Wave Equation and Characteristics of Waves Questions in English

(C) The velocity of a wave $(v)$ is defined as the product of its frequency $(n)$ and its wavelength $(\lambda)$.
Mathematically,this is expressed as: $v = n \times \lambda$.
To find the frequency $(n)$,we rearrange the formula:
$n = \frac{v}{\lambda}$.
Therefore,the correct relation is $n = v/\lambda$.

(A) The distance between two consecutive crests is defined as the wavelength $(\lambda)$.
Given,$\lambda = 5 \ cm$.
The number of complete waves passing through a point per second is the frequency ($n$ or $f$).
Given,$n = 2 \ Hz$ (or $2 \ waves/sec$).
The velocity of the wave $(v)$ is given by the formula: $v = n \times \lambda$.
Substituting the values: $v = 2 \times 5 = 10 \ cm/sec$.

(C) 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 $\Delta \phi = 1.6 \pi$ and $\Delta x = 40 \; cm = 0.4 \; m$.
Substituting these values: $1.6 \pi = \frac{2\pi}{\lambda} \times 0.4$.
Solving for wavelength $(\lambda)$: $\lambda = \frac{2 \times 0.4}{1.6} = \frac{0.8}{1.6} = 0.5 \; m$.
Using the wave equation $v = f \lambda$,where $v = 330 \; m/s$ and $\lambda = 0.5 \; m$:
$330 = f \times 0.5$.
Therefore,$f = \frac{330}{0.5} = 660 \; Hz$.

(A) For a wave,the phase change over a full wavelength $\lambda$ is $2\pi$ radians.
Since the phase changes linearly with distance,the phase difference $(\Delta \phi)$ for a path difference $(\Delta x)$ is given by the ratio of the path difference to the wavelength,multiplied by the total phase change of $2\pi$.
Therefore,the formula is $\Delta \phi = \frac{2\pi}{\lambda} \Delta x$.

(A) In a sinusoidal wave,the motion from maximum displacement (amplitude) to zero displacement corresponds to one-quarter of the total time period $(T)$.
Therefore,the time taken is $t = \frac{T}{4}$.
Since the frequency ($n$ or $f$) is the reciprocal of the time period,$T = \frac{1}{n}$.
Substituting this into the equation,we get $t = \frac{1}{4n}$.
Rearranging for frequency,$n = \frac{1}{4t}$.
Given $t = 0.170\,s$,we calculate $n = \frac{1}{4 \times 0.170} = \frac{1}{0.680} \approx 1.47\,Hz$.

(B) The number of waves per unit length is defined as the wave number.
It is the reciprocal of the wavelength $(\lambda)$.
Mathematically,it is expressed as $\overline{n} = \frac{1}{\lambda}$.
Therefore,the correct option is $B$.

(D) The relationship between velocity $(v)$, frequency $(f)$, and wavelength $(\lambda)$ is given by the equation $v = f \lambda$. These three parameters are intrinsically linked to the wave propagation and the medium. Amplitude, however, represents the maximum displacement of particles from their mean position and is independent of the wave's velocity, frequency, or wavelength. Therefore, amplitude is different from the others.

(C) Given: Path difference $\Delta x = 1 \ m$,Frequency $f = 120 \ Hz$,Phase difference $\phi = 90^o = \frac{\pi}{2} \text{ radians}$.
The relationship between phase difference and path difference is given by $\phi = \frac{2\pi}{\lambda} \Delta x$.
Substituting the values: $\frac{\pi}{2} = \frac{2\pi}{\lambda} \times 1$.
Solving for wavelength $\lambda$: $\lambda = 4 \ m$.
The wave velocity $v$ is given by $v = f \lambda$.
$v = 120 \times 4 = 480 \ m/s$.

(D) Given:
Speed of the wave,$v = 960\, m/s$.
Number of waves,$N = 3600$.
Time taken,$t = 1\, minute = 60\, s$.
First,calculate the frequency $(n)$ of the wave:
$n = \frac{N}{t} = \frac{3600}{60} = 60\, Hz$.
Now,use the wave equation $v = n \lambda$ to find the wavelength $(\lambda)$:
$\lambda = \frac{v}{n} = \frac{960}{60} = 16\, m$.
Therefore,the wavelength is $16\, m$.

(C) The frequency $n$ is given by the number of waves per second.
Given,$54$ waves per minute.
$n = \frac{54}{60} \, Hz = 0.9 \, Hz$.
The wavelength $\lambda = 10 \, m$.
The wave velocity $v$ is calculated using the formula $v = n \times \lambda$.
$v = 0.9 \times 10 = 9 \, m/s$.
Therefore,the wave velocity is $9 \, m/s$.

(B) The frequency $n$ of the wave is the number of waves passing through a point per unit time.
Given: Number of waves = $3600$, Time $t = 2\, \text{minutes} = 2 \times 60 = 120\, \text{seconds}$.
Frequency $n = \frac{3600}{120} = 30\, \text{Hz}$.
The relationship between wave speed $v$, frequency $n$, and wavelength $\lambda$ is given by $v = n \lambda$.
Therefore, $\lambda = \frac{v}{n} = \frac{760}{30} = 25.33\, \text{m}$.
Rounding to one decimal place, the wavelength is $25.3\, \text{m}$.

(C) The wave number $\bar{\nu}$ is defined as the reciprocal of the wavelength $\lambda$.
$\bar{\nu} = \frac{1}{\lambda}$
Given,$\lambda = 6000 \mathring{A} = 6000 \times 10^{-10} \ m = 6 \times 10^{-7} \ m$.
Substituting the value of $\lambda$ in the formula:
$\bar{\nu} = \frac{1}{6 \times 10^{-7} \ m}$
$\bar{\nu} = \frac{1}{6} \times 10^7 \ m^{-1}$
$\bar{\nu} = 0.1666 \times 10^7 \ m^{-1} = 1.66 \times 10^6 \ m^{-1}$.
Therefore,the correct option is $C$.

(C) The relationship between path difference $(\Delta x)$ and phase difference $(\phi)$ is given by the formula: $\Delta x = \frac{\lambda}{2\pi} \times \phi$.
Given phase difference $\phi = 60^{\circ}$.
Convert the phase difference into radians: $\phi = 60^{\circ} \times \frac{\pi}{180^{\circ}} = \frac{\pi}{3} \text{ radians}$.
Substitute the value of $\phi$ into the formula: $\Delta x = \frac{\lambda}{2\pi} \times \frac{\pi}{3}$.
Simplifying the expression: $\Delta x = \frac{\lambda}{6}$.
Therefore,the path difference is $\lambda / 6$.

(C) The distance between two successive crests in a wave is equal to one wavelength,denoted by $\lambda$.
The relationship between phase difference $\phi$ and path difference $\Delta x$ is given by the formula $\phi = \frac{2\pi}{\lambda} \times \Delta x$.
Substituting the path difference $\Delta x = \lambda$ into the formula,we get:
$\phi = \frac{2\pi}{\lambda} \times \lambda = 2\pi$.
Therefore,the phase difference between two successive crests is $2\pi$ radians.

(B) Given frequency $n = 500 \, Hz$ and velocity $v = 360 \, m/s$.
First,calculate the wavelength $\lambda$ using the formula $v = n\lambda$:
$\lambda = \frac{v}{n} = \frac{360}{500} = 0.72 \, m$.
The phase difference $\phi$ is given as $60^o$. Converting this to radians:
$\phi = 60^o \times \frac{\pi}{180^o} = \frac{\pi}{3} \, \text{radians}$.
The path difference $\Delta x$ is related to the phase difference $\phi$ by the formula:
$\Delta x = \frac{\lambda}{2\pi} \times \phi$.
Substituting the values:
$\Delta x = \frac{0.72}{2\pi} \times \frac{\pi}{3} = \frac{0.72}{6} = 0.12 \, m$.
Converting the distance to centimeters:
$\Delta x = 0.12 \times 100 = 12 \, cm$.

(D) The frequency $n$ is the number of waves per second.
Given,number of waves per minute $= 54$.
Therefore,frequency $n = \frac{54}{60} \ Hz = 0.9 \ Hz$.
The wavelength $\lambda = 10 \ m$.
The velocity of the wave is given by the formula $v = n \lambda$.
Substituting the values,$v = 0.9 \times 10 = 9 \ m/s$.

(D) The standard equation of a progressive wave is given by $y = A \sin(kx - \omega t)$.
Comparing the given equation $y = 2 \sin \pi (0.5x - 200t)$ with the standard form,we get:
$y = 2 \sin (0.5\pi x - 200\pi t)$.
Here,the wave number $k = 0.5\pi$ and the angular frequency $\omega = 200\pi$.
The wave velocity $v$ is given by the ratio of the coefficient of $t$ to the coefficient of $x$:
$v = \frac{\omega}{k} = \frac{200\pi}{0.5\pi} = 400 \ cm/sec$.

(B) The time interval between two successive crests is defined as the time period $T$ of the wave.
Given,$T = 0.2 \ s$.
The frequency $f$ (or $n$) of the wave is the reciprocal of the time period:
$f = \frac{1}{T} = \frac{1}{0.2 \ s} = 5 \ Hz$.
Therefore,the frequency of the wave is $5 \ Hz$.

(C) The standard equation of a transverse wave is given by $y = A \sin(kx - \omega t)$.
Comparing the given equation $y = 10 \sin \pi (0.01x - 2t)$ with the standard form,we get:
$y = 10 \sin (0.01\pi x - 2\pi t)$.
Here,the angular frequency $\omega = 2\pi \text{ rad/s}$.
We know that the frequency $f$ is related to angular frequency by the formula $\omega = 2\pi f$.
Substituting the value of $\omega$:
$2\pi = 2\pi f$.
Therefore,$f = 1 \text{ sec}^{-1}$.

(D) The given wave equation is $y = 4\sin \frac{\pi }{2}\left( {8t - \frac{x}{8}} \right)$.
Expanding the term inside the sine function,we get:
$y = 4\sin \left( {4\pi t - \frac{{\pi x}}{{16}}} \right)$.
The standard form of a travelling wave is $y = A\sin(\omega t - kx)$.
Here,the angular frequency $\omega = 4\pi$ and the wave number $k = \frac{\pi}{16}$.
The velocity of the wave is given by $v = \frac{\omega}{k}$.
Substituting the values: $v = \frac{4\pi}{\pi/16} = 4\pi \times \frac{16}{\pi} = 64\, cm/s$.
Since the sign between $\omega t$ and $kx$ is negative,the wave is travelling in the $+x$ direction.

(D) The first wave is given by $y_1 = a \sin(\omega t - kx)$.
The second wave is given by $y_2 = a \cos(\omega t - kx)$.
Using the trigonometric identity $\cos(\theta) = \sin(\theta + \frac{\pi}{2})$,we can rewrite the second wave as:
$y_2 = a \sin(\omega t - kx + \frac{\pi}{2})$.
Comparing the phase of the two waves,the phase of $y_1$ is $(\omega t - kx)$ and the phase of $y_2$ is $(\omega t - kx + \frac{\pi}{2})$.
Therefore,the phase difference $\Delta \phi$ is $(\omega t - kx + \frac{\pi}{2}) - (\omega t - kx) = \frac{\pi}{2}$.

(C) The general equation for a plane wave traveling in the $y$-direction is given by $x = A \sin(\omega t \pm ky)$.
Comparing the given equation $x = 1.2 \sin(314t + 12.56y)$ with the standard form,we identify the wave number $k = 12.56 \ rad/m$.
The presence of the '$+$' sign between the $t$ term and the $y$ term indicates that the wave is traveling in the negative $y$-direction ($-ve \ y$ direction).
The wave number $k$ is related to the wavelength $\lambda$ by the formula $k = \frac{2\pi}{\lambda}$.
Substituting the values: $12.56 = \frac{2 \times 3.14}{\lambda}$.
$\lambda = \frac{6.28}{12.56} = 0.5 \ m$.
Thus,the wave has a wavelength of $0.5 \ m$ and travels in the $-ve \ y$ direction.

(C) The standard equation for a simple harmonic wave is $y = a \sin(\omega t - kx)$.
Comparing the given equation $y = \frac{10}{\pi} \sin \left( 2000\pi t - \frac{\pi x}{17} \right)$ with the standard equation,we get:
Amplitude $a = \frac{10}{\pi} \, cm = \frac{0.1}{\pi} \, m$ (since $1 \, cm = 10^{-2} \, m$).
Angular frequency $\omega = 2000\pi \, rad/s$.
$1$. Maximum velocity of the particles $(v_{\max})$:
$v_{\max} = a\omega = \left( \frac{0.1}{\pi} \right) \times (2000\pi) = 0.1 \times 2000 = 200 \, m/s$.
$2$. Periodic time $(T)$:
We know $\omega = \frac{2\pi}{T}$.
$2000\pi = \frac{2\pi}{T} \implies T = \frac{2\pi}{2000\pi} = \frac{1}{1000} = 10^{-3} \, s$.
Therefore,the periodic time is $10^{-3} \, s$ and the maximum velocity is $200 \, m/s$.

(B) The standard equation of a travelling wave is given by $y = a \cos(\omega t - kx)$.
Comparing the given equation $y = 3 \cos \pi(100t - x)$ with the standard form,we rewrite it as $y = 3 \cos(100\pi t - \pi x)$.
Here,the wave number $k$ is the coefficient of $x$,so $k = \pi$.
We know that the relationship between wave number $k$ and wavelength $\lambda$ is $k = \frac{2\pi}{\lambda}$.
Substituting the value of $k$,we get $\pi = \frac{2\pi}{\lambda}$.
Solving for $\lambda$,we find $\lambda = 2 \ cm$.

(B) Comparing the given equation $Y = Y_0 \sin 2\pi \left( ft - \frac{x}{\lambda} \right)$ with the standard wave equation $y = a \sin(\omega t - kx)$:
We identify the amplitude $a = Y_0$,angular frequency $\omega = 2\pi f$,and wave number $k = \frac{2\pi}{\lambda}$.
The maximum particle velocity is given by $(v_{\max})_{\text{particle}} = a\omega = Y_0 \times 2\pi f$.
The wave velocity (phase velocity) is given by $v_{\text{wave}} = \frac{\omega}{k} = \frac{2\pi f}{2\pi / \lambda} = f\lambda$.
According to the problem,the maximum particle velocity is four times the wave velocity:
$(v_{\max})_{\text{particle}} = 4 v_{\text{wave}}$
$Y_0 \times 2\pi f = 4 f\lambda$
Dividing both sides by $4f$,we get:
$\lambda = \frac{2\pi Y_0}{4} = \frac{\pi Y_0}{2}$.

(D) Comparing the given equation $y = 10^4 \sin(60t + 2x)$ with the standard wave equation $y = a \sin(\omega t + kx)$:
$1$. Since the sign between $\omega t$ and $kx$ is positive,the wave is travelling in the negative $X$-direction.
$2$. The angular frequency is $\omega = 60 \, rad/s$. The frequency $f$ is given by $f = \frac{\omega}{2\pi} = \frac{60}{2\pi} = \frac{30}{\pi} \, Hz$.
$3$. The wave number is $k = 2 \, m^{-1}$. The wavelength $\lambda$ is given by $k = \frac{2\pi}{\lambda} \implies \lambda = \frac{2\pi}{k} = \frac{2\pi}{2} = \pi \, m$.
$4$. The wave velocity $v$ is given by $v = \frac{\omega}{k} = \frac{60}{2} = 30 \, m/s$.
Since all the statements are correct,the correct option is $(d)$.

(B) The general equation for a wave traveling in the negative $x$-direction is given by $y(x, t) = A \sin(kx + \omega t + \phi)$ or $A \cos(kx + \omega t + \phi)$.
Given: Amplitude $A = 0.5\, m$,Wavelength $\lambda = 1\, m$,Frequency $f = 2\, Hz$.
Calculate the angular wave number $k = \frac{2\pi}{\lambda} = \frac{2\pi}{1} = 2\pi\, rad/m$.
Calculate the angular frequency $\omega = 2\pi f = 2\pi(2) = 4\pi\, rad/s$.
Substituting these values into the wave equation: $y(x, t) = 0.5 \cos(2\pi x + 4\pi t)$.
Thus,option $B$ is correct.

(B) The standard wave equation is given by $y = A \sin(\omega t - kx)$.
Comparing this with the given equation $y = 5 \times 10^{-4} \sin(100t - 50x)$,we identify the angular frequency $\omega = 100 \, rad/s$ and the wave number $k = 50 \, rad/m$.
The wave velocity $v$ is defined as the ratio of the coefficient of $t$ to the coefficient of $x$:
$v = \frac{\omega}{k} = \frac{100}{50} = 2 \, m/s$.
Therefore,the velocity of the wave is $2 \, m/s$.

(D) travelling wave must be a function of the form $y = f(x \pm vt)$.
In options $A$,$B$,and $C$,the arguments are of the form $(x - vt)$,$(x + vt)$,and $(x - vt)$ respectively,which satisfy the condition for a travelling wave.
In option $D$,the argument is $(x^2 - vt^2)$. This does not represent a travelling wave because it cannot be expressed as a function of $(x \pm vt)$.
Therefore,$y = f(x^2 - vt^2)$ does not represent a travelling wave.

(D) The standard wave equation for a wave traveling in the negative $x$-direction is $y = A\sin(kx + \omega t + \phi_0)$.
Comparing the given equation $Y = A\sin(10\pi x + 15\pi t + \frac{\pi}{3})$ with the standard form,we identify the angular frequency $\omega = 15\pi\,rad/s$ and the wave number $k = 10\pi\,rad/m$.
The wave velocity $v$ is given by $v = \frac{\omega}{k} = \frac{15\pi}{10\pi} = 1.5\,m/s$. Since the signs of the $x$ and $t$ terms are both positive,the wave travels in the negative $x$-direction.
The wavelength $\lambda$ is given by $\lambda = \frac{2\pi}{k} = \frac{2\pi}{10\pi} = 0.2\,m$.
Thus,both statements $(b)$ and $(c)$ are correct.

(A) The given wave equation is $y = a \cos(kx - \omega t + \phi)$.
Comparing this with the standard equation $y = 3 \cos \left( \frac{x}{4} - 10t - \frac{\pi}{2} \right)$,we get the amplitude $a = 3$ and the angular frequency $\omega = 10$.
The maximum velocity of the particles of the medium is given by the formula $v_{\max} = a \omega$.
Substituting the values,we get $v_{\max} = 3 \times 10 = 30$.

(B) Given the two wave equations:
$y_1 = a_1 \sin \left( \omega t - \frac{2\pi x}{\lambda} \right)$
$y_2 = a_2 \cos \left( \omega t - \frac{2\pi x}{\lambda} + \phi \right)$
To compare the phases,convert the cosine function to a sine function using the identity $\cos(\theta) = \sin(\theta + \frac{\pi}{2})$:
$y_2 = a_2 \sin \left( \omega t - \frac{2\pi x}{\lambda} + \phi + \frac{\pi}{2} \right)$
The phase difference $\Delta \Phi$ is the difference between the arguments of the sine functions:
$\Delta \Phi = \left( \omega t - \frac{2\pi x}{\lambda} + \phi + \frac{\pi}{2} \right) - \left( \omega t - \frac{2\pi x}{\lambda} \right) = \phi + \frac{\pi}{2}$
The relationship between path difference $\Delta x$ and phase difference $\Delta \Phi$ is given by $\Delta x = \frac{\lambda}{2\pi} \Delta \Phi$.
Substituting the phase difference:
$\Delta x = \frac{\lambda}{2\pi} \left( \phi + \frac{\pi}{2} \right)$.

(A) The general equation of a traveling wave is $y = a \sin(\omega t \pm kx + \phi)$.
For the first wave,$y_1 = a \sin(\omega t - kx)$,the negative sign before $kx$ indicates that the wave is traveling in the positive $x$-direction.
For the second wave,$y_2 = a \sin(\omega t + kx)$,the positive sign before $kx$ indicates that the wave is traveling in the negative $x$-direction.
Since the waves are traveling in opposite directions,option $A$ is correct.

(A) The standard equation of a travelling wave is given by $y = A \sin(\omega t - kx)$.
Comparing the given equation $y = 0.5 \sin(10t - x)$ with the standard equation,we get:
Angular frequency $\omega = 10 \ rad/s$
Wave number $k = 1 \ rad/m$
The velocity of the wave $v$ is given by the ratio of the coefficient of $t$ to the coefficient of $x$:
$v = \frac{\omega}{k} = \frac{10}{1} = 10 \ m/s$.

(A) The general equation of a travelling wave is given by $y = A \sin(\omega t - kx + \phi)$.
Comparing the given equation $y = 7 \sin(7\pi t - 0.04\pi x + \frac{\pi}{3})$ with the general equation,we identify the angular frequency $\omega = 7\pi \, rad/s$ and the wave number $k = 0.04\pi \, rad/m$.
The speed of the wave $v$ is given by the ratio of the coefficient of $t$ to the coefficient of $x$ (ignoring the sign),which is $v = \frac{\omega}{k}$.
Substituting the values: $v = \frac{7\pi}{0.04\pi} = \frac{7}{0.04} = \frac{700}{4} = 175 \, m/s$.

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

(D) The given wave equation is $y = y_0 \sin \frac{2\pi}{\lambda} (vt - x)$.
Comparing this with the standard wave equation $y = a \sin \frac{2\pi}{\lambda} (vt - x)$,we identify the amplitude $a = y_0$ and wave velocity $v_{wave} = v$.
The particle velocity $v_p$ is given by the derivative of displacement with respect to time: $v_p = \frac{\partial y}{\partial t} = y_0 \cdot \frac{2\pi v}{\lambda} \cos \frac{2\pi}{\lambda} (vt - x)$.
The maximum particle velocity is $(v_{max})_{particle} = y_0 \cdot \frac{2\pi v}{\lambda}$.
According to the problem,$(v_{max})_{particle} = 2 \cdot v_{wave}$.
Substituting the values: $\frac{y_0 \cdot 2\pi v}{\lambda} = 2v$.
Canceling $v$ from both sides: $\frac{2\pi y_0}{\lambda} = 2$.
Solving for $\lambda$: $\lambda = \pi y_0$.

(A) The displacement of the particle in the string is given by $y = A\sin (kx - \omega t)$.
To find the particle velocity,we differentiate the displacement $y$ with respect to time $t$:
$v = \frac{dy}{dt} = \frac{d}{dt} [A\sin (kx - \omega t)]$
$v = A \cos (kx - \omega t) \cdot (-\omega)$
$v = -A\omega \cos (kx - \omega t)$
The maximum value of the particle velocity occurs when the magnitude of the cosine function is $1$.
Therefore,$v_{\max} = | -A\omega | = A\omega$.

(A) The standard equation for a traveling wave is given by $y(x, t) = a \sin(\omega t - kx)$.
Given equation: $y(x, t) = 0.03 \sin(2\pi t - 0.01\pi x)$.
Comparing the given equation with the standard form,we identify the wave number $k$ as the coefficient of $x$:
$k = 0.01\pi$.
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 $k$:
$0.01\pi = \frac{2\pi}{\lambda}$.
Solving for $\lambda$:
$\lambda = \frac{2\pi}{0.01\pi} = \frac{2}{0.01} = 200 \ m$.

(B) The phase difference $\Delta \phi$ between the vibrations of two particles of a medium separated by a distance $\Delta x$ is given by the formula:
$\Delta \phi = \frac{2 \pi}{\lambda} \Delta x$
where $\lambda$ is the wavelength of the wave.
From this expression,it is clear that the phase difference depends directly on the path difference $\Delta x$,which is the distance separating the two particles. Therefore,the phase difference varies with the distance separating them.

(D) Comparing the given equation $y = 3 \sin 2\pi \left( \frac{t}{0.04} - \frac{x}{0.01} \right)$ with the standard wave equation $y = a \sin 2\pi \left( \frac{t}{T} - \frac{x}{\lambda} \right)$:
$1$. The time period $T = 0.04 \, s$.
$2$. The frequency $f = \frac{1}{T} = \frac{1}{0.04} = 25 \, Hz$.
$3$. The amplitude $a = 3 \, cm$.
$4$. The angular frequency $\omega = \frac{2\pi}{T} = \frac{2\pi}{0.04} = 50\pi \, rad/s$.
$5$. The maximum acceleration of a particle is given by $a_{max} = \omega^2 a$.
$a_{max} = (50\pi)^2 \times 3 = 2500 \times \pi^2 \times 3 = 7500 \times 9.8696 \approx 7.4 \times 10^4 \, cm/s^2$.

(D) The standard form of a sinusoidal wave equation is $y = a \cos(\omega t + kx + \phi)$.
Comparing the given equation $y = 0.40 \cos(2000t + 0.80x)$ with the standard form,we identify the angular frequency $\omega = 2000 \text{ rad/s}$.
The relationship between angular frequency $\omega$ and frequency $f$ is given by $\omega = 2\pi f$.
Substituting the value of $\omega$,we get $2000 = 2\pi f$.
Solving for $f$,we find $f = \frac{2000}{2\pi} = \frac{1000}{\pi} \text{ Hz}$.

(D) The general equation of a simple harmonic progressive wave is given by $y = A \sin (\omega t - kx + \phi)$ or $y = A \sin (kx - \omega t + \phi)$.
Option $D$,$Y = A \sin (at - bx + c)$,represents a progressive wave because it is a function of $(at - bx)$,which satisfies the wave equation $\frac{\partial^2 y}{\partial t^2} = v^2 \frac{\partial^2 y}{\partial x^2}$,where the wave speed $v = \frac{a}{b}$.
Options $A$,$B$,and $C$ do not represent a traveling wave as they lack the necessary dependence on both space $(x)$ and time $(t)$ in the required functional form.

(A) The standard equation of a transverse wave is given by $y = A \sin(kz - \omega t)$.
Comparing the given equation $y = 100 \sin \pi (0.04z - 2t)$ with the standard form,we distribute $\pi$ inside the parentheses:
$y = 100 \sin (0.04\pi z - 2\pi t)$.
Here,the angular frequency $\omega$ is the coefficient of $t$,which is $\omega = 2\pi \text{ rad/s}$.
The relationship between angular frequency $\omega$ and frequency $f$ is $\omega = 2\pi f$.
Substituting the value of $\omega$:
$2\pi = 2\pi f$.
Therefore,$f = 1 \text{ Hz}$.

(A) The standard equation of a plane progressive wave is given by $y = a \sin (\omega t + kx)$.
Comparing the given equation $y = 0.025 \sin (100t + 0.25x)$ with the standard equation,we identify the angular frequency $\omega = 100 \text{ rad/s}$.
We know that the relationship between angular frequency $\omega$ and frequency $n$ is given by $\omega = 2\pi n$.
Substituting the value of $\omega$,we get $100 = 2\pi n$.
Solving for $n$,we find $n = \frac{100}{2\pi} = \frac{50}{\pi} \text{ Hz}$.

(B) The standard equation of a traveling wave is given by $y = A \sin (kx + \omega t + \phi)$.
Comparing the given equation $y = 0.0015 \sin (62.4x + 316t)$ with the standard form,we identify the wave number $k$ as $62.4 \, \text{rad/unit}$.
The relationship between the wave number $k$ and the wavelength $\lambda$ is given by $k = \frac{2\pi}{\lambda}$.
Substituting the value of $k$,we get $62.4 = \frac{2 \times 3.14}{\lambda}$.
Solving for $\lambda$: $\lambda = \frac{6.28}{62.4} \approx 0.1 \, \text{unit}$.

(B) The given equation of the progressive wave is $Y = a \sin(\omega t - kx)$.
Comparing this with the given equation $Y = 0.5 \sin(10\pi t - 5x)$,we get the amplitude $a = 0.5 \text{ cm}$ and the angular frequency $\omega = 10\pi \text{ rad/s}$.
The maximum velocity of a particle in a simple harmonic wave is given by the formula $v_{\max} = a\omega$.
Substituting the values,we get $v_{\max} = 0.5 \times 10\pi = 5\pi \text{ cm/s}$.