Progressive Waves Questions in English

(C) Comparing the given equation $y = 0.2 \cos \pi (0.04t + 0.02x - \frac{\pi}{6})$ with the standard wave equation $y = a \cos (\omega t + kx - \phi)$,we distribute $\pi$ inside the bracket:
$y = 0.2 \cos (0.04\pi t + 0.02\pi x - \frac{\pi^2}{6})$.
Here,the wave number $k = 0.02\pi$.
We know that $k = \frac{2\pi}{\lambda}$,so $0.02\pi = \frac{2\pi}{\lambda}$.
Solving for wavelength $\lambda$: $\lambda = \frac{2\pi}{0.02\pi} = 100 \ cm$.
The relationship between phase difference $\Delta \phi$ and path difference $\Delta x$ is given by $\Delta \phi = \frac{2\pi}{\lambda} \Delta x$.
Given $\Delta \phi = \frac{\pi}{2}$,we have $\frac{\pi}{2} = \frac{2\pi}{100} \Delta x$.
$\Delta x = \frac{\pi}{2} \times \frac{100}{2\pi} = 25 \ cm$.

(D) The phase of a wave is given by $\phi = (\omega t - kx + \phi_0)$.
The phase difference due to a spatial separation $\Delta x$ is $\Delta \phi_x = \frac{2\pi}{\lambda} \Delta x$.
Given $\Delta x = 15 \ cm$ and $\lambda = 60 \ cm$,the phase change due to distance is $\Delta \phi_x = \frac{2\pi}{60} \times 15 = \frac{\pi}{2}$.
Since the particle is ahead in the direction of wave propagation,its phase is $\phi_1 = \phi_0 - \Delta \phi_x = \frac{\pi}{3} - \frac{\pi}{2} = -\frac{\pi}{6}$.
The phase change due to a time interval $\Delta t = \frac{T}{2}$ is $\Delta \phi_t = \omega \Delta t = \frac{2\pi}{T} \times \frac{T}{2} = \pi$.
The new phase is $\phi_{new} = \phi_1 + \Delta \phi_t = -\frac{\pi}{6} + \pi = \frac{5\pi}{6}$.

(C) The standard equation of a progressive wave is given by $y = a \sin(\omega t - kx)$,where $\omega$ is the angular frequency and $k$ is the wave number.
Comparing the given equation $y = a \sin(628t - 31.4x)$ with the standard form,we get:
$\omega = 628 \, rad/s$
$k = 31.4 \, rad/cm$
The wave velocity $v$ is given by the ratio of the coefficient of $t$ to the coefficient of $x$:
$v = \frac{\omega}{k} = \frac{628}{31.4} = 20 \, cm/sec$.

(B) The standard equation of a progressive wave is $y = a \sin (\omega t - kx + \phi)$.
Comparing the given equation $y = 4 \sin \left( \frac{\pi t}{5} - \frac{\pi x}{9} + \frac{\pi}{6} \right)$ with the standard form:
Amplitude $a = 4$ units (assuming units are in $m$,$a = 4 \, m$).
Angular frequency $\omega = \frac{\pi}{5} \, rad/s$.
Wave number $k = \frac{\pi}{9} \, rad/m$.
Frequency $n = \frac{\omega}{2\pi} = \frac{\pi/5}{2\pi} = 0.1 \, Hz$.
Wave length $\lambda = \frac{2\pi}{k} = \frac{2\pi}{\pi/9} = 18 \, m$.
Wave speed $v = \frac{\omega}{k} = \frac{\pi/5}{\pi/9} = 1.8 \, m/s$.
Thus,option $B$ is correct.

(C) The standard equation of a progressive wave is $y = a \sin (\omega t - kx)$.
Given equation: $y = a \sin \pi (40t - x) = a \sin (40\pi t - \pi x)$.
Comparing this with the standard equation,we get the angular frequency $\omega = 40\pi$ and the wave number $k = \pi$.
The velocity of the wave $v$ is given by the formula $v = \frac{\omega}{k}$.
Substituting the values: $v = \frac{40\pi}{\pi} = 40 \, m/s$.
Alternatively,for a wave equation of the form $y = f(vt - x)$,the velocity is the ratio of the coefficient of $t$ to the coefficient of $x$ inside the function: $v = \frac{40\pi}{\pi} = 40 \, m/s$.

(A) The standard equation of a progressive wave is $y = A \sin (\omega t - kx)$.
Given equation: $y = 0.2 \sin \left( \frac{2\pi t}{0.01} - \frac{2\pi x}{0.3} \right)$.
Comparing this with the standard form,the angular frequency $\omega = \frac{2\pi}{0.01} \text{ rad/s}$ and the wave number $k = \frac{2\pi}{0.3} \text{ rad/m}$.
The velocity of propagation $v$ is given by $v = \frac{\omega}{k}$.
Substituting the values: $v = \frac{2\pi / 0.01}{2\pi / 0.3} = \frac{0.3}{0.01} = 30 \text{ m/s}$.

(D) The given equation is $y = 4\sin 2\pi \left( \frac{t}{0.02} - \frac{x}{100} \right)$.
Comparing this with the standard wave equation $y = a\sin 2\pi \left( \frac{t}{T} - \frac{x}{\lambda} \right)$:
$1$. Amplitude $a = 4 \ cm$.
$2$. Time period $T = 0.02 \ sec$,so frequency $f = \frac{1}{T} = \frac{1}{0.02} = 50 \ \text{Hz}$ (or $\text{cycles/sec}$).
$3$. Wavelength $\lambda = 100 \ cm$.
$4$. Wave velocity $v = f \lambda = 50 \times 100 = 5000 \ \text{cm/sec} = 5 \times 10^3 \ \text{cm/sec}$.
Comparing this with option $(d)$,$50 \times 10^3 \ \text{cm/sec}$ is incorrect. Thus,statement $(d)$ is not true.

(B) The given wave equation is $y = a \sin \pi [\frac{t}{2} - \frac{x}{4}]$.
Comparing this with the standard wave equation $y = a \sin (\omega t - kx)$,we have $\omega = \frac{\pi}{2}$ and $k = \frac{\pi}{4}$.
The wave velocity $v$ is given by $v = \frac{\omega}{k} = \frac{\pi/2}{\pi/4} = 2 \ m/s$.
The distance $d$ traveled by the wave in time $t = 8 \ s$ is $d = v \times t$.
$d = 2 \ m/s \times 8 \ s = 16 \ m$.

(B) The equation of wave $A$ is given by $y_A = A \sin(\omega t)$.
From the graph,wave $B$ starts at its maximum positive value at $t=0$,so its equation is $y_B = A \sin(\omega t + \pi/2)$. This means wave $B$ is ahead of wave $A$ by a phase angle of $\pi/2$.
Wave $C$ starts at its minimum value at $t=0$,so its equation is $y_C = A \sin(\omega t - \pi/2)$. This means wave $C$ lags behind wave $A$ by a phase angle of $\pi/2$.
Wave $D$ starts at $t=0$ with a negative slope,representing a phase shift of $\pi$ relative to $A$,so $y_D = A \sin(\omega t - \pi)$.
Thus,wave $C$ lags behind by a phase angle of $\pi/2$ and wave $B$ is ahead by a phase angle of $\pi/2$.

(D) The particle velocity $v_p$ of a point on a string is given by the relation $v_p = -v \times (\text{slope of the graph at that point})$, where $v$ is the wave speed.
Since the wave is traveling from left to right, the slope determines the direction of the particle's motion.
At point $1$: The slope of the curve is positive. Therefore, $v_p = -v \times (\text{positive}) = \text{negative}$, which means the velocity is downward $(\downarrow)$.
At point $2$: The slope of the curve is negative. Therefore, $v_p = -v \times (\text{negative}) = \text{positive}$, which means the velocity is upward $(\uparrow)$.
At point $3$: The slope of the curve is positive. Therefore, $v_p = -v \times (\text{positive}) = \text{negative}$, which means the velocity is downward $(\downarrow)$.
Thus, the directions are $\downarrow, \uparrow, \downarrow$ respectively.

(A) The standard equation of a progressive wave is $y = A \sin(\omega t + kx + \phi)$.
Comparing the given equation $y = 0.5 \sin(10t + x)$ with the standard equation,we get:
Angular frequency $\omega = 10 \text{ rad/s}$.
Wave number $k = 1 \text{ rad/m}$.
The wave velocity $v$ is given by the formula $v = \frac{\omega}{k}$.
Substituting the values,$v = \frac{10}{1} = 10 \text{ m/s}$.

(C) The standard equation for a progressive wave is $x = A \sin(\omega t + ky)$.
Comparing the given equation $x = 1.2 \sin(314t + 12.56y)$ with the standard form,we get the wave number $k = 12.56 \, \text{rad/m}$.
The wavelength $\lambda$ is given by the formula $\lambda = \frac{2\pi}{k}$.
Substituting the value of $k$ and $\pi \approx 3.14$:
$\lambda = \frac{2 \times 3.14}{12.56} = \frac{6.28}{12.56} = 0.5 \, \text{m}$.
Since the term inside the sine function is $(314t + 12.56y)$,the positive sign between the $t$ and $y$ terms indicates that the wave is traveling in the negative direction of the axis associated with the variable $y$.
Therefore,the wave is traveling in the $-ve \, y$-direction.

(D) The given wave equation is $y = 4 \sin \frac{\pi}{2} (8t - \frac{x}{8})$.
Expanding the equation,we get $y = 4 \sin (4\pi t - \frac{\pi x}{16})$.
Comparing this with the standard wave equation $y = A \sin (\omega t - kx)$,we identify the angular frequency $\omega = 4\pi \, rad/s$ and the wave number $k = \frac{\pi}{16} \, rad/cm$.
The wave velocity $v$ is given by the formula $v = \frac{\omega}{k}$.
Substituting the values,$v = \frac{4\pi}{\pi/16} = 4\pi \times \frac{16}{\pi} = 64 \, cm/s$.
Since the sign between the $t$ and $x$ terms is negative,the wave is traveling in the positive $x$ direction.

(C) The equation of the wave is $y = 2 \sin(2\pi x - 100\pi t + \pi/3)$.
For the particle to be at the mean position,the displacement $y$ must be $0$.
So,$2 \sin(2\pi x - 100\pi t + \pi/3) = 0$.
This implies the argument of the sine function must be an integer multiple of $\pi$,i.e.,$2\pi x - 100\pi t + \pi/3 = n\pi$,where $n$ is an integer.
Given $x = 4 \text{ m}$,we substitute this into the equation:
$2\pi(4) - 100\pi t + \pi/3 = n\pi$
$8\pi - 100\pi t + \pi/3 = n\pi$
$100\pi t = 8\pi + \pi/3 - n\pi = \frac{25\pi}{3} - n\pi$.
To find the first time $t > 0$,we choose $n$ such that $t$ is positive and minimal. For $n = 8$:
$100\pi t = \frac{25\pi}{3} - 8\pi = \frac{\pi}{3}$.
$t = \frac{\pi/3}{100\pi} = \frac{1}{300} \text{ s}$.

(B) $1$. The incident wave travels in the positive $x$-direction,so the reflected wave will travel in the negative $x$-direction. This changes the sign of the $x$ term inside the sine function from $-x/2$ to $+x/2$.
$2$. When a wave reflects from a denser medium,it undergoes a phase change of $\pi$ radians $(180^{\circ})$,which introduces a negative sign to the amplitude.
$3$. The new amplitude is given as $2/3$ of the incident amplitude: $A' = (2/3) \times 0.6 = 0.4$.
$4$. Combining these,the equation of the reflected wave is $y = -0.4 \sin 2\pi (t + x/2)$.

(A) In a plane progressive mechanical wave,particles of the medium perform $S.H.M.$ with the same amplitude but different phases depending on their position.
Statement $A$ is incorrect because particles at different positions along the direction of wave propagation vibrate with different phases.
Statements $B$,$C$,and $D$ are standard properties of a plane progressive wave and are therefore correct.

(B) The incident wave is traveling in the positive $x$-direction,given by $y = 0.6 \sin 2\pi (t - x/2)$.
Upon reflection from a denser medium,the wave travels in the opposite direction (negative $x$-direction),so the term $(t - x/2)$ becomes $(t + x/2)$.
Additionally,reflection from a denser medium introduces a phase change of $\pi$ $(180^{\circ})$,which is equivalent to a negative sign.
The new amplitude is given as $A' = \frac{2}{3} \times 0.6 = 0.4$.
Combining these,the equation of the reflected wave is $y = -0.4 \sin 2\pi (t + x/2)$.

(D) The particle velocity $v_p$ is given by the relation $v_p = -v \times (\text{slope of the graph at that point})$, where $v$ is the wave velocity.
Since the wave is traveling from left to right, $v$ is positive.
At point $1$: The slope of the curve is positive. Therefore, the particle velocity $v_p = -v \times (\text{positive slope})$ is negative, which means the direction is downward $(\downarrow)$.
At point $2$: The slope of the curve is negative. Therefore, the particle velocity $v_p = -v \times (\text{negative slope})$ is positive, which means the direction is upward $(\uparrow)$.
At point $3$: The slope of the curve is positive. Therefore, the particle velocity $v_p = -v \times (\text{positive slope})$ is negative, which means the direction is downward $(\downarrow)$.
Thus, the correct directions are $1 - \downarrow, 2 - \uparrow, 3 - \downarrow$.

(B) From the given figure,we can observe the phase relationship between the waves.
Wave $A$ is represented by $y = a \sin(\omega t)$.
Wave $B$ starts at $t=0$ with a negative slope and reaches its mean position at $t=0$,which corresponds to a phase shift of $+\pi/2$ relative to $A$ (since it is a negative sine wave,$y_B = a \sin(\omega t + \pi/2) = a \cos(\omega t)$).
Wave $C$ reaches its mean position at $t = \pi/2\omega$,which is $\pi/2$ later than wave $A$. Thus,wave $C$ lags behind wave $A$ by a phase angle of $\pi/2$ $(y_C = a \sin(\omega t - \pi/2))$.
Therefore,wave $C$ lags behind by a phase angle of $\pi/2$ and wave $B$ is ahead by a phase angle of $\pi/2$.

(A) In a plane progressive mechanical wave,particles of the medium perform $SHM$ with the same amplitude and frequency,but they vibrate with different phases depending on their position. The wave velocity is determined by the properties of the medium (such as elasticity and density). Therefore,the statement that 'all particles are vibrating in the same phase' is incorrect,as the phase varies with the position $x$ according to the equation $\phi = kx - \omega t$.

(C) travelling wave is represented by a function of the form $f(x \pm vt)$ or a linear combination of harmonic terms $k x \pm \omega t$.
Option $A$ represents a standing wave because the spatial and temporal parts are separated.
Option $B$ is not a valid wave function as it does not satisfy the wave equation $\frac{\partial^2 y}{\partial x^2} = \frac{1}{v^2} \frac{\partial^2 y}{\partial t^2}$.
Option $C$ can be written as $y = R \sin(5x - 0.5t + \phi)$,which is a standard form of a travelling wave.
Option $D$ represents a superposition of two standing waves.
Therefore,only option $C$ represents a travelling wave.

(B) The standard equation of a progressive wave is $y = A \sin (\omega t - kx)$.
Given equation is $y = 0.02 \sin \left( \frac{2\pi t}{0.01} - \frac{2\pi x}{0.30} \right)$.
Comparing the two,we get angular frequency $\omega = \frac{2\pi}{0.01} \text{ rad/s}$ and wave number $k = \frac{2\pi}{0.30} \text{ rad/m}$.
The velocity of propagation $v$ is given by $v = \frac{\omega}{k}$.
Substituting the values: $v = \frac{2\pi / 0.01}{2\pi / 0.30} = \frac{0.30}{0.01} = 30 \text{ m/s}$.

(A) The standard equation for a progressive wave is $y(x, t) = a \sin (\omega t + kx + \phi)$.
Comparing this with $y(x, t) = 3.0 \sin (36t + 0.018x + \pi/4)$:
$(a)$ Since the equation is of the form $f(ax + bt)$,it represents a travelling wave. Because the sign between $t$ and $x$ is positive,it propagates from right to left (negative $x$-direction). The speed $v = \omega/k = 36 / 0.018 = 2000 \, cm/s = 20 \, m/s$.
$(b)$ Amplitude $a = 3.0 \, cm$. Frequency $\nu = \omega / (2\pi) = 36 / (2 \times 3.1416) \approx 5.73 \, Hz$.
$(c)$ The initial phase at the origin $(x=0, t=0)$ is $\phi = \pi/4 \, rad$.
$(d)$ The distance between two successive crests is the wavelength $\lambda = 2\pi / k = 2 \times 3.1416 / 0.018 \approx 349 \, cm = 3.49 \, m$.

(N/A) $1$. Progressive Wave: $A$ wave that travels continuously in a medium in a specific direction without any change in its form is called a progressive or traveling wave.
$2$. Wave Equation: $A$ mathematical expression that describes the displacement of particles of the medium as a function of both position $(x)$ and time $(t)$ is called a wave equation.
$3$. Wave Equation for a Progressive Harmonic Transverse Wave: The displacement $y$ of a particle at position $x$ and time $t$ is given by $y(x, t) = A \sin(kx - \omega t + \phi)$,where $A$ is the amplitude,$k = \frac{2\pi}{\lambda}$ is the angular wave number,$\omega = 2\pi f$ is the angular frequency,and $\phi$ is the initial phase constant.
$4$. Graphical Explanation:
- $y-x$ graph (at constant $t$): Represents the snapshot of the wave at a specific instant,showing the spatial variation of displacement.
- $y-t$ graph (at constant $x$): Represents the oscillation of a single particle at a fixed position over time.

(B) progressive wave (or traveling wave) is a wave that moves continuously in a medium without change in its shape.
It is characterized by the continuous transfer of energy and momentum from one point to another in the medium.
Examples include sound waves,water waves,and light waves.
In a progressive wave,each particle of the medium executes simple harmonic motion about its mean position with the same amplitude and frequency,but with a phase difference relative to its neighbors.

(N/A) Wave speed: The distance covered by a wave in unit time is called wave speed. Its $SI$ unit is $m/s$.
To find the speed of a progressive wave,we consider the motion of a point on the wave (e.g.,a crest). As shown in the figure,a point on the wave maintains its displacement relative to the wave pattern as it moves.
If the wave covers a displacement $\Delta x$ in time $\Delta t$,the wave speed is given by:
$v = \frac{\Delta x}{\Delta t}$
For a progressive wave represented by $y(x, t) = A \sin(kx - \omega t + \phi)$,the phase of the wave remains constant for a specific point on the wave pattern:
$kx - \omega t = \text{constant}$
As the wave moves,for a point to maintain the same phase at a later time $(t + \Delta t)$,its position must change to $(x + \Delta x)$:
$k(x + \Delta x) - \omega(t + \Delta t) = kx - \omega t$
$kx + k\Delta x - \omega t - \omega\Delta t = kx - \omega t$
$k\Delta x = \omega\Delta t$
$\frac{\Delta x}{\Delta t} = \frac{\omega}{k}$
Since $v = \frac{\Delta x}{\Delta t}$,we get $v = \frac{\omega}{k}$.
We know that $\omega = \frac{2\pi}{T}$ and $k = \frac{2\pi}{\lambda}$.
Substituting these values:
$v = \frac{(2\pi / T)}{(2\pi / \lambda)} = \frac{\lambda}{T}$.

(A) The speed of a wave $(v)$ is given by the product of its wavelength $(\lambda)$ and frequency $(\nu)$: $v = \nu \lambda$.
Distance travelled by the wave in time $t$ is given by: $\text{Distance} = \text{speed} \times \text{time}$.
Substituting the value of speed: $\text{Distance} = (\nu \lambda) \times t = \lambda \nu t$.

(B) The standard equation for a progressive wave is given by $y = A \sin (kx - \omega t)$.
Comparing the given equation $y = 5 \sin (0.01x - 2t)$ with the standard equation,we identify the wave number $k$ and the angular frequency $\omega$:
$k = 0.01 \ rad/cm$
$\omega = 2 \ rad/s$
The speed of propagation $v$ of a wave is defined by the ratio of angular frequency to the wave number:
$v = \frac{\omega}{k}$
Substituting the values:
$v = \frac{2}{0.01} = 200 \ cm/s$
Thus,the speed of propagation is $200 \ cm/s$.

(N/A) Comparing the given equation $y = 5 \sin(100\pi t + 0.4\pi x)$ with the standard wave equation $y = a \sin(\omega t + kx + \phi)$:
$(a)$ Amplitude $a = 5 \ m$.
$(b)$ Wave number $k = 0.4\pi$. Since $k = \frac{2\pi}{\lambda}$,we have $0.4\pi = \frac{2\pi}{\lambda} \Rightarrow \lambda = \frac{2}{0.4} = 5 \ m$.
$(c)$ Angular frequency $\omega = 100\pi$. Since $\omega = 2\pi f$,we have $100\pi = 2\pi f \Rightarrow f = 50 \ Hz$.
$(d)$ Wave velocity $v = \frac{\omega}{k} = \frac{100\pi}{0.4\pi} = 250 \ m/s$.
$(e)$ Particle velocity amplitude $v_{\max} = a\omega = 5 \times 100\pi = 500\pi \ m/s$.

(B) plane progressive wave traveling in the positive $x$-direction is mathematically represented by the equation $y = A \sin(\omega t - kx)$.
Here,$A$ is the amplitude,$\omega$ is the angular frequency,$t$ is time,$k$ is the wave number,and $x$ is the position.
The negative sign between $\omega t$ and $kx$ indicates propagation in the positive $x$-direction,while a positive sign would indicate propagation in the negative $x$-direction.

(C) The given wave equation is $y = A \cos 240(t - x/12)$.
Expanding this,we get $y = A \cos(240t - 240x/12) = A \cos(240t - 20x)$.
Comparing this with the standard wave equation $y = A \cos(\omega t - kx)$,we identify the wave number $k = 20 \, rad/m$.
The phase difference $\Delta \phi$ between two points separated by a distance $\Delta x$ is given by the formula $\Delta \phi = k \cdot \Delta x$.
Given $\Delta x = 0.5 \, m$ and $k = 20 \, rad/m$,we calculate:
$\Delta \phi = 20 \times 0.5 = 10 \, rad$.
Thus,the phase difference is $10 \, rad$.

(D) Two points are in the same state of vibration (same phase) if they have the same displacement and are moving in the same direction. This occurs at points separated by an integer multiple of the wavelength $\lambda$.
Looking at the diagram,point $B$ is on the downward slope of the first crest. Point $E$ is on the downward slope of the third crest.
The distance between $B$ and $E$ is exactly one full wavelength $\lambda$.
Therefore,points $B$ and $E$ are in the same phase.

(B) The general equation for a progressive wave is $Y = a \sin (\omega t - kx)$.
Comparing this with the given equation $Y = a \sin (2 \pi n t - \frac{2 \pi x}{5})$:
Angular frequency $\omega = 2 \pi n$.
Wave number $k = \frac{2 \pi}{5}$.
Wave velocity $v = \frac{\omega}{k} = \frac{2 \pi n}{2 \pi / 5} = 5n$.
Particle velocity $v_p = \frac{dY}{dt} = a \omega \cos (\omega t - kx)$.
Maximum particle velocity $v_m = a \omega = a (2 \pi n) = 2 \pi n a$.
The ratio of maximum particle velocity to wave velocity is $\frac{v_m}{v} = \frac{2 \pi n a}{5n} = \frac{2 \pi a}{5}$.

(C) The relationship between phase difference $\Delta \phi$ and time interval $\Delta t$ is given by:
$\Delta \phi = \omega \Delta t = (2\pi f) \Delta t$
Given frequency $f = 50 \,Hz$ and time interval $\Delta t = 0.01 \,s$.
Substituting the values:
$\Delta \phi = 2 \times \pi \times 50 \times 0.01$
$\Delta \phi = 100 \pi \times 0.01$
$\Delta \phi = \pi \,rad$.

(C) The given equation of the progressive wave is $y = 12 \sin(5t - 4x)$.
Comparing this with the standard wave equation $y = A \sin(\omega t - kx)$,we get the wave number $k = 4 \ rad/cm$.
The phase difference $\Delta \phi$ between two points separated by a distance $\Delta x$ is given by the relation $\Delta \phi = k \cdot \Delta x$.
We are given the phase difference $\Delta \phi = 90^{\circ} = \frac{\pi}{2} \ rad$.
Substituting the values,we have $\frac{\pi}{2} = 4 \cdot \Delta x$.
Therefore,$\Delta x = \frac{\pi}{2 \times 4} = \frac{\pi}{8} \ cm$.

(B) Given:
Frequency $f = 400 \ Hz$
Velocity $v = 336 \ m/s$
Phase difference $\Delta \phi = 60^{\circ} = \frac{\pi}{3} \text{ radians}$.
First,calculate the wavelength $\lambda$ using the formula $v = f \lambda$:
$\lambda = \frac{v}{f} = \frac{336}{400} = 0.84 \ m$.
The relationship between phase difference $\Delta \phi$ and path difference $\Delta x$ is given by:
$\Delta \phi = \frac{2\pi}{\lambda} \Delta x$.
Rearranging for $\Delta x$:
$\Delta x = \frac{\Delta \phi \cdot \lambda}{2\pi}$.
Substituting the values:
$\Delta x = \frac{(\pi/3) \cdot 0.84}{2\pi} = \frac{0.84}{6} = 0.14 \ m$.
Therefore,the two points are $0.14 \ m$ apart.

(A) The standard equation of a progressive wave is $Y = A \sin \left( \omega t - kx + \phi \right)$.
Comparing the given equation $Y = 3 \sin \left[ \pi \left( \frac{t}{3} - \frac{x}{5} \right) + \frac{\pi}{4} \right]$ with the standard form:
$Y = 3 \sin \left( \frac{\pi t}{3} - \frac{\pi x}{5} + \frac{\pi}{4} \right)$.
Here,the amplitude $A = 3 \ m$ (since $x$ and $y$ are in meters).
The angular frequency $\omega = \frac{\pi}{3} \ rad/s$.
The wave number $k = \frac{\pi}{5} \ rad/m$.
$1$. Wavelength $\lambda = \frac{2\pi}{k} = \frac{2\pi}{\pi/5} = 10 \ m$.
$2$. Velocity $v = \frac{\omega}{k} = \frac{\pi/3}{\pi/5} = \frac{5}{3} \approx 1.67 \ m/s$.
$3$. Frequency $f = \frac{\omega}{2\pi} = \frac{\pi/3}{2\pi} = \frac{1}{6} \approx 0.167 \ Hz$.
Comparing these results with the options,option $A$ is correct.

(B) The given equation of the progressive wave is $y = 12 \sin (5t - 4x)$.
The standard form of a progressive wave is $y = A \sin (\omega t - kx)$,where $k = \frac{2\pi}{\lambda}$.
Comparing the given equation with the standard form,we get the wave number $k = 4$.
The relationship between phase difference $(\Delta \phi)$ and path difference $(\Delta x)$ is given by $\Delta \phi = k \cdot \Delta x$.
Given phase difference $\Delta \phi = 90^{\circ} = \frac{\pi}{2}$ radians.
Substituting the values: $\frac{\pi}{2} = 4 \cdot \Delta x$.
Therefore,$\Delta x = \frac{\pi}{2 \cdot 4} = \frac{\pi}{8}$.

(C) simple harmonic progressive wave is a wave that travels through a medium without any permanent displacement of the medium particles.
Mathematically,a progressive wave is represented by a function of the form $y = f(vt \pm x)$ or $y = a \sin(\omega t \pm kx + \phi)$.
In this expression,$y$ is the displacement,$a$ is the amplitude,$\omega$ is the angular frequency,$t$ is time,$k$ is the wave number,and $x$ is the position.
Option $A$ represents simple harmonic motion at a fixed position.
Option $B$ represents a standing wave.
Option $C$ represents a simple harmonic progressive wave because it contains both space $(x)$ and time $(t)$ variables in the phase argument.
Option $D$ represents a stationary spatial variation.

(B) The given equation of a progressive wave is $y=3 \sin \pi\left(\frac{t}{2}-\frac{x}{4}\right)$.
Comparing this with the standard wave equation $y=A \sin 2\pi \left(\frac{t}{T}-\frac{x}{\lambda}\right)$,we rewrite the given equation as $y=3 \sin 2\pi \left(\frac{t}{4}-\frac{x}{8}\right)$.
From this,we identify the time period $T=4 \,s$ and the wavelength $\lambda=8 \,m$.
The wave velocity $v$ is given by $v = \frac{\lambda}{T} = \frac{8}{4} = 2 \,m/s$.
The distance travelled by the wave in time $t=5 \,s$ is $s = v \times t = 2 \times 5 = 10 \,m$.

(B) progressive wave is represented by a function of the form $y = f(ax \pm bt)$.
Option $(A)$ represents a standing wave because it is a product of spatial and temporal functions.
Option $(B)$ is not a periodic function.
Option $(C)$ can be rewritten using the identity $A \sin \theta + B \cos \theta = R \sin(\theta + \phi)$,where $R = \sqrt{3^2 + 4^2} = 5$. Thus,$y = 5 \sin(5x - 0.5t + \phi)$,which is a standard progressive wave equation.
Option $(D)$ is a superposition of two different waves with different frequencies and wave numbers,not a single progressive wave.
Therefore,only $(C)$ represents a single progressive wave.

(C) The standard equation of a progressive wave is given by $x = A \cos(\omega t - ky)$.
Comparing the given equation $x = 0.4 \cos \left(8t - \frac{y}{2}\right)$ with the standard equation,we get:
Angular frequency $\omega = 8 \ rad \cdot s^{-1}$
Wave number $k = \frac{1}{2} \ m^{-1}$
The speed of the wave $v$ is given by the formula $v = \frac{\omega}{k}$.
Substituting the values,$v = \frac{8}{1/2} = 8 \times 2 = 16 \ m \cdot s^{-1}$.
Therefore,the speed of the wave is $16 \ m \cdot s^{-1}$.

(C) Given: Frequency $f = 500 \,Hz$, Velocity $v = 360 \,ms^{-1}$, Phase difference $\Delta\phi = 60^{\circ}$.
First, calculate the wavelength $\lambda$ using the formula $\lambda = \frac{v}{f}$.
$\lambda = \frac{360}{500} = 0.72 \,m$.
The relationship between path difference $\Delta x$ and phase difference $\Delta\phi$ is given by $\Delta\phi = \frac{2\pi}{\lambda} \Delta x$.
Converting $60^{\circ}$ to radians: $60^{\circ} = \frac{\pi}{3} \,rad$.
Substituting the values: $\frac{\pi}{3} = \frac{2\pi}{\lambda} \Delta x$.
$\Delta x = \frac{\lambda}{6} = \frac{0.72}{6} = 0.12 \,m$.
Therefore, the distance between the two points is $0.12 \,m$.

(A) The general equation for a wave travelling in the positive $x$-direction is $y = A \sin(\omega t - kx + \phi)$.
At $t=0$,the equation is $y = 3 \sin(-2\pi x / 3)$,which implies $k = 2\pi / 3$.
Given the wave velocity $v = 4 \ m/s$,we use the relation $v = \omega / k$.
Thus,$\omega = v \cdot k = 4 \times (2\pi / 3) = 8\pi / 3$.
The general wave equation is $y = 3 \sin \left(\frac{8\pi}{3}t - \frac{2\pi}{3}x\right)$.
At $t=4 \ s$,substituting the value of $t$:
$y = 3 \sin \left(\frac{8\pi}{3}(4) - \frac{2\pi}{3}x\right)$
$y = 3 \sin \left(\frac{32\pi}{3} - \frac{2\pi}{3}x\right)$
Factoring out $2\pi$:
$y = 3 \sin 2\pi \left(\frac{16}{3} - \frac{x}{3}\right) = 3 \sin 2\pi \left(-\frac{x-16}{3}\right)$.

(A) The given wave equation is $y = A \sin \left[ \frac{2 \pi}{\lambda} (vt - x) \right]$.
Comparing this with the standard form $y = A \sin (\omega t - kx)$,we have $\omega = \frac{2 \pi v}{\lambda}$ and $k = \frac{2 \pi}{\lambda}$.
The particle velocity $v_p$ is given by $\frac{\partial y}{\partial t} = A \omega \cos \left[ \frac{2 \pi}{\lambda} (vt - x) \right]$.
The maximum particle velocity is $(v_p)_{\max} = A \omega = A \left( \frac{2 \pi v}{\lambda} \right)$.
According to the problem,$(v_p)_{\max} = 3v$.
Therefore,$A \left( \frac{2 \pi v}{\lambda} \right) = 3v$.
Canceling $v$ from both sides,we get $\frac{2 \pi A}{\lambda} = 3$.
Solving for $\lambda$,we find $\lambda = \frac{2 \pi A}{3}$.

(D) The standard equation of a plane progressive wave is $y = A \cos(\omega t - kx)$.
Given equation: $y = 5 \cos(200\pi t - \frac{\pi x}{150})$.
Comparing the given equation with the standard form,we identify the angular frequency $\omega = 200\pi \text{ rad/s}$ and the wave number $k = \frac{\pi}{150} \text{ rad/cm}$.
The wave velocity $v$ is given by the formula $v = \frac{\omega}{k}$.
Substituting the values: $v = \frac{200\pi}{\pi/150} = 200 \times 150 = 30000 \text{ cm/s}$.
To convert the velocity into m/s,we divide by $100$: $v = \frac{30000}{100} = 300 \text{ m/s}$.