Wave Equation and Characteristics of Waves Questions in English

(A) The motion of a point having a constant phase is described by the phase velocity.
For a wave represented by the equation $y(x, t) = A \sin(kx - \omega t + \phi)$,the phase is given by $\Phi = kx - \omega t + \phi$.
For a constant phase,$\frac{d\Phi}{dt} = 0$.
Therefore,$k \frac{dx}{dt} - \omega = 0$,which implies $\frac{dx}{dt} = \frac{\omega}{k} = v_p$.
Here,$v_p$ is the phase velocity,which represents the speed at which the wave pattern (or a point of constant phase) propagates through space.

(N/A) The wave speed $(v)$ is defined as the distance traveled by the wave per unit time.
For a periodic wave,the distance traveled in one time period $(T)$ is equal to one wavelength $(\lambda)$.
Therefore,the wave speed is given by the ratio of wavelength to time period:
$v = \frac{\lambda}{T}$
where:
$v$ = wave speed
$\lambda$ = wavelength
$T$ = time period

(N/A) The wave speed $(v)$ of a traveling wave is defined as the product of its frequency $(f)$ and wavelength $(\lambda)$,i.e.,$v = f \lambda$.
We know that angular frequency is given by $\omega = 2\pi f$,which implies $f = \frac{\omega}{2\pi}$.
We also know that the angular wave number is given by $k = \frac{2\pi}{\lambda}$,which implies $\lambda = \frac{2\pi}{k}$.
Substituting these expressions into the wave speed formula:
$v = \left( \frac{\omega}{2\pi} \right) \times \left( \frac{2\pi}{k} \right)$.
Simplifying the expression,we get the relation:
$v = \frac{\omega}{k}$.

(N/A) The wave speed $(v)$ is defined as the distance traveled by a wave per unit time.
For a periodic wave,the distance traveled in one time period $(T)$ is equal to one wavelength $(\lambda)$.
Therefore,the wave speed is given by the ratio of wavelength to the time period: $v = \frac{\lambda}{T}$.
Since the frequency $(f)$ is the reciprocal of the time period $(f = \frac{1}{T})$,we can substitute this into the equation.
Thus,the equation for wave speed in terms of wavelength and frequency is: $v = f \lambda$.

(A) When a wave is reflected from a free end (or open boundary), the particles at the boundary are free to move.
In this case, the reflected wave is in phase with the incident wave.
Therefore, the phase change is $0$ radians.

(C) When a wave travels from one medium to another,the frequency of the wave depends only on the source of the wave.
Since the source remains the same,the frequency $(f)$ remains constant.
Consequently,the angular frequency $(\omega = 2\pi f)$ and the periodic time $(T = 1/f)$ also remain constant.
However,the velocity $(v)$ and wavelength $(\lambda)$ change depending on the refractive index or properties of the new medium.

(A) In a wave,a crest represents the point of maximum positive displacement,and the subsequent trough represents the point of maximum negative displacement.
These two points are separated by half a wavelength,which corresponds to a distance of $\frac{\lambda}{2}$.
Since a full wavelength $\lambda$ corresponds to a phase change of $2\pi \ rad$,a distance of $\frac{\lambda}{2}$ corresponds to a phase difference of $\frac{2\pi}{2} = \pi \ rad$.

(N/A) Consider a long string held horizontally with one end fixed. If we move the free end of the string up and down in a periodic manner,a wave is generated propagating in the $+x$-direction as shown in figure $(a)$.
The curves represent the displacement of the string at $t=0$ and at $t=\Delta t$ respectively as the sinusoidal wave propagates in the $+x$-direction.
In figure $(b)$,the curve represents the time variation of the displacement at $x=0$ for the same wave.
The displacement of the wave in the $+x$-direction occurs in the $y$-direction,so its equation is given by:
$y(x, t) = a \sin(kx - \omega t)$
where $a$ is the amplitude of the wave,$\omega = 2\pi\nu$ is the angular frequency,and $k = \frac{2\pi}{\lambda}$ is the wave vector.
According to this equation,the displacement of the string particles (in the $y$-direction) is at right angles to the direction of propagation of the wave,making it a transverse wave. Since the displacement is restricted to the $y$-direction,it is called a $y$-polarized wave.
Definition: $A$ wave is called linearly polarized if the displacement of the particles of the medium is confined to a single straight line perpendicular to the direction of wave propagation. Since each point on the string moves along a straight line,this is a linearly polarized wave. Because the string remains confined to the $xy$-plane,it is also called a plane-polarized wave.

(A) $(i)$ True. In longitudinal waves,particles oscillate parallel or antiparallel to the direction of wave propagation,so the angle is $0^{\circ}$ or $180^{\circ}$.
$(ii)$ False. In transverse waves,particles oscillate perpendicular to the direction of wave propagation,so the angle is $\frac{\pi}{2} \text{ rad}$ $(90^{\circ})$.
$(iii)$ False. The distance between two consecutive particles having the same phase is called the wavelength. If the phase difference is not specified as $2\pi \text{ rad}$,the statement is incomplete.
$(iv)$ False. When a wave is reflected from a denser medium,its phase increases by $\pi \text{ rad}$. Reflection from a rarer medium involves no phase change.

(N/A) The wave equation $y = 100 \cos (100\pi t + 0.5x)$ represents a wave propagating along the $-x$ direction because the coefficients of $t$ and $x$ have the same sign.
$(b)$ The stationary wave is $y = 5 \cos (4x) \sin (20t)$ because it is in the form $y = A \cos (kx) \sin (\omega t)$,which represents a standing wave.
$(c)$ The equation $y = 10 \cos (252\pi t) \cos (250\pi t)$ represents beats,as it is the product of two cosine functions with slightly different frequencies,which is the characteristic form of the beat phenomenon.
$(d)$ The equation $y = 4 \sin (5x - t/2) + 3 \cos (5x - t/2)$ represents a travelling wave along the $+x$ direction because the coefficients of $t$ and $x$ have opposite signs.

(A) The given wave equation is $y = 5 \cos(20\pi t - 0.016\pi x + 7\pi)$.
Comparing this with the standard form $y = a \cos(\omega t - kx + \phi)$,we get:
$k = 0.016\pi \ rad/cm$ and $\omega = 20\pi \ rad/s$.
$(a)$ For $\Delta x = 4 \ m = 400 \ cm$,the phase difference is $\Delta \phi = k \Delta x = 0.016\pi \times 400 = 6.4\pi \ rad$.
$(b)$ For $\Delta x = 0.5 \ m = 50 \ cm$,the phase difference is $\Delta \phi = k \Delta x = 0.016\pi \times 50 = 0.8\pi \ rad$.
$(c)$ For $\Delta x = \frac{\lambda}{2}$,the phase difference is $\Delta \phi = \frac{2\pi}{\lambda} \times \frac{\lambda}{2} = \pi \ rad$.
$(d)$ For $\Delta x = \frac{3\lambda}{4}$,the phase difference is $\Delta \phi = \frac{2\pi}{\lambda} \times \frac{3\lambda}{4} = 1.5\pi \ rad$.
$(e)$ For a particle at a fixed position,the phase difference between two times $t_1$ and $t_2$ is $\Delta \phi = \omega(t_2 - t_1)$.
Given $t_1 = T = \frac{2\pi}{\omega} = \frac{2\pi}{20\pi} = 0.1 \ s$ and $t_2 = 5 \ s$,
$\Delta \phi = 20\pi(5 - 0.1) = 20\pi(4.9) = 98\pi \ rad$.

(D) The distance between two consecutive crests is the wavelength $\lambda$. However,if the wave is not necessarily consecutive,the distance between two crests is $n_2 \lambda$,where $n_2$ is an integer.
Given $n_2 \lambda = 5 \, m \implies \lambda = \frac{5}{n_2}$.
The distance between a crest and a trough is given by $(2n_1 + 1) \frac{\lambda}{2}$,where $n_1$ is an integer.
Given $(2n_1 + 1) \frac{\lambda}{2} = 1.5 \, m \implies (2n_1 + 1) \lambda = 3 \, m \implies \lambda = \frac{3}{2n_1 + 1}$.
Equating the two expressions for $\lambda$: $\frac{5}{n_2} = \frac{3}{2n_1 + 1} \implies 5(2n_1 + 1) = 3n_2 \implies 10n_1 + 5 = 3n_2$.
For $n_1 = 1$,$3n_2 = 15 \implies n_2 = 5$,so $\lambda = \frac{5}{5} = 1 \, m$.
For $n_1 = 4$,$3n_2 = 45 \implies n_2 = 15$,so $\lambda = \frac{5}{15} = \frac{1}{3} \, m$.
For $n_1 = 7$,$3n_2 = 75 \implies n_2 = 25$,so $\lambda = \frac{5}{25} = \frac{1}{5} \, m$.
Thus,the possible wavelengths are $1, \frac{1}{3}, \frac{1}{5}, \dots$.

(A) The general equation for a travelling wave is $Y(x, t) = A \sin(kx - \omega t)$.
Given frequency $f = 245 \,Hz$,speed $v = 300 \,ms^{-1}$,and total path length (peak-to-peak) $= 6 \,cm$.
Amplitude $A = \frac{6 \,cm}{2} = 3 \,cm = 0.03 \,m$.
Angular frequency $\omega = 2 \pi f = 2 \times 3.14 \times 245 \approx 1538.6 \,rad/s \approx 1.5 \times 10^{3} \,rad/s$.
Wave number $k = \frac{\omega}{v} = \frac{1538.6}{300} \approx 5.12 \,m^{-1} \approx 5.1 \,m^{-1}$.
Substituting these values,the equation is $Y(x, t) = 0.03 \sin(5.1 x - 1.5 \times 10^{3} t)$.

(A) travelling wave is represented by a function of the form $y = f(x \pm vt)$.
For a wave to be a travelling wave,the variables $x$ and $t$ must appear in the combination $(x \pm vt)$ within the function.
In option $A$,$y = A \sin (15 x - 2 t)$,which can be rewritten as $y = A \sin [15(x - \frac{2}{15}t)]$. This matches the form $f(x - vt)$ where $v = \frac{2}{15}$.
Options $B$,$C$,and $D$ do not represent a travelling wave because they do not satisfy the condition of being a function of $(x \pm vt)$.

(A) The general equation for a wave traveling in the positive $x$-direction without changing its shape is given by $y(x, t) = f(x - vt)$,where $v$ is the wave velocity.
At $t=0$,the equation is $y = f(x) = \frac{1}{1+x^2}$.
At $t=1 \text{ s}$,the equation is $y = f(x - v(1)) = \frac{1}{1+(x-v)^2}$.
We are given that at $t=1 \text{ s}$,the equation is $y = \frac{1}{1+(x-2)^2}$.
Comparing the two expressions for $t=1 \text{ s}$,we get $x - v = x - 2$.
Therefore,$v = 2 \text{ m/s}$.

(B) The wave equation is given by $y = A \sin 2 \pi (nt - \frac{x}{\lambda})$.
Here,the amplitude $A = 10 \, cm$.
The maximum particle velocity is given by $V_{p, \text{max}} = \omega A = (2 \pi n) A$.
The wave velocity is given by $V_{\text{wave}} = n \lambda = \frac{\omega}{k} = \frac{2 \pi n}{2 \pi / \lambda} = n \lambda$.
According to the problem,$V_{p, \text{max}} = 4 V_{\text{wave}}$.
Substituting the expressions: $(2 \pi n) A = 4 (n \lambda)$.
Simplifying,we get $2 \pi A = 4 \lambda$.
Therefore,$\lambda = \frac{2 \pi A}{4}$.
Substituting $A = 10 \, cm$: $\lambda = \frac{2 \pi (10)}{4} = \frac{20 \pi}{4} = 5 \pi \, cm$.

(A) The given wave equation is $y = A \sin(\omega t - kx)$,where amplitude $A = 2 \ cm$.
The maximum particle velocity is given by $v_{p,max} = A\omega$.
The wave velocity (phase velocity) is given by $v_w = \frac{\omega}{k}$.
According to the problem,the wave velocity is equal to the maximum particle velocity:
$v_w = v_{p,max}$
$\frac{\omega}{k} = A\omega$
Canceling $\omega$ from both sides,we get:
$\frac{1}{k} = A$
Since $k = \frac{2\pi}{\lambda}$,we substitute this into the equation:
$\frac{\lambda}{2\pi} = A$
$\lambda = 2\pi A$
Given $A = 2 \ cm$,we have:
$\lambda = 2\pi(2) = 4\pi \ cm$.

(C) The standard form of a traveling wave equation is $y = A \sin(\omega t - kx)$.
Given the equation: $y = 0.5 \sin \left( \frac{2 \pi}{\lambda} (400 t - x) \right) = 0.5 \sin \left( \frac{800 \pi}{\lambda} t - \frac{2 \pi}{\lambda} x \right)$.
Comparing this with the standard form,we identify the angular frequency $\omega$ and the wave number $k$:
$\omega = \frac{800 \pi}{\lambda}$
$k = \frac{2 \pi}{\lambda}$
The velocity of the wave $v$ is given by the ratio of angular frequency to wave number:
$v = \frac{\omega}{k} = \frac{800 \pi / \lambda}{2 \pi / \lambda} = \frac{800 \pi}{2 \pi} = 400 \, m/s$.
Therefore,the velocity of the wave is $400 \, m/s$.

(B) The general equation for a travelling wave is $y = a \sin(\omega t - kx + \phi_0)$.
At $t=0$ and $x=0$,$y=0$,which implies $\phi_0 = 0$.
Thus,the wave equation is $y = a \sin(\omega t - kx)$.
Given frequency $f = 500 \,Hz$ and speed $v = 100 \,m/s$,we calculate:
$\omega = 2\pi f = 2\pi \times 500 = 1000\pi \,rad/s$.
$k = \frac{\omega}{v} = \frac{1000\pi}{100} = 10\pi \,m^{-1}$.
So,$y = a \sin(1000\pi t - 10\pi x)$.
At $t=0$ and $x=0.25 \,m$,$y = 0.02 \,m$:
$0.02 = a \sin(0 - 10\pi \times 0.25) = a \sin(-2.5\pi) = a \sin(-2\pi - 0.5\pi) = -a \sin(0.5\pi) = -a(1)$.
Thus,$a = -0.02 \,m$.
The wave equation is $y = -0.02 \sin(1000\pi t - 10\pi x)$.
At $t = 5 \times 10^{-4} \,s$ and $x = 0.2 \,m$:
$y = -0.02 \sin(1000\pi \times 5 \times 10^{-4} - 10\pi \times 0.2) = -0.02 \sin(0.5\pi - 2\pi) = -0.02 \sin(-1.5\pi) = -0.02 \sin(0.5\pi) = -0.02 \times 1 = -0.02 \,m$.

(C) The general equation of a traveling wave is given by $y = f(ax \pm bt)$.
Comparing the given equation $y = \frac{5}{(4x + 6t)^2}$ with the standard form,we identify the coefficients of $x$ and $t$.
The velocity $v$ of a wave pulse is given by the ratio of the coefficient of $t$ to the coefficient of $x$,i.e.,$v = \left| \frac{\text{coefficient of } t}{\text{coefficient of } x} \right|$.
Here,the coefficient of $t$ is $6$ and the coefficient of $x$ is $4$.
Therefore,$v = \frac{6}{4} = 1.5 \, m/s$.
The negative sign in the direction indicates the pulse is traveling in the negative $x$-direction.

(A) The given wave equation is $x = 4 \cos \left(8t - \frac{y}{2}\right)$.
Comparing this with the standard wave equation form $x = A \cos (\omega t - ky)$,we get the angular frequency $\omega = 8 \text{ rad/s}$.
We know that the relationship between angular frequency $\omega$ and frequency $f$ is given by $\omega = 2\pi f$.
Substituting the value of $\omega$:
$8 = 2\pi f$
$f = \frac{8}{2\pi} = \frac{4}{\pi} \text{ s}^{-1}$.
Thus,the frequency of the wave is $\frac{4}{\pi} \text{ s}^{-1}$.

(D) For a traveling wave represented by $y = A \sin(kx - \omega t)$,the instantaneous velocity of a particle $u$ is given by the partial derivative of displacement with respect to time:
$u = \frac{\partial y}{\partial t} = -A\omega \cos(kx - \omega t)$
The slope of the waveform is given by the partial derivative of displacement with respect to position:
$\text{Slope} = \frac{\partial y}{\partial x} = Ak \cos(kx - \omega t)$
The wave velocity $v$ is given by $v = \frac{\omega}{k}$,which implies $\omega = vk$.
Substituting $\omega = vk$ into the expression for $u$:
$u = -A(vk) \cos(kx - \omega t)$
$u = -v \times [Ak \cos(kx - \omega t)]$
$u = -v \times (\text{slope of waveform})$
Taking the magnitude on both sides:
$|u| = |v| \times |\text{slope of waveform}|$
Thus,option $D$ is correct.

(A) The displacement of a particle in a simple harmonic wave is given by $y = A \sin(\omega t - kx)$.
The velocity of the particle is $v = \frac{dy}{dt} = A\omega \cos(\omega t - kx)$.
Consider two particles at positions $x_1$ and $x_2$. Their velocities are $v_1 = A\omega \cos(\omega t - kx_1)$ and $v_2 = A\omega \cos(\omega t - kx_2)$.
For the speeds to be equal,$|v_1| = |v_2|$,which implies $|\cos(\omega t - kx_1)| = |\cos(\omega t - kx_2)|$.
This condition is satisfied when the arguments differ by $\pi/2$ or $\pi$. The minimum distance corresponds to a phase difference of $\pi/2$.
Thus,$k(x_2 - x_1) = \frac{2\pi}{\lambda} \Delta x = \frac{\pi}{2}$.
Solving for $\Delta x$,we get $\Delta x = \frac{\lambda}{4}$.

(C) The frequency of a wave is determined by the source of the wave and remains constant regardless of the medium through which it travels or the interface it encounters.
When a wave travels from medium $1$ to medium $2$,both reflection and refraction occur.
While the speed and wavelength of the wave may change depending on the refractive indices of the media,the frequency remains unchanged.
Therefore,the frequency of the transmitted (refracted) wave is the same as the frequency of the incident wave.
Option $(c)$ is correct.

(A) function $y = f(x, t)$ represents a travelling wave if and only if it can be expressed in the form $f(x \pm vt)$,where $v$ is the wave speed.
$(a)$ $y = (x^2 - vt)^2$: This cannot be written in the form $f(x \pm vt)$ because the $x$ term is squared while the $vt$ term is linear. Thus,it cannot represent a travelling wave.
$(b)$ $y = \log \left[ \frac{x+vt}{x_0} \right]$: This is of the form $f(x+vt)$. It represents a travelling wave.
$(c)$ $y = e^{-\left( \frac{x+vt}{x_0} \right)^2}$: This is of the form $f(x+vt)$. It represents a travelling wave.
$(d)$ $y = \frac{1}{x+vt}$: This is of the form $f(x+vt)$. It represents a travelling wave.
Therefore,only $(a)$ cannot represent a travelling wave.

(C) Given: Frequency $f = 500 \,Hz$,Speed $v = 350 \,m/s$,Distance $\Delta x = 1 \,m$.
The wavelength $\lambda$ is given by $\lambda = \frac{v}{f} = \frac{350}{500} = 0.7 \,m$.
The phase difference $\Delta \phi$ between two points separated by distance $\Delta x$ is given by $\Delta \phi = \frac{2\pi}{\lambda} \Delta x$.
Substituting the values: $\Delta \phi = \frac{2\pi}{0.7} \times 1 = \frac{20\pi}{7} \approx 2.857\pi$.
Since the phase difference is periodic with $2\pi$,we calculate $\Delta \phi \pmod{2\pi} = \frac{20\pi}{7} - 2\pi = \frac{6\pi}{7} \approx 0.857\pi$.
Re-evaluating the question context: If the distance is $0.35 \,m$,the phase would be $\pi$. Given the options provided and standard textbook problems of this type,there is often a typo in the distance value. Assuming the intended distance was $0.35 \,m$,the phase difference is $\pi$.

(C) The given equation of the travelling wave is $y = a \sin 2 \pi (p t - \frac{x}{5})$.
Comparing this with the standard wave equation $y = a \sin (\omega t - kx)$,we have $\omega = 2 \pi p$ and $k = \frac{2 \pi}{5}$.
The maximum particle velocity is given by $v_{max} = a \omega = a (2 \pi p) = 2 \pi a p$.
The wave velocity is given by $v = \frac{\omega}{k} = \frac{2 \pi p}{2 \pi / 5} = 5 p$.
The ratio of the maximum particle velocity to the wave velocity is $\frac{v_{max}}{v} = \frac{2 \pi a p}{5 p} = \frac{2 \pi a}{5}$.

(A) The general equation of a travelling wave pulse is given by $y = f(ax + bt)$.
Given the equation: $y = \frac{4}{3x^2 + 48t^2 + 24xt + 2}$.
We can rewrite the denominator by factoring out the terms involving $x$ and $t$:
$3x^2 + 24xt + 48t^2 = 3(x^2 + 8xt + 16t^2) = 3(x + 4t)^2$.
Substituting this back into the equation:
$y = \frac{4}{3(x + 4t)^2 + 2}$.
This is in the form $f(x + vt)$,where the wave function represents a pulse moving in the negative $x$-direction.
Comparing $(x + 4t)$ with $(x - vt)$,we identify the velocity magnitude as $v = 4 \, m/s$.

(A) The given equation is $y(x, t) = e^{-(ax^2 + bt^2 + 2\sqrt{ab}xt)}$.
This can be rewritten as $y(x, t) = e^{-(\sqrt{a}x + \sqrt{b}t)^2}$.
$A$ wave function representing a travelling wave is generally of the form $f(kx + \omega t)$ or $f(kx - \omega t)$.
Comparing this with the form $f(kx + \omega t)$,where $k = \sqrt{a}$ and $\omega = \sqrt{b}$,we see that the wave is moving in the $-x$ direction.
The speed of the wave $v$ is given by $v = \frac{\omega}{k} = \frac{\sqrt{b}}{\sqrt{a}} = \sqrt{\frac{b}{a}}$.
Therefore,the wave is moving in the $-x$ direction with a speed of $\sqrt{\frac{b}{a}}$.

(C) The standard equation of a travelling wave is given by $y(x, t) = A \sin(kx - \omega t)$.
Comparing this with the given equation $y(x, t) = 0.05 \sin(8x - 4t)$:
We identify the wave number $k = 8 \; m^{-1}$ and the angular frequency $\omega = 4 \; rad/s$.
The velocity of the wave $v$ is given by the ratio of angular frequency to wave number:
$v = \frac{\omega}{k}$
Substituting the values:
$v = \frac{4}{8} = 0.5 \; m/s$.
Therefore,the velocity of the wave is $0.5 \; m/s$.

(D) The phase difference $\Delta \phi$ is related to the path difference $\Delta x$ by the formula: $\Delta \phi = \frac{2\pi}{\lambda} \Delta x$.
Given $\Delta \phi = 60^{\circ} = \frac{\pi}{3}$ radians and $\Delta x = 6.0\,m$.
Substituting these values: $\frac{\pi}{3} = \frac{2\pi}{\lambda} \times 6$.
Solving for wavelength $\lambda$: $\lambda = 2 \times 6 \times 3 = 36\,m$.
The wave velocity $V$ is given by $V = f \lambda$.
Given frequency $f = 500\,Hz$,we have $V = 500\,Hz \times 36\,m = 18000\,m/s$.
Converting to $km/s$: $V = \frac{18000}{1000}\,km/s = 18\,km/s$.

(D) The standard equation of a transverse harmonic wave is $y(x, t) = A \sin(\omega t + kx)$.
Comparing this with the given equation $y(x, t) = 5 \sin(6t + 0.003x)$,we get:
Angular frequency $\omega = 6\,rad/s$.
Wave number $k = 0.003\,cm^{-1}$.
To convert $k$ into $m^{-1}$,we multiply by $100$ because $1\,cm^{-1} = 100\,m^{-1}$.
So,$k = 0.003 \times 100 = 0.3\,m^{-1}$.
The wave velocity $v$ is given by the formula $v = \frac{\omega}{k}$.
Substituting the values,$v = \frac{6}{0.3} = 20\,ms^{-1}$.

(B) The standard wave equation is $Y = A \sin (\omega t - kx + \phi)$.
Comparing the given equation $Y = 10^{-2} \sin 2 \pi (160 t - 0.5 x + \frac{\pi}{4})$ with the standard form,we have:
$\omega = 2 \pi \times 160 \, rad/s$
$k = 2 \pi \times 0.5 \, m^{-1}$
The speed of the wave $v$ is given by $v = \frac{\omega}{k} = \frac{2 \pi \times 160}{2 \pi \times 0.5} = \frac{160}{0.5} = 320 \, m/s$.
To convert the speed from $m/s$ to $km/h$,we multiply by $\frac{18}{5}$:
$v = 320 \times \frac{18}{5} = 64 \times 18 = 1152 \, km/h$.

(B) The given equation of the plane progressive wave is $y = 2 \cos 2 \pi (330 t - x) \ m$.
The standard equation for a plane progressive wave is $y = A \cos (\omega t - kx)$.
By comparing the given equation with the standard form,we have:
$y = 2 \cos (2 \pi \times 330 t - 2 \pi x)$
Here,the angular frequency $\omega = 2 \pi \times 330 \ rad/s$.
We know that $\omega = 2 \pi f$,where $f$ is the frequency of the wave.
Equating the two expressions for $\omega$:
$2 \pi f = 2 \pi \times 330$
$f = 330 \ Hz$.
Therefore,the frequency of the wave is $330 \ Hz$.

(B) Given: Wave speed $v = 10 \text{ cm/s} = 0.1 \text{ m/s}$,wavelength $\lambda = 0.5 \text{ m}$,amplitude $A = 10 \text{ cm} = 0.1 \text{ m}$.
Angular frequency $\omega = \frac{2\pi v}{\lambda} = \frac{2\pi \times 0.1}{0.5} = 0.4\pi \text{ rad/s} = \frac{2\pi}{5} \text{ rad/s}$.
The displacement equation is $y = A \sin(kx - \omega t + \phi)$.
At displacement $y = 5 \text{ cm} = 0.05 \text{ m}$,we have $0.05 = 0.1 \sin(\theta)$,so $\sin(\theta) = 0.5$,which means $\theta = 30^{\circ}$ or $150^{\circ}$.
From the figure,point $P$ is on the downward slope,so its velocity $v_y = \frac{dy}{dt}$ must be negative.
The particle velocity is $v_y = A\omega \cos(\theta)$.
Since the wave moves in the positive $x$-direction,a particle on the downward slope has a negative vertical velocity.
Thus,$v_y = -(0.1) \times (0.4\pi) \times \cos(30^{\circ}) = -0.04\pi \times \frac{\sqrt{3}}{2} = -0.02\pi\sqrt{3} \text{ m/s} = -\frac{\sqrt{3}\pi}{50} \hat{j} \text{ m/s}$.

(C) The standard equation of a progressive wave is $y(x, t) = A \sin(kx + \omega t + \phi)$.
Comparing the given equation $y(x, t) = 4.0 \sin(20 \times 10^{-3} x + 600 t)$ with the standard form:
Angular frequency $\omega = 600 \ rad/s$.
Wave number $k = 20 \times 10^{-3} \ mm^{-1} = 20 \ m^{-1}$.
The wave velocity $v$ is given by $v = -\frac{\omega}{k}$.
The negative sign indicates that the wave is travelling in the negative $x$-direction.
$v = -\frac{600}{20} = -30 \ m/s$.

(A) The wave speed $v$ is given by $v = \frac{\text{distance}}{\text{time}} = \frac{1.2 \ cm}{0.3 \ s} = 4 \ cm/s$.
The angular wave number $k$ is $k = \frac{2 \pi}{\lambda} = \frac{2 \pi}{7.5 \ cm} \approx 0.837 \ rad/cm \approx 0.83 \ rad/cm$.
The angular frequency $\omega$ is $\omega = v k = 4 \ cm/s \times 0.837 \ rad/cm \approx 3.35 \ rad/s$.
Since the crest is at $x = 0$ at $t = 0$,the wave is represented by a cosine function: $y = A \cos(kx - \omega t)$.
Substituting the values,$y = 2 \cos(0.83 x - 3.35 t) \ cm$.

(C) The given equation for the displacement of the wave is $x(t) = 5 \cos \left(628 t + \frac{\pi}{2}\right)$.
Comparing this with the standard form $x(t) = A \cos(\omega t + \phi)$, we get the angular frequency $\omega = 628 \text{ rad/s}$.
We know that $\omega = 2 \pi f$, where $f$ is the frequency.
So, $2 \times 3.14 \times f = 628$.
$6.28 f = 628$.
$f = \frac{628}{6.28} = 100 \text{ Hz}$.
The relationship between velocity $(v)$, frequency $(f)$, and wavelength $(\lambda)$ is given by $v = f \lambda$.
Given $v = 300 \text{ m/s}$, we have $300 = 100 \times \lambda$.
Therefore, $\lambda = \frac{300}{100} = 3 \text{ m}$.

(A) The general equation of a wave is $y = A \sin(kx + \omega t + \phi)$.
Comparing this with the given equation $y = \sin(20 \pi x + 10 \pi t)$,we get the wave number $k = 20 \pi \ rad/m$.
The wavelength $\lambda$ is given by $\lambda = \frac{2 \pi}{k} = \frac{2 \pi}{20 \pi} = 0.1 \ m = 10 \ cm$.
Points having the same oscillating speed in a wave are separated by a distance of $\frac{\lambda}{2}$ (or multiples thereof).
Therefore,the minimum distance is $\frac{\lambda}{2} = \frac{10 \ cm}{2} = 5 \ cm$.

(A) The maximum particle velocity $V_{p_{\max}}$ is given by $a\omega = \frac{v}{10}$.
Given $a = 10^{-3} \ m$ and $v = 10 \ ms^{-1}$,we have $10^{-3} \times \omega = \frac{10}{10} = 1$.
Thus,$\omega = 1000 \ rad/s$.
The frequency $f$ is given by $f = \frac{\omega}{2\pi} = \frac{1000}{2\pi} = \frac{10^3}{2\pi} \ Hz$.
The wave speed $v$ is related to wavelength $\lambda$ by $v = f\lambda = \frac{\omega}{2\pi} \lambda$.
Rearranging for $\lambda$,we get $\lambda = \frac{2\pi v}{\omega} = \frac{2\pi \times 10}{1000} = 2\pi \times 10^{-2} \ m$.

(C) For a wave represented by $y = A \sin(\omega t - kx)$,the wave velocity is given by $V = \frac{\omega}{k}$.
We know that the wave number $k = \frac{2 \pi}{\lambda}$.
Thus,$V = \frac{\omega}{2 \pi / \lambda} = \frac{\omega \lambda}{2 \pi}$.
The maximum particle velocity in $S.H.M.$ is given by $\nu = A \omega$.
From this,we have $\omega = \frac{\nu}{A}$.
Substituting the value of $\omega$ in the expression for $V$:
$V = \frac{(\nu / A) \lambda}{2 \pi} = \frac{\lambda \nu}{2 \pi A}$.
Rearranging for $\nu$:
$\nu = \frac{2 \pi A}{\lambda} V$.
Therefore,the correct relation is $\nu = \frac{2 \pi A}{\lambda} V$.

(A) Given the displacement equations:
$y_1 = a_1 \sin \left(\frac{2 \pi x}{\lambda} - \omega t\right) \quad \dots(i)$
$y_2 = a_2 \cos \left(\frac{2 \pi x}{\lambda} - \omega t + \phi\right) \quad \dots(ii)$
To compare the phases,we convert the cosine function into a sine function using the identity $\cos(\theta) = \sin(\theta + \frac{\pi}{2})$:
$y_2 = a_2 \sin \left(\frac{2 \pi x}{\lambda} - \omega t + \phi + \frac{\pi}{2}\right) \quad \dots(iii)$
Comparing the phase of $y_1$ (which is $\frac{2 \pi x}{\lambda} - \omega t$) with the phase of $y_2$ (which is $\frac{2 \pi x}{\lambda} - \omega t + \phi + \frac{\pi}{2}$),the phase difference is:
$\Delta \phi = \left(\frac{2 \pi x}{\lambda} - \omega t + \phi + \frac{\pi}{2}\right) - \left(\frac{2 \pi x}{\lambda} - \omega t\right)$
$\Delta \phi = \phi + \frac{\pi}{2}$

(D) The given wave equation is $Y = 3 \sin 2 \pi \left( \frac{t}{0.04} - \frac{x}{0.01} \right)$.
Comparing this with the standard wave equation $Y = A \sin 2 \pi \left( ft - \frac{x}{\lambda} \right)$,we get the frequency $f = \frac{1}{0.04} = 25 \ Hz$.
The angular frequency is $\omega = 2 \pi f = 2 \pi \times 25 = 50 \pi \ rad/s$.
The maximum acceleration $a_{\max}$ is given by $a_{\max} = \omega^2 A$.
Substituting the values: $a_{\max} = (50 \pi)^2 \times 3 = 2500 \times \pi^2 \times 3$.
Given $\pi^2 = 10$,we have $a_{\max} = 2500 \times 10 \times 3 = 75000 \ cm/s^2 = 7.5 \times 10^4 \ cm/s^2$.

(B) The general equation of a simple harmonic progressive wave is $y = A \sin(\omega t - kx)$.
Comparing this with the given equation $y = A \sin(100 \pi t - 3x)$,we get the propagation constant $k = 3 \text{ rad/m}$.
The relationship between phase difference $(\Delta \phi)$ and path difference $(\Delta x)$ is given by $\Delta \phi = k \cdot \Delta x$.
Given the phase difference $\Delta \phi = \frac{\pi}{18}$.
Substituting the values,we get $\frac{\pi}{18} = 3 \cdot \Delta x$.
Therefore,the distance between the two particles is $\Delta x = \frac{\pi}{18 \cdot 3} = \frac{\pi}{54} \text{ m}$.
Thus,the correct option is $B$.

(D) The given wave equation is $Y = A \sin 2 \pi (n t - \frac{x}{\lambda})$.
Comparing this with the standard wave equation $Y = A \sin (\omega t - kx)$,we have $\omega = 2 \pi n$ and $k = \frac{2 \pi}{\lambda}$.
The maximum particle velocity is given by $v_{p, \text{max}} = A \omega = A (2 \pi n)$.
The wave velocity is given by $v_w = \frac{\omega}{k} = \frac{2 \pi n}{2 \pi / \lambda} = n \lambda$.
According to the problem,$v_{p, \text{max}} = 4 v_w$.
Substituting the expressions,we get $A (2 \pi n) = 4 (n \lambda)$.
Simplifying the equation,$2 \pi A n = 4 n \lambda$.
Dividing both sides by $2n$,we get $\lambda = \pi A$.

(C) The given wave equation is $Y = a \sin(2 \pi b t - 2 \pi c x)$.
Comparing this with the standard wave equation $Y = A \sin(\omega t - k x)$,we get angular frequency $\omega = 2 \pi b$ and wave number $k = 2 \pi c$.
The maximum particle velocity is given by $(V_{max})_p = \omega A = (2 \pi b) a$.
The wave velocity is given by $V_w = \frac{\omega}{k} = \frac{2 \pi b}{2 \pi c} = \frac{b}{c}$.
According to the problem,the maximum particle velocity is twice the wave velocity:
$(V_{max})_p = 2 V_w$.
Substituting the values: $2 \pi b a = 2 \left( \frac{b}{c} \right)$.
Dividing both sides by $2b$,we get $\pi a = \frac{1}{c}$,which implies $c = \frac{1}{\pi a}$.

(D) The given equation is $y=5(\sin 4 \pi t+\sqrt{3} \cos 4 \pi t)$.
Multiplying by $5$,we get $y=5 \sin 4 \pi t+5 \sqrt{3} \cos 4 \pi t$.
This is in the form $y=A_1 \sin \omega t+A_2 \cos \omega t$,where $A_1=5$ and $A_2=5 \sqrt{3}$.
The resultant amplitude $A$ is given by $A=\sqrt{A_1^2+A_2^2}$.
Substituting the values,$A=\sqrt{(5)^2+(5 \sqrt{3})^2}$.
$A=\sqrt{25+75} = \sqrt{100}$.
Thus,$A=10$ units.

(D) The standard equation of a simple harmonic wave is given by $y = A \sin(\omega t - kx)$.
Comparing the given equation $y = 5 \sin \frac{\pi}{2}(100t - x)$ with the standard form:
$y = 5 \sin(50\pi t - \frac{\pi}{2}x)$.
Here,the angular frequency $\omega = 50\pi \ rad/s$.
We know that the time period $T$ is related to angular frequency by the formula $T = \frac{2\pi}{\omega}$.
Substituting the value of $\omega$:
$T = \frac{2\pi}{50\pi} = \frac{1}{25} \ s$.
$T = 0.04 \ s$.
Therefore,the correct option is $D$.

(C) Key Idea: The distance between two consecutive crests in a wave is called the wavelength $(\lambda)$.
When a boat is rocked by waves,it completes one full bounce (up and down) as one full wavelength passes by the boat.
Given:
Wavelength $(\lambda)$ $= 100 \,m$
Velocity of wave $(v)$ $= 25 \,m/s$
The time period $(T)$ of the bounce is the time taken for one wavelength to pass a fixed point.
Using the formula: $T = \frac{\lambda}{v}$
$T = \frac{100 \,m}{25 \,m/s} = 4 \,s$
Therefore,the boat bounces up once every $4 \,s$.

(B) Given waves are $Y_1 = a_1 \sin \left(\omega t - \frac{2 \pi x}{\lambda}\right)$ and $Y_2 = a_2 \cos \left(\omega t - \frac{2 \pi x}{\lambda} + \phi\right)$.
To compare the phases,we convert the cosine function to sine: $\cos(\theta) = \sin(\theta + \frac{\pi}{2})$.
Thus,$Y_2 = a_2 \sin \left(\omega t - \frac{2 \pi x}{\lambda} + \phi + \frac{\pi}{2}\right)$.
The phase difference $\delta$ is the difference between the arguments of the sine functions: $\delta = \left(\omega t - \frac{2 \pi x}{\lambda} + \phi + \frac{\pi}{2}\right) - \left(\omega t - \frac{2 \pi x}{\lambda}\right) = \phi + \frac{\pi}{2}$.
The relationship between path difference $\Delta x$ and phase difference $\delta$ is $\Delta x = \frac{\lambda}{2 \pi} \delta$.
Substituting the value of $\delta$,we get $\Delta x = \frac{\lambda}{2 \pi} \left(\phi + \frac{\pi}{2}\right)$.