Wave Equation and Characteristics of Waves Questions in English

(B) The wavelength $\lambda$ of the wave is given by the formula $\lambda = \frac{v}{f}$.
Given $v = 300\, m/s$ and $f = 500\, Hz$,we have $\lambda = \frac{300}{500} = 0.6\, m$.
$A$ phase difference of $360^{\circ}$ corresponds to a path difference of one full wavelength $\lambda$.
Therefore,a phase difference of $\Delta\phi = 60^{\circ}$ corresponds to a path difference $\Delta x$ given by $\Delta x = \frac{\Delta\phi}{360^{\circ}} \times \lambda$.
Substituting the values,$\Delta x = \frac{60^{\circ}}{360^{\circ}} \times 0.6\, m = \frac{1}{6} \times 0.6\, m = 0.1\, m$.

(C) Given frequency $f = \frac{1}{8} \, Hz$.
Therefore,time period $T = \frac{1}{f} = 8 \, s$.
Given wavelength $\lambda = 24 \, m$.
The wave speed $v = f \lambda = \frac{1}{8} \times 24 = 3 \, m/s$.
The time taken by the wave to reach the particle at distance $x = 9 \, m$ is $t_1 = \frac{x}{v} = \frac{9}{3} = 3 \, s$.
The particle starts its motion from the mean position (moving up first). To reach the lower extreme for the first time,the particle must complete $\frac{3}{4}$ of its oscillation cycle.
The time taken by the particle to reach the lower extreme from the moment the wave reaches it is $t_2 = \frac{3T}{4} = \frac{3}{4} \times 8 = 6 \, s$.
Total time taken = $t_1 + t_2 = 3 \, s + 6 \, s = 9 \, s$.

(C) The standard wave equation is $y = A \sin(\omega t + kx)$.
Comparing the given equation $y = 10^{-4} \sin(60t + 2x)$ with the standard form:
Angular frequency $\omega = 60 \ rad/s$ and wave number $k = 2 \ m^{-1}$.
Since $\omega = 2\pi f$,we have $60 = 2\pi f$,which gives $f = \frac{30}{\pi} \ Hz$.
Wave velocity $v = \frac{\omega}{k} = \frac{60}{2} = 30 \ m/s$. Since the sign between $\omega t$ and $kx$ is positive,the wave travels in the negative $x$-direction.
Wavelength $\lambda = \frac{2\pi}{k} = \frac{2\pi}{2} = \pi \ m$.
Comparing these results with the options,option $C$ is correct.

(A) The standard equation of a traveling wave is $y(x, t) = A \cos(kx - \omega t)$.
Here,$k = \alpha$ (wave number) and $\omega = \beta$ (angular frequency).
The wave number is given by $k = \frac{2\pi}{\lambda}$.
Given $\lambda = 0.08 \ m$,therefore $\alpha = \frac{2\pi}{0.08} = 25\pi \ rad/m$.
The angular frequency is given by $\omega = \frac{2\pi}{T}$.
Given $T = 2.0 \ s$,therefore $\beta = \frac{2\pi}{2.0} = \pi \ rad/s$.
Thus,the correct option is $A$.

(A) The given wave equation is $y = A \sin^2 (\omega t - kx)$.
Using the trigonometric identity $\sin^2 \theta = \frac{1 - \cos(2\theta)}{2}$,we can rewrite the equation as:
$y = \frac{A}{2} [1 - \cos(2\omega t - 2kx)]$.
The particle velocity $v_p$ is given by the derivative of $y$ with respect to time $t$:
$v_p = \frac{\partial y}{\partial t} = \frac{A}{2} [2\omega \sin(2\omega t - 2kx)] = A\omega \sin(2\omega t - 2kx)$.
The maximum particle velocity is $(v_p)_{\max} = A\omega$.
The wave velocity $v$ is given by the ratio of the coefficient of $t$ to the coefficient of $x$ in the argument of the function:
$v = \frac{2\omega}{2k} = \frac{\omega}{k}$.
Since $k = \frac{2\pi}{\lambda}$,we have $v = \frac{\omega}{2\pi / \lambda} = \frac{\omega \lambda}{2\pi}$.
According to the problem,$(v_p)_{\max} = v$:
$A\omega = \frac{\omega \lambda}{2\pi}$.
Solving for $A$,we get $A = \frac{\lambda}{2\pi}$.

(C) The frequency of a wave is determined solely by the source of the wave. When a wave travels from one medium to another,its frequency remains constant,while its velocity and wavelength change based on the properties of the medium. Therefore,frequency is independent of the medium's properties and the other wave parameters.

(A) The given wave equation is $y = a \sin \left( \frac{2\pi}{T}t - \frac{2\pi}{\lambda}x \right)$,where amplitude $a = \frac{10}{\pi} \ cm$.
The wave velocity $v$ is given by $v = \frac{\omega}{k} = \frac{2\pi/T}{2\pi/\lambda} = \frac{\lambda}{T} = f\lambda$.
The maximum particle velocity $v_{p,max}$ is given by $v_{p,max} = a\omega = a \left( \frac{2\pi}{T} \right)$.
According to the problem,the wave velocity is twice the maximum particle velocity:
$v = 2 v_{p,max}$
Substituting the expressions:
$\frac{\lambda}{T} = 2 \left( a \cdot \frac{2\pi}{T} \right)$
Canceling $T$ from both sides:
$\lambda = 4\pi a$
Substituting the value of $a = \frac{10}{\pi} \ cm$:
$\lambda = 4\pi \left( \frac{10}{\pi} \right) = 40 \ cm$.

(B) The standard equation for a traveling wave is given by $y(x, t) = A \cos(kx - \omega t)$,where $k = \frac{\omega}{v}$.
For the first wave $y_1 = 0.05 \cos(0.50 \pi x - 100 \pi t)$:
Comparing with the standard equation,we have angular frequency $\omega_1 = 100 \pi \text{ rad/s}$ and wave number $k_1 = 0.50 \pi \text{ rad/m}$.
The velocity $v_1$ is given by $v_1 = \frac{\omega_1}{k_1} = \frac{100 \pi}{0.50 \pi} = 200 \text{ m/s}$.
For the second wave $y_2 = 0.05 \cos(0.46 \pi x - 92 \pi t)$:
Comparing with the standard equation,we have angular frequency $\omega_2 = 92 \pi \text{ rad/s}$ and wave number $k_2 = 0.46 \pi \text{ rad/m}$.
The velocity $v_2$ is given by $v_2 = \frac{\omega_2}{k_2} = \frac{92 \pi}{0.46 \pi} = 200 \text{ m/s}$.
Since both waves have the same velocity,the velocity of the waves is $200 \text{ m/s}$.

(B) The distance between a crest and the neighbouring trough at the same instant is half the wavelength $(\lambda/2)$.
Given $\lambda/2 = 4.0 \, cm$,so $\lambda = 8.0 \, cm$.
The distance between a crest and a trough at the same place is twice the amplitude $(2a)$.
Given $2a = 1.0 \, cm$,so $a = 0.5 \, cm$.
The time interval between two consecutive crests at the same place is the time period $(T)$.
Given $T = 0.4 \, s$.
The angular frequency is $\omega = \frac{2\pi}{T} = \frac{2\pi}{0.4} = 5\pi \, rad/s$.
The maximum speed of the vibrating particles is given by $v_{max} = a\omega$.
Substituting the values: $v_{max} = 0.5 \times 5\pi = 2.5\pi = \frac{5\pi}{2} \, cm/s$.

(C) The general equation of a wave propagating in the positive $x$-direction 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 any time $t$,the equation becomes $y = f(x - vt) = \frac{1}{1 + (x - vt)^2}$.
Given that at $t = 2 \ s$,the equation is $y = \frac{1}{1 + (x - 1)^2}$.
Comparing the two expressions for $t = 2 \ s$:
$vt = 1$
Since $t = 2 \ s$,we have $v(2) = 1$.
Therefore,$v = \frac{1}{2} = 0.5 \ m/s$.

(C) The general equation for a travelling harmonic wave is $y(x, t) = A \sin(\omega t + kx + \phi)$.
Comparing this with the given equation $y(x, t) = 10^{-3} \sin(50t + 2x)$:
Here,$\omega = 50 \ rad/s$ and $k = 2 \ rad/m$.
Since the sign between the $t$ and $x$ terms is positive $(+)$,the wave is propagating along the negative $x$-axis.
The wave speed $v$ is given by the ratio of angular frequency to the wave number: $v = \frac{\omega}{k}$.
Substituting the values: $v = \frac{50}{2} = 25 \ m/s$.
Therefore,the wave is propagating along the negative $x$-axis with a speed of $25 \ m/s$.

(A) The given wave equation is $y(x, t) = A \sin(kx - \omega t + \phi)$.
At $t = 0$,the equation becomes $y(x, 0) = A \sin(kx + \phi)$.
From the given graph at $t = 0$,we observe that at $x = 0$,the displacement $y(0, 0) = 0$.
Substituting these values into the equation: $0 = A \sin(k(0) + \phi) \Rightarrow \sin(\phi) = 0$.
This implies $\phi = 0$ or $\phi = \pi$.
To determine the correct value,we check the slope of the wave at $x = 0$. The slope is given by $\frac{\partial y}{\partial x} = Ak \cos(kx + \phi)$.
At $x = 0$,the slope is $Ak \cos(\phi)$.
From the graph,at $x = 0$,the wave is moving downwards as $x$ increases,meaning the slope is negative.
If $\phi = 0$,the slope is $Ak \cos(0) = Ak$,which is positive.
If $\phi = \pi$,the slope is $Ak \cos(\pi) = -Ak$,which is negative.
Therefore,the phase $\phi$ must be $\pi$.

(D) The particle velocity is given by $v_p = \frac{dy}{dt} = -v \left( \frac{dy}{dx} \right)$,where $v$ is the wave velocity and $\frac{dy}{dx}$ is the slope of the wave at that point.
$(1)$ For upward velocity,$v_p$ must be positive. This requires the slope $\frac{dy}{dx}$ to be negative. Looking at the graph,the slope is negative at points $D, E,$ and $F$. Thus,these points move upwards.
$(2)$ For downward velocity,$v_p$ must be negative. This requires the slope $\frac{dy}{dx}$ to be positive. Looking at the graph,the slope is positive at points $A, B,$ and $H$. Thus,these points move downwards.
$(3)$ For zero velocity,the slope $\frac{dy}{dx}$ must be zero. This occurs at the crest and trough,which are points $C$ and $G$.
$(4)$ The magnitude of the particle velocity is maximum where the slope $| \frac{dy}{dx} |$ is maximum. This occurs at points $A, E,$ and the point between $G$ and $H$. Points $A$ and $E$ have maximum velocity,not minimum. Therefore,statement $(D)$ is wrong.

(C) The standard wave equation is given by $y = A \sin(\omega t + kx)$.
Comparing the given equation $y = 10^{-4} \sin(60t + 2x)$ with the standard equation,we get:
Angular frequency $\omega = 60 \text{ rad/s}$ and wave number $k = 2 \text{ m}^{-1}$.
Since the sign between $\omega t$ and $kx$ is positive,the wave travels in the negative $x$-direction.
Frequency $f = \omega / (2\pi) = 60 / (2\pi) = 30/\pi \text{ Hz}$.
Wave velocity $v = \omega / k = 60 / 2 = 30 \text{ m/s}$.
Wavelength $\lambda = 2\pi / k = 2\pi / 2 = \pi \text{ m}$.
Comparing these results with the options,option $C$ is correct.

(A) The general equation for a wave travelling in the $-z$ direction is given by $x = A \sin(\omega t + kz)$.
Given:
Amplitude $A = 1 \ m$.
Wavelength $\lambda = \pi \ m$.
Frequency $f = \frac{1}{\pi} \ Hz$.
Angular frequency $\omega = 2\pi f = 2\pi \times \frac{1}{\pi} = 2 \ rad/s$.
Wave number $k = \frac{2\pi}{\lambda} = \frac{2\pi}{\pi} = 2 \ m^{-1}$.
Substituting these values into the general equation,we get:
$x = 1 \sin(2t + 2z)$.
Thus,the correct option is $A$.

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

(A) The given wave equation is $y = 0.25 \sin(10\pi x - 2\pi t)$.
Comparing this with the standard wave equation $y = A \sin(kx - \omega t)$:
$1$. The amplitude $A = 0.25 \, m$.
$2$. The wave number $k = 10\pi$. Since $k = \frac{2\pi}{\lambda}$,we have $\lambda = \frac{2\pi}{10\pi} = 0.2 \, m$.
$3$. The angular frequency $\omega = 2\pi$. Since $\omega = 2\pi f$,we have $f = \frac{2\pi}{2\pi} = 1 \, Hz$.
$4$. Since the sign between $kx$ and $\omega t$ is negative,the wave travels in the positive $x$-direction.
Therefore,the wave travels in the $+ve$ $x$-direction with frequency $1 \, Hz$ and wavelength $\lambda = 0.2 \, m$.

(D) Two points are in the same phase if they have the same displacement from the mean position and are moving in the same direction.
In the given sinusoidal wave,points $C$ and $E$ are both below the $x$-axis,have the same vertical displacement,and are both moving in the upward direction (towards the equilibrium position). Therefore,points $C$ and $E$ are in the same phase.

(A) The first wave is given by $y_1 = a \sin(\omega t + kx + 0.57)$.
Therefore,the phase of the first wave is $\phi_1 = (\omega t + kx + 0.57)$.
The second wave is given by $y_2 = a \cos(\omega t + kx)$.
Using the trigonometric identity $\cos(\theta) = \sin(\theta + \pi/2)$,we can write $y_2 = a \sin(\omega t + kx + \pi/2)$.
Therefore,the phase of the second wave is $\phi_2 = (\omega t + kx + \pi/2)$.
The phase difference $\Delta \phi$ is given by $\phi_2 - \phi_1$.
$\Delta \phi = (\omega t + kx + \pi/2) - (\omega t + kx + 0.57)$.
$\Delta \phi = \pi/2 - 0.57$.
Since $\pi \approx 3.14$,$\pi/2 \approx 1.57$.
$\Delta \phi = 1.57 - 0.57 = 1.0 \ radian$.

(B) Given: Wavelength in medium $A$,$\lambda_{A} = 100 \, mm = 0.10 \, m$.
Wavelength in medium $B$,$\lambda_{B} = 0.25 \, m$.
Velocity in medium $A$,$v_{A} = 80 \, cm s^{-1} = 0.80 \, m s^{-1}$.
Since the frequency $f$ of the wave remains constant when it travels from one medium to another,we have $f = \frac{v_{A}}{\lambda_{A}} = \frac{v_{B}}{\lambda_{B}}$.
Rearranging for $v_{B}$,we get $v_{B} = v_{A} \times \frac{\lambda_{B}}{\lambda_{A}}$.
Substituting the values: $v_{B} = 0.80 \times \frac{0.25}{0.10}$.
$v_{B} = 0.80 \times 2.5 = 2.0 \, m s^{-1}$.

(D) Given: Frequency of the wave,$f = 150 \, Hz$.
From the figure,the distance of $20 \, cm$ covers $2.5$ wavelengths (or $\frac{5}{2}$ cycles).
So,$\frac{5}{2} \lambda = 20 \, cm$.
$\lambda = \frac{20 \times 2}{5} \, cm = 8 \, cm = 0.08 \, m$.
Velocity of the wave is given by $v = f \lambda$.
$v = 150 \, Hz \times 0.08 \, m = 12 \, ms^{-1}$.
Thus,the wavelength is $0.08 \, m$ and the velocity is $12 \, ms^{-1}$.

(C) To determine the direction of wave propagation,we look at the motion of the particle at $x = x_1$.
At $x = x_1$,the particle is on the rising slope of the wave and is moving upward.
For a wave traveling in the $+x$ direction,the particles on the rising slope (where the slope is positive) move upward.
Conversely,for a wave traveling in the $-x$ direction,the particles on the rising slope move downward.
Since the particle at $x = x_1$ is moving upward,the wave must be propagating in the $+x$ direction.

(B) The general equation for a wave propagating in the positive $X-$ direction without changing its shape is given by $y = f(x - vt)$.
At $t = 0$,the equation is $y = f(x) = 1/(1 + x^2)$.
At $t = 2$ seconds,the equation is $y = f(x - 2v) = 1/[1 + (x - 2v)^2]$.
Comparing this with the given equation at $t = 2$ seconds,which is $y = 1/[1 + (x - 1)^2]$,we get:
$x - 2v = x - 1$
$2v = 1$
$v = 0.5 \, ms^{-1}$.

(C) The given wave equation is $y = A \sin(\omega t - kx)$.
To find the slope of the curve at any point $x$ at a given time $t$,we calculate the partial derivative of $y$ with respect to $x$:
$\frac{\partial y}{\partial x} = A \cos(\omega t - kx) \cdot (-k) = -kA \cos(\omega t - kx)$.
The magnitude of the slope is $|\frac{\partial y}{\partial x}| = |kA \cos(\omega t - kx)|$.
At point $B$,the displacement $y = 0$,which implies $\sin(\omega t - kx) = 0$. This occurs when $(\omega t - kx) = n\pi$ (where $n$ is an integer).
At these points,$\cos(\omega t - kx) = \pm 1$.
Therefore,the magnitude of the slope at $B$ is $|-kA(\pm 1)| = kA$.

(D) 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 $y = 5 \sin 2\pi \left[ \frac{t}{0.04} - \frac{x}{50} \right]$ with the standard equation,we identify the term $\frac{x}{\lambda} = \frac{x}{50}$.
Therefore,the wavelength $\lambda = 50 \, cm$.

(A) The given equation is of the form $y = A_1 \sin \omega t + A_2 \cos \omega t$,where $A_1 = \frac{1}{\sqrt{a}}$ and $A_2 = \frac{1}{\sqrt{b}}$.
For an equation of the form $y = A_1 \sin \omega t + A_2 \cos \omega t$,the resultant amplitude $A$ is given by $A = \sqrt{A_1^2 + A_2^2}$.
Substituting the values of $A_1$ and $A_2$:
$A = \sqrt{\left(\frac{1}{\sqrt{a}}\right)^2 + \left(\frac{1}{\sqrt{b}}\right)^2}$
$A = \sqrt{\frac{1}{a} + \frac{1}{b}}$
$A = \sqrt{\frac{b + a}{ab}}$
$A = \frac{\sqrt{a + b}}{\sqrt{ab}}$

(C) The wavelength $\lambda$ is the distance between two consecutive crests,so $\lambda = 100 \ m$.
Given the wave velocity $v = 25 \ m/s$.
The frequency $f$ of the wave is given by the formula $v = f \lambda$,which implies $f = \frac{v}{\lambda}$.
Substituting the values,$f = \frac{25}{100} = 0.25 \ Hz$.
The time period $T$ is the time taken for one complete bounce,which is the reciprocal of the frequency: $T = \frac{1}{f}$.
Therefore,$T = \frac{1}{0.25} = 4 \ s$.

(C) When a wave travels from one medium to another, its velocity $(v)$ and wavelength $(\lambda)$ change depending on the properties of the medium (such as refractive index or density).
However, the frequency $(f)$ of the wave is determined solely by the source of the wave and remains constant regardless of the medium through which it propagates.
Therefore, frequency is independent of the medium's properties, while velocity and wavelength are dependent on them.

(B) The standard equation of a travelling wave is given by $y = A \sin (\omega t - kx)$.
Comparing the given equation $x = A \sin (2t - 0.1x)$ with the standard form,we identify the wave number $k$ as $k = 0.1$.
The wave number $k$ is related to the wavelength $\lambda$ by the formula $k = \frac{2\pi}{\lambda}$.
Substituting the value of $k$:
$0.1 = \frac{2\pi}{\lambda}$
$\lambda = \frac{2\pi}{0.1} = 20\pi$.
Therefore,the wavelength of the wave is $20\pi$.

(A) The wavelength $(\lambda)$ is defined as the distance between two nearest particles in the same phase.
The time period $(T)$ is the time taken by the wave to travel a distance equal to one wavelength.
By definition,the speed of a wave $(v)$ is the distance traveled per unit time.
Therefore,$v = \frac{\lambda}{T}$,which is $\text{Speed of wave} = \frac{\text{wavelength}}{\text{time period}}$.
Since the reason correctly defines wavelength and provides the basis for the relationship in the assertion,the reason is the correct explanation of the assertion.

(N/A) The standard wave equation is $y(x, t) = a \sin (kx - \omega t)$.
$(a)$ Comparing the given equation with the standard form,the amplitude $a = 0.005 \, m = 5 \, mm$.
$(b)$ The angular wave number $k = 80.0 \, m^{-1}$. The wavelength $\lambda$ is given by $\lambda = 2\pi / k = 2\pi / 80.0 \approx 0.0785 \, m = 7.85 \, cm$.
$(c)$ The angular frequency $\omega = 3.0 \, rad \, s^{-1}$. The period $T = 2\pi / \omega = 2\pi / 3.0 \approx 2.09 \, s$. The frequency $\nu = 1 / T = 3.0 / 2\pi \approx 0.48 \, Hz$.
For displacement at $x = 30.0 \, cm = 0.3 \, m$ and $t = 20 \, s$:
$y = 0.005 \sin (80.0 \times 0.3 - 3.0 \times 20) = 0.005 \sin (24 - 60) = 0.005 \sin (-36 \, rad)$.
Since $\sin(-36 \, rad) \approx \sin(-36 + 12\pi) \approx \sin(1.699 \, rad) \approx 0.992$,
$y \approx 0.005 \times 0.992 \approx 0.00496 \, m \approx 5 \, mm$.

(NONE) The converse is not true. $A$ function $y=f(x \pm vt)$ represents a travelling wave only if it remains finite for all values of $x$ and $t$.
$(a)$ For $y = (x-vt)^2$: As $x \to \infty$ or $t \to \infty$,$y \to \infty$. Since the function does not remain finite for all $x$ and $t$,it does not represent a physically possible travelling wave.
$(b)$ For $y = \log[(x+vt)/x_0]$: As $x+vt \to 0$,$y \to -\infty$. Since the function does not remain finite for all $x$ and $t$,it does not represent a physically possible travelling wave.
$(c)$ For $y = 1/(x+vt)$: As $x+vt \to 0$,$y \to \infty$. Since the function does not remain finite for all $x$ and $t$,it does not represent a physically possible travelling wave.

(N/A) The given transverse harmonic wave equation is $y(x, t) = 3.0 \sin (36 t + 0.018 x + \pi / 4)$.
For $x = 0, 2,$ and $4 \; cm$,the equations are:
$y(0, t) = 3.0 \sin (36 t + \pi / 4)$
$y(2, t) = 3.0 \sin (36 t + 0.036 + \pi / 4)$
$y(4, t) = 3.0 \sin (36 t + 0.072 + \pi / 4)$
The angular frequency is $\omega = 36 \; rad/s$,so the time period is $T = 2\pi / \omega = \pi / 18 \; s$.
The shapes of these graphs are sinusoidal. In a travelling wave,the amplitude and frequency remain constant for all points,but the phase changes from one point to another due to the $kx$ term in the wave equation.

(N/A) The equation for a travelling harmonic wave is given by $y(x, t) = 2.0 \cos 2 \pi(10 t - 0.0080 x + 0.35)$.
Comparing this with the standard wave equation $y = a \cos(2 \pi \nu t - kx + \phi_0)$,we get:
Propagation constant $k = 0.0160 \pi \, rad/cm$.
The phase difference $\Delta \phi$ between two points separated by distance $\Delta x$ is given by $\Delta \phi = k \Delta x$.
$(a)$ For $\Delta x = 4 \, m = 400 \, cm$:
$\Delta \phi = (0.0160 \pi \, rad/cm) \times (400 \, cm) = 6.4 \pi \, rad$.
$(b)$ For $\Delta x = 0.5 \, m = 50 \, cm$:
$\Delta \phi = (0.0160 \pi \, rad/cm) \times (50 \, cm) = 0.8 \pi \, rad$.
$(c)$ For $\Delta x = \lambda / 2$:
Since $k = 2 \pi / \lambda$,$\Delta \phi = (2 \pi / \lambda) \times (\lambda / 2) = \pi \, rad$.
$(d)$ For $\Delta x = 3 \lambda / 4$:
$\Delta \phi = (2 \pi / \lambda) \times (3 \lambda / 4) = 1.5 \pi \, rad$.

(A) The given harmonic wave is $y(x, t) = 7.5 \sin (0.0050 x + 12 t + \pi / 4)$.
$(a)$ For $x = 1 \; cm$ and $t = 1 \; s$:
$y(1, 1) = 7.5 \sin (0.0050 + 12 + \pi / 4) = 7.5 \sin (12.0050 + 0.785) = 7.5 \sin (12.79 \; rad)$.
Converting to degrees: $12.79 \times (180 / \pi) \approx 732.81^{\circ}$.
$y = 7.5 \sin (732.81^{\circ}) = 7.5 \sin (8 \times 90^{\circ} + 12.81^{\circ}) = 7.5 \sin (12.81^{\circ}) \approx 7.5 \times 0.2217 \approx 1.663 \; cm$.
The velocity of oscillation is $v = \frac{\partial y}{\partial t} = 7.5 \times 12 \cos (0.0050 x + 12 t + \pi / 4) = 90 \cos (0.0050 x + 12 t + \pi / 4)$.
At $x = 1 \; cm, t = 1 \; s$: $v = 90 \cos (12.79 \; rad) = 90 \cos (12.81^{\circ}) \approx 90 \times 0.975 = 87.75 \; cm/s$.
The wave propagation velocity $v_p = \omega / k = 12 / 0.0050 = 2400 \; cm/s$.
Since $87.75 \; cm/s \neq 2400 \; cm/s$,the oscillation velocity is not equal to the wave propagation velocity.
$(b)$ The wavelength $\lambda = 2 \pi / k = 2 \pi / 0.0050 = 1256 \; cm = 12.56 \; m$.
Points with the same displacement and velocity are separated by integer multiples of the wavelength $\lambda$. Thus,the points are $x = (1 \; cm \pm n \lambda)$,where $n = 1, 2, 3, \dots$.

(N/A) The equation of a travelling wave propagating along the positive $x$-direction is given by $y(x, t) = a \sin(\omega t - kx + \phi)$.
Given $y(0, 0) = 0$,we have $a \sin(\phi) = 0$,so $\phi = 0$.
Linear mass density $\mu = 8.0 \times 10^{-3} \; kg \; m^{-1}$.
Frequency $v = 256 \; Hz$.
Amplitude $a = 5.0 \; cm = 0.05 \; m$.
Tension $T = mg = 90 \times 9.8 = 882 \; N$.
Wave velocity $v = \sqrt{T/\mu} = \sqrt{882 / (8.0 \times 10^{-3})} = \sqrt{110250} \approx 332 \; m/s$.
Angular frequency $\omega = 2\pi v = 2 \times 3.14159 \times 256 \approx 1608 \; rad/s$.
Propagation constant $k = \omega / v = 1608 / 332 \approx 4.84 \; m^{-1}$.
Substituting these values,the displacement equation is $y(x, t) = 0.05 \sin(1608t - 4.84x) \; m$.

(N/A) The magnitude of the maximum displacement of a particle participating in the propagation of a wave is called the amplitude of the wave. It is represented by the symbol $a$ or $A$.
According to the wave equation:
$y = a \sin(kx - \omega t + \phi)$
Since the value of $\sin(kx - \omega t + \phi)$ has extreme values $\pm 1$,we can write:
$y_{\max} = a(\pm 1) = \pm a$
Therefore,the amplitude of the wave is $|y_{\max}| = a$.
The amplitude of a wave is always positive. Its $SI$ unit is $m$ and its dimensional formula is $[M^0 L^1 T^0]$.
Initial phase $(\phi)$: If we know the initial position of a particle at the origin of the wave and the direction of its motion at time $t = 0$,we can find the value of the initial phase $\phi$ using:
$y(x, t) = a \sin(\omega t - kx + \phi)$
Putting $x = 0$ and $t = 0$,we get $y(0, 0) = a \sin \phi$.
By knowing $\sin \phi$,we can determine the initial phase $\phi$.
In the wave equation $y = y(x, t) = a \sin(\omega t - kx + \phi)$,the argument of the sine function $(\omega t - kx + \phi)$ is called the total phase of the wave at time $t$ at a distance $x$ from its source. It represents the total phase of oscillation of a particle at distance $x$ from the origin of the wave at time $t$.

(N/A) Wavelength: The linear distance between any two consecutive points or particles having a phase difference of $2 \pi \text{ rad}$ is called the wavelength $(\lambda)$ of the wave. Its $SI$ unit is metre $(m)$.
Wave Number: The number of waves per unit distance is called the wave number $(\bar{\nu})$. It is defined as the reciprocal of the wavelength.
Equation for Wavelength: For a wave represented by $y(x, t) = a \sin(kx - \omega t)$,the displacement repeats after a distance $\lambda = \frac{2 \pi}{k}$.
Equation for Wave Number: $\bar{\nu} = \frac{1}{\lambda} = \frac{k}{2 \pi}$.
Units:
- Wavelength $(\lambda)$: $SI$ unit is metre $(m)$.
- Wave Number $(\bar{\nu})$: $SI$ unit is metre$^{-1}$ $(m^{-1})$.

(N/A) Time period: The time taken by any string element to complete one full oscillation is called the Time period $(T)$.
In the equation of displacement of a wave,$y(x, t) = a \sin (kx - \omega t + \phi)$. If $\phi = 0$,then observing the motion of an element at $x = 0$,we get $y(0, t) = a \sin(-\omega t) = -a \sin(\omega t)$. This represents simple harmonic motion $(SHM)$ with amplitude $a$ and time period $T$. Since the sine function repeats at $2\pi$ intervals,$\omega T = 2\pi$,which gives $\omega = \frac{2\pi}{T}$.
Angular frequency: The angular frequency of oscillations of particles of the medium in a wave is called the angular frequency of the wave. Its symbol is $\omega$,its $SI$ unit is $\text{rad } s^{-1}$,and its dimensional formula is $[M^0 L^0 T^{-1}]$.
Frequency: The number of oscillations performed by a particle of the medium in one second is known as the frequency of oscillation. Its symbol is $\nu$ or $f$. Since $f = \frac{1}{T}$,the $SI$ unit of frequency is $s^{-1}$ or $\text{Hz}$ (Hertz),and its dimensional formula is $[M^0 L^0 T^{-1}]$.

(N/A) The displacement function for a harmonic wave is given by the equation: $s(x, t) = a \sin(kx - \omega t + \phi)$.
Here,$s(x, t)$ represents the displacement of an element of the medium at position $x$ and time $t$.
$a$ is the amplitude of the wave.
$k$ is the angular wave number,defined as $k = \frac{2\pi}{\lambda}$.
$\omega$ is the angular frequency,defined as $\omega = 2\pi f$.
$\phi$ is the initial phase constant.

(N/A) Wave speed: The distance covered by a wave in unit time is called wave speed.
Its unit is $m/s$ and its dimensional formula is $[M^0 L^1 T^{-1}]$.
Let $\lambda$ be the wavelength (distance covered by the wave in one time period $T$).
By definition,wave speed $v = \frac{\text{distance}}{\text{time}} = \frac{\lambda}{T}$.
Since frequency $f = \frac{1}{T}$,we have $v = f \lambda$.
Multiplying and dividing by $2\pi$,we get $v = \frac{2\pi f \lambda}{2\pi}$.
Using the relations $\omega = 2\pi f$ (angular frequency) and $k = \frac{2\pi}{\lambda}$ (wave number),we substitute these into the equation:
$v = \frac{\omega}{k}$.

(N/A) wave equation is a second-order linear partial differential equation that describes the propagation of various types of waves,such as sound waves,light waves,and water waves.
For a one-dimensional wave traveling along the $x$-axis with a wave speed $v$,the wave equation is given by:
$\frac{\partial^2 y}{\partial x^2} = \frac{1}{v^2} \frac{\partial^2 y}{\partial t^2}$
where $y(x, t)$ represents the displacement of the wave at position $x$ and time $t$.

(N/A) Definition: The amplitude of a wave is defined as the maximum displacement of the particles of the medium from their mean or equilibrium position during the propagation of the wave.
Unit: The $SI$ unit of amplitude is the meter $(m)$.
Dimensional Formula: Since amplitude represents a length,its dimensional formula is $[M^0 L^1 T^0]$.

(N/A) Wavelength is defined as the distance between two consecutive crests or two consecutive troughs in a transverse wave,or the distance between two consecutive compressions or two consecutive rarefactions in a longitudinal wave.
It represents the distance covered by the wave in one complete oscillation.
The $SI$ unit of wavelength is the meter $(m)$.

(N/A) The angular wave number,also known as the propagation constant,is defined as the phase change per unit distance of a wave.
It represents the number of radians of phase change per unit length.
Mathematically,the angular wave number $k$ is given by:
$k = \frac{2\pi}{\lambda}$
where $\lambda$ is the wavelength of the wave.
The $SI$ unit of angular wave number is $\text{rad/m}$ or $\text{m}^{-1}$.

(N/A) The time period of a wave is defined as the time taken by a particle of the medium to complete one full oscillation or vibration as the wave passes through it.
Alternatively,it is the time taken by the wave to travel a distance equal to one wavelength $(\lambda)$.
It is denoted by the symbol $T$ and its $SI$ unit is the second $(s)$.

(N/A) The angular frequency $(\omega)$ of a wave is defined as the rate of change of the phase of the wave with respect to time.
It represents the number of radians the wave oscillates per unit of time.
Mathematically, it is given by $\omega = 2\pi f$, where $f$ is the frequency of the wave.
The $SI$ unit of angular frequency is radians per second $(rad/s)$.

(N/A) The wave speed is defined as the distance traveled by a wave per unit time. It is the speed at which the wave disturbance propagates through a medium. Mathematically, it is given by the product of the wave frequency $(f)$ and the wavelength $(\lambda)$: $v = f \lambda$. In terms of angular frequency $(\omega)$ and wave number $(k)$, it is expressed as $v = \frac{\omega}{k}$.

(N/A) The general equation of a traveling wave is given by $y(x, t) = A \sin(kx - \omega t + \phi)$.
Here,$k$ is the angular wave number and $\omega$ is the angular frequency.
The phase of the wave is given by $\Phi = kx - \omega t$.
For a constant phase (the position of a specific point on the wave),we have $kx - \omega t = \text{constant}$.
Differentiating both sides with respect to time $t$,we get $k \frac{dx}{dt} - \omega = 0$.
Since the wave velocity $v$ is defined as $v = \frac{dx}{dt}$,we substitute this into the equation:
$kv - \omega = 0$.
Therefore,the relation is $v = \frac{\omega}{k}$.

(N/A) Wave speed is defined as the distance traveled by a wave per unit time. It represents the speed at which the wave profile or the disturbance moves through the medium.
Mathematically, it is given by the product of the frequency $(f)$ and the wavelength $(\lambda)$: $v = f \lambda$.
The $SI$ unit of wave speed is meters per second $(m/s)$.