction (i.e.,$-a_0$) is $Q$.

(A) The relationship between angular frequency $(\omega)$ and frequency $(\nu)$ is given by $\omega = 2\pi \nu$.
We know that frequency $(\nu)$ is related to wave number $(\bar \nu)$ by the equation $\nu = c \bar \nu$,where $c$ is the speed of light.
Substituting this into the first equation,we get $\omega = 2\pi c \bar \nu$.
Since $2\pi$ and $c$ are constants,this equation is of the form $y = mx$,which represents a straight line passing through the origin.
Therefore,the graph between angular frequency $(\omega)$ and wave number $(\bar \nu)$ is a straight line passing through the origin.

(B) Given the two wave equations:
$y_1 = a \sin(\omega t)$
$y_2 = b \cos(\omega t)$
We can rewrite the second equation in terms of the sine function using the trigonometric identity $\cos(\theta) = \sin(\theta + \frac{\pi}{2})$:
$y_2 = b \sin(\omega t + \frac{\pi}{2})$
Comparing both equations with the general form $y = A \sin(\omega t + \phi)$,we find the phase of the first wave is $\phi_1 = 0$ and the phase of the second wave is $\phi_2 = \frac{\pi}{2}$.
The phase difference is $\Delta \phi = \phi_2 - \phi_1 = \frac{\pi}{2} - 0 = \frac{\pi}{2}$.

(C) The relationship between path difference $(\Delta x)$ and phase difference $(\Delta \phi)$ is given by the formula: $\Delta \phi = \frac{2\pi}{\lambda} \Delta x$.
For a full cycle,a phase difference of $2\pi$ corresponds to a path difference of $\lambda$.
Therefore,for a phase difference of $\phi$,the path difference $\Delta x$ is calculated as:
$\Delta x = \frac{\lambda}{2\pi} \times \phi$.
Thus,the correct option is $(c)$.

(D) Given equations are $y_1 = a \sin \omega t$ and $y_2 = a \cos \omega t$.
We can rewrite $y_2$ using the trigonometric identity $\cos \theta = \sin(\theta + \frac{\pi}{2})$.
So,$y_2 = a \sin(\omega t + \frac{\pi}{2})$.
Comparing the phases,the phase of $y_1$ is $\omega t$ and the phase of $y_2$ is $\omega t + \frac{\pi}{2}$.
The phase difference is $\phi = (\omega t + \frac{\pi}{2}) - \omega t = \frac{\pi}{2}$.
Since $y_2$ has a phase that is $\frac{\pi}{2}$ greater than $y_1$,$y_1$ lags behind $y_2$ by $\frac{\pi}{2}$.

(D) Given equations are $y_1 = a \sin \omega t$ and $y_2 = a \cos \omega t$.
We can rewrite $y_2$ as $y_2 = a \sin(\omega t + \pi / 2)$.
The phase of the first wave is $\phi_1 = \omega t$.
The phase of the second wave is $\phi_2 = \omega t + \pi / 2$.
The phase difference is $\Delta \phi = \phi_1 - \phi_2 = \omega t - (\omega t + \pi / 2) = -\pi / 2$.
Since the phase difference is negative,the first wave lags behind the second wave by a phase of $\pi / 2$.

(A) The relationship between phase difference $(\Delta \phi)$ and path difference $(\Delta x)$ for a wave is given by the formula:
$\Delta \phi = \frac{2\pi}{\lambda} \Delta x$
Given that the path difference is $\Delta x = x$,we substitute this into the formula:
$\Delta \phi = \frac{2\pi}{\lambda} x$
Therefore,the corresponding phase difference is $\frac{2\pi x}{\lambda}$.

(D) Given: Wave speed $v = 960 \, m/s$.
Number of waves passing in $1 \, minute$ $(60 \, s)$ is $3600$.
Frequency $n = \frac{\text{Total number of waves}}{\text{Total time}} = \frac{3600}{60} = 60 \, Hz$.
The relationship between wave speed,frequency,and wavelength is $v = n \lambda$.
Therefore,wavelength $\lambda = \frac{v}{n} = \frac{960}{60} = 16 \, m$.

(D) The given wave equation is $y = 8 \sin 2\pi (0.1x - 2t) \, cm$.
Comparing this with the standard wave equation $y = a \sin 2\pi (\frac{x}{\lambda} - \frac{t}{T})$,we get the wave number term $k = \frac{2\pi}{\lambda} = 2\pi(0.1) = 0.2\pi$.
Thus,$\lambda = \frac{2\pi}{0.2\pi} = 10 \, cm$.
The phase difference $\Delta \phi$ is given by the formula $\Delta \phi = \frac{2\pi}{\lambda} \times \Delta x$.
Given path difference $\Delta x = 2 \, cm$.
Substituting the values: $\Delta \phi = \frac{2\pi}{10} \times 2 = \frac{4\pi}{10} = 0.4\pi \, radians$.
To convert radians to degrees,multiply by $\frac{180^\circ}{\pi}$:
$\Delta \phi = 0.4\pi \times \frac{180^\circ}{\pi} = 0.4 \times 180^\circ = 72^\circ$.

(A) The given wave equation is $y = 10 \sin \pi (0.01x - 2.00t) \text{ cm}$.
Expanding the equation,we get $y = 10 \sin (0.01\pi x - 2\pi t) \text{ cm}$.
Comparing this with the standard wave equation $y = A \sin (kx - \omega t)$,we identify the amplitude $A = 10 \text{ cm}$ and the angular frequency $\omega = 2\pi \text{ rad/sec}$.
The maximum particle velocity $v_{\text{max}}$ is given by the formula $v_{\text{max}} = A\omega$.
Substituting the values,$v_{\text{max}} = 10 \times 2\pi = 20\pi \text{ cm/sec}$.
Using $\pi \approx 3.14159$,we get $v_{\text{max}} \approx 20 \times 3.14159 = 62.83 \text{ cm/sec}$.
Rounding to the nearest integer,the maximum velocity is $63 \text{ cm/sec}$.

(D) The standard equation for a simple harmonic progressive wave is given by $x = A \cos(\omega t + \phi)$.
Comparing the given equation $x = 0.05 \cos \left( 4\pi t + \frac{\pi}{4} \right)$ with the standard equation,we get the angular frequency $\omega = 4\pi \, rad/s$.
The relationship between frequency $f$ and angular frequency $\omega$ is $\omega = 2\pi f$.
Therefore,$f = \frac{\omega}{2\pi} = \frac{4\pi}{2\pi} = 2 \, Hz$.

(A) The standard equation of a traveling wave is given by $Y = A \sin(\omega t - kx + \phi_0)$.
Comparing the given equation $Y = 7 \sin(7\pi t - 0.04\pi x + \frac{\pi}{3})$ with the standard form,we get:
Angular frequency,$\omega = 7\pi \, rad/s$.
Wave number,$k = 0.04\pi \, rad/m$.
The wave speed $v$ is given by the formula $v = \frac{\omega}{k}$.
Substituting the values,$v = \frac{7\pi}{0.04\pi} = \frac{7}{0.04} = \frac{700}{4} = 175 \, m/s$.