Wave Equation and Characteristics of Waves Questions in English

(C) Given: Velocity $v = 330 \,m/s$, path difference $\Delta x = 40 \,cm = 0.4 \,m$, and phase difference $\Delta \phi = 1.6 \pi$.
We know the relationship between phase difference and path difference is given by $\Delta \phi = \frac{2 \pi}{\lambda} \Delta x$.
Substituting the values: $1.6 \pi = \frac{2 \pi}{0.4} \Delta x$ is not needed directly, instead use $\lambda = \frac{v}{f}$.
Thus, $\Delta \phi = \frac{2 \pi f}{v} \Delta x$.
Rearranging for frequency $f$: $f = \frac{\Delta \phi \cdot v}{2 \pi \cdot \Delta x}$.
Substituting the values: $f = \frac{1.6 \pi \cdot 330}{2 \pi \cdot 0.4}$.
$f = \frac{1.6 \cdot 330}{0.8} = 2 \cdot 330 = 660 \,Hz$.

(A) Given: Phase difference $\phi = 60^{\circ} = \frac{\pi}{3} \text{ radians}$.
Path difference $x = 0.6 \text{ m}$.
Frequency $n = 500 \text{ Hz}$.
The relationship between path difference $(x)$ and phase difference $(\phi)$ is given by $x = \frac{\lambda}{2\pi} \times \phi$.
Substituting the values: $0.6 = \frac{\lambda}{2\pi} \times \frac{\pi}{3}$.
$0.6 = \frac{\lambda}{6} \implies \lambda = 3.6 \text{ m}$.
The velocity of the wave is $v = n \lambda$.
$v = 500 \times 3.6 = 1800 \text{ m/s}$.
Converting to $\text{km/s}$: $v = 1.8 \text{ km/s}$.

(B) The general equation of a simple harmonic progressive wave is $y=A \sin 2 \pi \left( \frac{t}{T} - \frac{x}{\lambda} \right)$.
Given equation: $y=a \sin 2 \pi(b t-c x)$.
Comparing the two,we get amplitude $A=a$,frequency $n = \frac{1}{T} = b$,and wave number $\frac{1}{\lambda} = c$.
The maximum particle velocity is given by $(v_p)_{\max} = A \omega = A(2 \pi n) = a(2 \pi b) = 2 \pi ab$.
The wave velocity is given by $v = n \lambda = \frac{n}{1/\lambda} = \frac{b}{c}$.
According to the problem,$(v_p)_{\max} = \frac{1}{2} v$.
Substituting the values: $2 \pi ab = \frac{1}{2} \left( \frac{b}{c} \right)$.
Canceling $b$ from both sides: $2 \pi a = \frac{1}{2c}$.
Solving for $c$: $c = \frac{1}{4 \pi a}$.

(D) The standard equation of a traveling wave is given by $Y = A \sin(\omega t - kx + \phi)$.
Comparing this with the given equation $Y = 6 \sin(12 \pi t - 0.02 \pi x + \frac{\pi}{3})$:
We have angular frequency $\omega = 12 \pi \ rad/s$ and wave number $k = 0.02 \pi \ rad/m$.
The velocity of the wave $v$ is given by the ratio of angular frequency to wave number:
$v = \frac{\omega}{k}$
$v = \frac{12 \pi}{0.02 \pi}$
$v = \frac{12}{0.02} = \frac{1200}{2} = 600 \ m/s$.

(B) Given equations are:
$y_1 = a_1 \sin \left(\omega t - \frac{2 \pi x}{\lambda}\right)$
$y_2 = a_2 \cos \left(\omega t - \frac{2 \pi x}{\lambda} + \phi\right)$
We know that $\cos \theta = \sin \left(\theta + \frac{\pi}{2}\right)$.
Therefore,$y_2$ can be rewritten as:
$y_2 = a_2 \sin \left(\omega t - \frac{2 \pi x}{\lambda} + \phi + \frac{\pi}{2}\right)$
The phase difference $\Delta \Phi$ between the two waves is the difference in their arguments:
$\Delta \Phi = \left(\omega t - \frac{2 \pi x}{\lambda} + \phi + \frac{\pi}{2}\right) - \left(\omega t - \frac{2 \pi x}{\lambda}\right) = \phi + \frac{\pi}{2}$
The relationship between path difference $\Delta x$ and phase difference $\Delta \Phi$ is given by:
$\Delta x = \frac{\lambda}{2 \pi} \times \Delta \Phi$
Substituting the value of $\Delta \Phi$:
$\Delta x = \frac{\lambda}{2 \pi} \left(\phi + \frac{\pi}{2}\right)$

(D) The given wave equation is $y = A \sin(2\omega t - 2kx)$.
Comparing this with the standard wave equation $y = A \sin(\omega' t - k' x)$,we have angular frequency $\omega' = 2\omega$ and wave number $k' = 2k$.
The velocity of the particle is $v_p = \frac{dy}{dt} = A \cdot (2\omega) \cos(2\omega t - 2kx)$.
The maximum particle velocity is $v_{p,max} = 2A\omega$.
The wave velocity is $v_w = \frac{\omega'}{k'} = \frac{2\omega}{2k} = \frac{\omega}{k}$.
Given that $v_{p,max} = v_w$,we have $2A\omega = \frac{\omega}{k}$.
Thus,$A = \frac{1}{2k}$.
Since the wave number $k = \frac{2\pi}{\lambda}$,we substitute this into the expression for $A$:
$A = \frac{1}{2(2\pi / \lambda)} = \frac{\lambda}{4\pi}$.

(C) The standard equation of a traveling wave is given by $y = A \sin (\omega t - kx + \phi)$ or $y = A \sin (kx - \omega t + \phi)$.
Comparing the given equation $Y = 5 \sin (10 \pi t - 0.02 \pi x + \pi / 3)$ with the standard form,we identify the angular frequency $\omega$ and the wave number $k$:
$\omega = 10 \pi \ rad/s$
$k = 0.02 \pi \ rad/m$
The velocity of the wave $v$ is given by the ratio of angular frequency to wave number:
$v = \frac{\omega}{k}$
Substituting the values:
$v = \frac{10 \pi}{0.02 \pi}$
$v = \frac{10}{0.02} = \frac{1000}{2} = 500 \ m/s$
Therefore,the velocity of the wave is $500 \ m/s$.

(D) The given wave equation is $y = 10 \sin \left(\frac{2 \pi t}{30} + \alpha\right)$.
At $t = 0 \ s$,the displacement $y = 5 \ cm$.
Substituting these values into the equation: $5 = 10 \sin \left(\frac{2 \pi (0)}{30} + \alpha\right) \implies 5 = 10 \sin \alpha \implies \sin \alpha = \frac{1}{2}$.
Thus,the initial phase constant $\alpha = \frac{\pi}{6} \ rad$.
The total phase angle $\phi$ at any time $t$ is given by $\phi = \frac{2 \pi t}{30} + \alpha$.
At $t = 7.5 \ s$,the total phase angle is $\phi = \frac{2 \pi (7.5)}{30} + \frac{\pi}{6}$.
$\phi = \frac{15 \pi}{30} + \frac{\pi}{6} = \frac{\pi}{2} + \frac{\pi}{6}$.
$\phi = \frac{3 \pi + \pi}{6} = \frac{4 \pi}{6} = \frac{2 \pi}{3} \ rad$.

(B) The given wave equation is $Y = Y_0 \sin(2 \pi n t - \frac{2 \pi x}{\lambda})$.
Comparing this with the standard form $Y = Y_0 \sin(\omega t - kx)$,we get angular frequency $\omega = 2 \pi n$.
The maximum particle velocity $v_{p, \text{max}} = Y_0 \omega = Y_0 (2 \pi n) = 2 \pi n Y_0$.
The wave velocity $v = n \lambda$.
According to the problem,the wave velocity is $(1/8)^{\text{th}}$ of the maximum particle velocity:
$v = \frac{1}{8} v_{p, \text{max}}$
$n \lambda = \frac{1}{8} (2 \pi n Y_0)$
$n \lambda = \frac{\pi n Y_0}{4}$
$\lambda = \frac{\pi Y_0}{4}$.

(D) The standard equation for a wave is $Y = A \sin 2 \pi \left(\frac{t}{T} - \frac{x}{\lambda}\right)$.
For the first wave: $Y_1 = 2 \sin 2 \pi \left(\frac{4t}{0.2} - \frac{4x}{2}\right) = 2 \sin 2 \pi \left(\frac{t}{0.05} - \frac{x}{0.5}\right)$.
Here,$T_1 = 0.05 \text{ s}$ and $\lambda_1 = 0.5 \text{ m}$.
Velocity $v_1 = \frac{\lambda_1}{T_1} = \frac{0.5}{0.05} = 10 \text{ m/s}$.
For the second wave: $Y_2 = 4 \sin 2 \pi \left(\frac{4t}{0.16} - \frac{4x}{1.6}\right) = 4 \sin 2 \pi \left(\frac{t}{0.04} - \frac{x}{0.4}\right)$.
Here,$T_2 = 0.04 \text{ s}$ and $\lambda_2 = 0.4 \text{ m}$.
Velocity $v_2 = \frac{\lambda_2}{T_2} = \frac{0.4}{0.04} = 10 \text{ m/s}$.
Since $v_1 = v_2 = 10 \text{ m/s}$,both waves have the same velocity.

(D) The given equation of the transverse wave is $y=2 \sin (0.01 x+30 t)$.
Comparing this with the standard wave equation $y=A \sin (kx+\omega t)$,we identify the wave number $k=0.01 \ rad/cm$ and the angular frequency $\omega=30 \ rad/s$.
The wave speed $v$ is calculated as $v = \frac{\omega}{k} = \frac{30}{0.01} = 3000 \ cm/s$.
Converting the speed to $SI$ units,$v = 30 \ m/s$.
The time taken to travel the length of the string is $t = 0.5 \ s$.
Therefore,the length of the string $L = v \times t = 30 \ m/s \times 0.5 \ s = 15 \ m$.

(B) Given: Wave speed $v = 960 \,m/s$.
Number of waves $n = 900$.
Time $t = 0.5 \,min = 30 \,s$.
The frequency $f$ is defined as the number of waves passing a point per unit time:
$f = \frac{n}{t} = \frac{900}{30} = 30 \,Hz$.
The relationship between wave speed, frequency, and wavelength is given by $v = f \lambda$.
Therefore, the wavelength $\lambda = \frac{v}{f} = \frac{960}{30} = 32 \,m$.

(B) The given equation of the $SHM$ wave is $y = 0.03 \sin \pi (2 t - 0.01 x) \ m$.
Expanding this,we get $y = 0.03 \sin (2 \pi t - 0.01 \pi x) \ m$.
Comparing this with the general wave equation $y = a \sin (\omega t - k x)$,we identify the wave number $k$ as $k = 0.01 \pi \ rad/m$.
The phase difference $\Delta \phi$ between two particles separated by a distance $\Delta x$ is given by the formula $\Delta \phi = k \Delta x$.
Given $\Delta x = 25 \ m$ and $k = 0.01 \pi \ rad/m$,we substitute these values into the formula:
$\Delta \phi = (0.01 \pi) \times 25 = 0.25 \pi = \frac{\pi}{4} \ rad$.
Thus,the phase difference is $\frac{\pi}{4} \ rad$.

(A) In a sine wave,the displacement of a particle is given by $y = A \sin(kx - \omega t)$.
The particle velocity is $v_p = \frac{\partial y}{\partial t} = -A\omega \cos(kx - \omega t)$.
The speed of the particle is $|v_p| = |A\omega \cos(kx - \omega t)|$.
Two particles have the same speed if their displacement magnitudes are equal,i.e.,$|y_1| = |y_2|$.
For a sine wave,points with the same speed are those that have the same magnitude of displacement from the mean position.
Specifically,the points at the crest $(A)$ and the trough $(B)$ both have zero velocity (speed = $0$). The distance between a crest and the adjacent trough is $\frac{\lambda}{2}$.
Alternatively,any two points symmetric about the mean position or the extremes will have the same speed. The minimum distance between two such points is $\frac{\lambda}{2}$.

(A) The given equation of the wave is $y=A \sin (100 \pi t-3 x)$.
Comparing this with the standard wave equation $y=A \sin (\omega t-kx)$,we get the wave number $k=3 \ m^{-1}$.
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 $\Delta \phi = \frac{\pi}{3}$.
Substituting the values,we have $\frac{\pi}{3} = 3 \cdot \Delta x$.
Solving for the distance $\Delta x$,we get $\Delta x = \frac{\pi}{3 \times 3} = \frac{\pi}{9} \ m$.

(A) Given the displacement of a particle in the wave is $y = A \sin (\omega t - k x)$.
The particle velocity $v_p$ is defined as the rate of change of displacement with respect to time:
$v_p = \frac{dy}{dt}$
Differentiating the displacement equation with respect to $t$:
$v_p = \frac{d}{dt} [A \sin (\omega t - k x)] = A \omega \cos (\omega t - k x)$
For the maximum particle velocity,the cosine term must be at its maximum value,which is $1$:
$v_{p, \text{max}} = A \omega (1) = A \omega$
Thus,the maximum particle velocity is $A \omega$.

(C) Given: Velocity of sound $v = 320 \,m/s$, Frequency $f = 480 \,Hz$, Number of vibrations $N = 180$.
The relationship between velocity, frequency, and wavelength is $v = f \lambda$.
Therefore, the wavelength $\lambda = \frac{v}{f} = \frac{320}{480} = \frac{2}{3} \,m$.
The total distance $D$ covered by the wave after $N$ vibrations is given by $D = N \times \lambda$.
Substituting the values: $D = 180 \times \frac{2}{3} = 120 \,m$.

(C) The relationship between phase difference $\phi$ and path difference $\Delta x$ is given by $\phi = \frac{2\pi}{\lambda} \Delta x$.
Given that the path difference $\Delta x = x \ cm = \frac{x}{100} \ m$ and phase difference $\phi = n\pi$.
Also,the relationship between velocity $V$,frequency $f$,and wavelength $\lambda$ is $V = f\lambda$,which implies $\lambda = \frac{V}{f}$.
Substituting these into the phase difference formula:
$n\pi = \frac{2\pi}{(V/f)} \cdot \frac{x}{100}$.
Note: In standard physics problems of this type,the unit conversion factor is often omitted if $x$ is treated as a variable in the same units as the wavelength. Assuming $x$ is in meters for the formula to hold:
$n\pi = \frac{2\pi f x}{V}$.
Solving for frequency $f$:
$f = \frac{nV}{2x}$.

(B) The given equation of the harmonic progressive wave is $y=A \sin (100 \pi t+3 x)$.
Comparing this with the standard wave equation $y=A \sin (\omega t+kx)$,we get the wave number $k=3 \ m^{-1}$.
The relationship between wave number $k$ and wavelength $\lambda$ is $k = \frac{2 \pi}{\lambda}$.
Substituting the value of $k$,we have $3 = \frac{2 \pi}{\lambda}$,which gives $\lambda = \frac{2 \pi}{3} \ m$.
The path difference $\Delta x$ and phase difference $\Delta \phi$ are related by the formula $\Delta x = \frac{\lambda}{2 \pi} \times \Delta \phi$.
Given the phase difference $\Delta \phi = \frac{\pi}{3}$ radian.
Substituting the values,we get $\Delta x = \frac{2 \pi / 3}{2 \pi} \times \frac{\pi}{3} = \frac{1}{3} \times \frac{\pi}{3} = \frac{\pi}{9} \ m$.

(B) The general equation of a plane progressive wave is given by $y = a \sin (kx + \omega t)$.
Given the equation $y = 0.0015 \sin (62.4 x + 316 t)$.
Comparing this with the general equation,we identify the wave number $k$ as $k = 62.4 \text{ rad/unit}$.
Since $k = \frac{2\pi}{\lambda}$,we have $\lambda = \frac{2\pi}{k}$.
Substituting the value of $k$ and taking $\pi \approx 3.14$,we get $\lambda = \frac{2 \times 3.14}{62.4} = \frac{6.28}{62.4} \approx 0.1 \text{ unit}$.

(B) The frequency $f$ of the waves is the number of waves per unit time.
Given that the observer counts $45$ waves in $1$ minute ($60$ seconds),the frequency is:
$f = \frac{45 \text{ waves}}{60 \text{ s}} = 0.75 \text{ Hz}$.
The velocity $v$ of a wave is given by the product of its frequency $f$ and wavelength $\lambda$:
$v = f \times \lambda$.
Given $\lambda = 7 \ m$,we have:
$v = 0.75 \text{ Hz} \times 7 \ m = 5.25 \ m/s$.
Therefore,the velocity of the waves is $5.25 \ m/s$.

(A) The given wave equation is $y = A \sin (\omega t - k x)$,where $A = 60 \ \mu m = 60 \times 10^{-6} \ m$,$\omega = 1200 \ rad/s$,and $k = 6 \ m^{-1}$.
The maximum particle velocity $(v_p)$ is given by $v_p = A \omega$.
$v_p = (60 \times 10^{-6} \ m) \times (1200 \ rad/s) = 72000 \times 10^{-6} \ m/s = 7.2 \times 10^{-2} \ m/s$.
The wave velocity $(v_w)$ is given by $v_w = \frac{\omega}{k}$.
$v_w = \frac{1200}{6} = 200 \ m/s$.
The ratio of maximum particle velocity to wave velocity is $\frac{v_p}{v_w} = \frac{A \omega}{\omega/k} = A k$.
$\frac{v_p}{v_w} = (60 \times 10^{-6} \ m) \times (6 \ m^{-1}) = 360 \times 10^{-6} = 3.6 \times 10^{-4}$.

(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 $\Delta \phi = 45^{\circ} = \frac{\pi}{4} \text{ radians}$ and $\Delta x = 4.0 \text{ m}$.
Substituting these values: $\frac{\pi}{4} = \frac{2\pi}{\lambda} \times 4.0$.
Solving for wavelength $\lambda$: $\lambda = 8 \times 4.0 = 32 \text{ m}$.
The wave velocity $v$ is given by $v = f \lambda$,where $f = 300 \text{ Hz}$.
$v = 300 \text{ Hz} \times 32 \text{ m} = 9600 \text{ m/s}$.
Converting to $\text{km/s}$: $v = \frac{9600}{1000} \text{ km/s} = 9.6 \text{ km/s}$.

(B) The standard equation for a progressive wave is $y = a \sin(\omega t + kx)$.
Comparing the given equation $y = 0.002 \sin(100t + x)$ with the standard form,we get:
Angular frequency $\omega = 100 \ rad/s$
Wave number $k = 1 \ rad/m$
Amplitude $a = 0.002 \ m$
Frequency $f = \frac{\omega}{2\pi} = \frac{100}{2\pi} = \frac{50}{\pi} \ Hz$.
Wave velocity $v = \frac{\omega}{k} = \frac{100}{1} = 100 \ m/s$.
Since the sign between $\omega t$ and $kx$ is positive,the wave is travelling in the negative $x$-direction.
Therefore,the wave is travelling with a velocity of $100 \ m/s$ in the negative $x$-direction.

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

(C) The wavelength $\lambda$ of the wave is given by the formula $\lambda = \frac{v}{f}$,where $v$ is the speed and $f$ is the frequency.
Substituting the given values,$\lambda = \frac{30}{250} = 0.12 \ m$.
The phase difference $\Delta\phi$ between two points separated by a path difference $\Delta x$ is given by $\Delta\phi = \frac{2\pi}{\lambda} \times \Delta x$.
Given the path difference $\Delta x = 10 \ cm = 0.1 \ m$.
Substituting the values,$\Delta\phi = \frac{2\pi}{0.12} \times 0.1$.
$\Delta\phi = \frac{0.2\pi}{0.12} = \frac{20\pi}{12} = \frac{5\pi}{3} \ radian$.

(D) The phase of a wave is given by the argument of the sine function,$\phi = kx - \omega t$.
For the first wave,$\phi_1 = 20x - 30t$.
For the second wave,$\phi_2 = 25x - 40t$.
Given $x = 5 \ cm$ and $t = 2 \ s$:
$\phi_1 = 20(5) - 30(2) = 100 - 60 = 40 \ \text{radians}$.
$\phi_2 = 25(5) - 40(2) = 125 - 80 = 45 \ \text{radians}$.
The phase difference $\Delta \phi = |\phi_2 - \phi_1| = |45 - 40| = 5 \ \text{radians}$.

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

(A) The standard equation of a travelling wave is given by $y = y_0 \sin(\omega t + kx + \phi)$.
Comparing the given equation $y = 10^{-4} \sin(100t + 20x + \frac{\pi}{3})$ with the standard form,we get the angular frequency $\omega = 100 \ rad/s$ and the wave number $k = 20 \ rad/m$.
The speed of the wave $v$ is related to the angular frequency and wave number by the formula $v = \frac{\omega}{k}$.
Substituting the values,we get $v = \frac{100}{20} = 5 \ m/s$.

(C) Given: Phase difference $\phi = \frac{\pi}{5} \text{ rad}$, Frequency $f = 25 \text{ Hz}$, Velocity $v = 75 \text{ m/s}$.
First, calculate the wavelength $\lambda$ using the formula $\lambda = \frac{v}{f}$:
$\lambda = \frac{75 \text{ m/s}}{25 \text{ Hz}} = 3 \text{ m}$.
The relationship between phase difference $\phi$ and path difference $x$ is given by:
$\phi = \frac{2\pi}{\lambda} x$.
Substituting the known values:
$\frac{\pi}{5} = \frac{2\pi}{3} x$.
Solving for $x$:
$x = \frac{\pi}{5} \times \frac{3}{2\pi} = \frac{3}{10} \text{ m} = 0.3 \text{ m}$.

(B) The first wave is given by $y_1 = A \sin (\omega t + kx + 0.57)$.
The second wave is $y_2 = A \cos (\omega t + kx)$.
Using the trigonometric identity $\cos \theta = \sin (\theta + \frac{\pi}{2})$,we can rewrite $y_2$ as:
$y_2 = A \sin (\omega t + kx + \frac{\pi}{2})$.
The phase of the first wave is $\phi_1 = \omega t + kx + 0.57$.
The phase of the second wave is $\phi_2 = \omega t + kx + \frac{\pi}{2}$.
The phase difference $\Delta \phi$ is given by $|\phi_2 - \phi_1| = |(\omega t + kx + \frac{\pi}{2}) - (\omega t + kx + 0.57)|$.
$\Delta \phi = \frac{\pi}{2} - 0.57$.
Since $\frac{\pi}{2} \approx 1.57$,we have:
$\Delta \phi = 1.57 - 0.57 = 1.0 \ \text{rad}$.

(C) The given wave equation is $Y=2 \sin (0.01 x+30 t)$.
Comparing this with the standard wave equation $Y=A \sin (kx+\omega t)$:
We get the wave number $k = 0.01 \ cm^{-1}$ and angular frequency $\omega = 30 \ rad/s$.
The wave velocity $v$ is given by $v = \frac{\omega}{k} = \frac{30}{0.01} = 3000 \ cm/s$.
Converting the velocity to meters per second: $v = 30 \ m/s$.
The length of the string $L$ is the distance traveled by the wave in time $t = 0.5 \ s$.
$L = v \times t = 30 \ m/s \times 0.5 \ s = 15 \ m$.

(D) The given wave equation is $Y = 10 \sin \left(\frac{2 \pi t}{30} + \alpha\right)$.
At $t = 0$,the displacement $Y = 5 \text{ cm}$.
Substituting these values: $5 = 10 \sin \left(\frac{2 \pi \times 0}{30} + \alpha\right)$.
$5 = 10 \sin \alpha \implies \sin \alpha = 0.5$.
Since $\sin 30^{\circ} = 0.5$,we have $\alpha = 30^{\circ} = \frac{\pi}{6} \text{ rad}$.
The total phase $\phi$ at any time $t$ is given by $\phi = \frac{2 \pi t}{30} + \alpha$.
At $t = 7.5 \text{ s}$:
$\phi = \frac{2 \pi \times 7.5}{30} + \frac{\pi}{6}$.
$\phi = \frac{15 \pi}{30} + \frac{\pi}{6} = \frac{\pi}{2} + \frac{\pi}{6}$.
$\phi = \frac{3 \pi + \pi}{6} = \frac{4 \pi}{6} = \frac{2 \pi}{3} \text{ rad}$.

(B) Given equation: $y=0.05 \sin \pi(2 t-0.02 x)$
Expanding the equation: $y=0.05 \sin (2 \pi t - 0.02 \pi x)$
Comparing this with the standard wave equation $y=a \sin (\omega t - k x)$:
Angular frequency $\omega = 2 \pi \ rad/s$
Wave number $k = 0.02 \pi \ m^{-1}$
The minimum distance between two particles in phase is the wavelength $\lambda$.
Since $k = \frac{2 \pi}{\lambda}$,we have $\lambda = \frac{2 \pi}{k} = \frac{2 \pi}{0.02 \pi} = \frac{2}{0.02} = 100 \ m$.
The wave velocity $v$ is given by $v = \frac{\omega}{k}$.
$v = \frac{2 \pi}{0.02 \pi} = \frac{2}{0.02} = 100 \ m/s$.

(B) The given wave equation is $y = 10 \sin \left( \frac{2 \pi}{45} t + \alpha \right)$.
At $t = 0$,the displacement $y = 5 \text{ cm}$.
Substituting these values: $5 = 10 \sin \left( \frac{2 \pi}{45} \times 0 + \alpha \right) = 10 \sin \alpha$.
Thus,$\sin \alpha = \frac{5}{10} = \frac{1}{2}$,which gives $\alpha = \frac{\pi}{6}$.
The total phase of the wave is given by $\phi = \frac{2 \pi}{45} t + \alpha$.
At $t = 7.5 \text{ s} = \frac{15}{2} \text{ s}$,the total phase is:
$\phi = \frac{2 \pi}{45} \times \frac{15}{2} + \frac{\pi}{6} = \frac{\pi}{3} + \frac{\pi}{6}$.
$\phi = \frac{2 \pi + \pi}{6} = \frac{3 \pi}{6} = \frac{\pi}{2}$.

(A) The equation of a simple harmonic wave is given by $y = a \sin \frac{2 \pi}{\lambda} (vt - x)$.
By differentiating the displacement $y$ with respect to time $t$,we obtain the particle velocity:
$v_p = \frac{dy}{dt} = a \cdot \frac{2 \pi v}{\lambda} \cos \frac{2 \pi}{\lambda} (vt - x)$.
The maximum particle velocity $(v_{p, \max})$ occurs when the cosine term is $1$:
$v_{p, \max} = \frac{2 \pi v a}{\lambda}$.
According to the problem,the maximum particle velocity is half the wave velocity $(v)$:
$v_{p, \max} = \frac{v}{2}$.
Equating the two expressions for $v_{p, \max}$:
$\frac{v}{2} = \frac{2 \pi v a}{\lambda}$.
Solving for the amplitude $a$:
$a = \frac{\lambda}{4 \pi}$.

(D) The standard equation of a progressive wave is $y = A \sin(2\pi ft - kx)$.
Comparing the given equation $y = 6 \sin 2\pi(2t - 0.1x)$ with the standard form,we get the wave number $k = 2\pi \times 0.1 = 0.2\pi \ rad/mm$.
The relationship between phase difference $\Delta \phi$ and path difference $\Delta x$ is given by $\Delta \phi = k \Delta x$.
Given the separation between particles $\Delta x = 2 \ mm$.
Substituting the values,$\Delta \phi = (0.2\pi) \times 2 = 0.4\pi \ rad$.
To convert radians to degrees,multiply by $\frac{180^{\circ}}{\pi}$:
$\Delta \phi = 0.4\pi \times \frac{180^{\circ}}{\pi} = 0.4 \times 180^{\circ} = 72^{\circ}$.

(B) When a wave travels from one medium to another,its frequency depends only on the source of the wave and does not change.
However,the velocity and wavelength of the wave change depending on the refractive index of the medium.
Therefore,the frequency of the wave remains constant.

(C) The displacement equation for a simple harmonic wave is given by:
$y = \frac{10}{\pi} \sin \left(2000 \pi t - \frac{\pi x}{17}\right) \text{ cm}$
Comparing this with the standard wave equation $y = a \sin(\omega t - kx)$:
Here,amplitude $a = \frac{10}{\pi} \text{ cm} = \frac{10}{\pi} \times 10^{-2} \text{ m}$.
Angular frequency $\omega = 2000 \pi \text{ rad/s}$.
$1$. Period $(T)$:
$\omega = \frac{2\pi}{T} = 2000 \pi$
$T = \frac{2\pi}{2000 \pi} = \frac{1}{1000} \text{ s} = 10^{-3} \text{ s}$.
$2$. Maximum velocity $(v_{\max})$:
$v_{\max} = \omega a$
$v_{\max} = (2000 \pi) \times \left(\frac{10}{\pi} \times 10^{-2}\right) \text{ m/s}$
$v_{\max} = 2000 \times 10 \times 10^{-2} = 200 \text{ m/s}$.
Thus,the period is $10^{-3} \text{ s}$ and the maximum velocity is $200 \text{ m/s}$.

(C) The phase difference between the two waves is given by the difference in their arguments.
Let $\phi_1 = \omega t - \frac{2 \pi x}{\lambda}$ and $\phi_2 = \omega t - \frac{2 \pi x}{\lambda} + \phi$.
The phase difference $\Delta \phi = \phi_2 - \phi_1 = \phi$.
The relationship between path difference $(\Delta x)$ and phase difference $(\Delta \phi)$ is given by the formula:
$\Delta \phi = \frac{2 \pi}{\lambda} \Delta x$.
Rearranging this formula to solve for the path difference $\Delta x$:
$\Delta x = \frac{\lambda}{2 \pi} \Delta \phi$.
Substituting $\Delta \phi = \phi$ into the equation:
$\Delta x = \frac{\lambda}{2 \pi} \phi$.

(B) Given the equations for the two waves are $x_1 = A \sin \left(\omega t + \frac{\pi}{6}\right)$ and $x_2 = A \cos \omega t$.
To find the phase difference,we express both waves in terms of the same trigonometric function (sine or cosine).
We know that $\cos \theta = \sin \left(\theta + \frac{\pi}{2}\right)$.
Therefore,$x_2 = A \sin \left(\omega t + \frac{\pi}{2}\right)$.
Now,the phase of $x_1$ is $\phi_1 = \omega t + \frac{\pi}{6}$ and the phase of $x_2$ is $\phi_2 = \omega t + \frac{\pi}{2}$.
The phase difference $\Delta \phi = \phi_2 - \phi_1 = \left(\omega t + \frac{\pi}{2}\right) - \left(\omega t + \frac{\pi}{6}\right)$.
$\Delta \phi = \frac{\pi}{2} - \frac{\pi}{6} = \frac{3\pi - \pi}{6} = \frac{2\pi}{6} = \frac{\pi}{3}$.

(A) The plane wave is given by $y=a \sin (\omega t-k x)$.
At $t=0$,the displacement is $y=a \sin (-k x) = -a \sin (k x)$.
The particle velocity $v_{pa}$ is the time derivative of displacement:
$v_{pa} = \frac{d y}{d t} = \frac{d}{d t} (-a \sin (k x)) = -a \cos (k x) \cdot \frac{d}{d t} (k x)$.
Since $k = \frac{\omega}{v_{wave}}$,this is not quite right. Let's differentiate correctly:
$v_{pa} = \frac{\partial y}{\partial t} = a \omega \cos (\omega t - k x)$.
At $t=0$,$v_{pa} = a \omega \cos (-k x) = a \omega \cos (k x)$.
Using $k = \frac{2 \pi}{\lambda}$,we get $v_{pa} = a \omega \cos \left( \frac{2 \pi x}{\lambda} \right)$.
At $x=0$,$v_{pa} = a \omega$.
At $x=\frac{\lambda}{4}$,$v_{pa} = a \omega \cos \left( \frac{\pi}{2} \right) = 0$.
At $x=\frac{\lambda}{2}$,$v_{pa} = a \omega \cos (\pi) = -a \omega$.
This corresponds to a cosine graph starting at a positive maximum. Looking at the provided options,the graph that matches $v_{pa} = a \omega \cos (kx)$ is option $A$.

(B) The given wave equation is $y = y_0 \sin 2 \pi \left( \nu t - \frac{x}{\lambda} \right)$.
Comparing this with the standard wave equation $y = y_0 \sin (\omega t - kx)$,we have $\omega = 2 \pi \nu$ and $k = \frac{2 \pi}{\lambda}$.
The wave velocity $v_w$ is given by $v_w = \frac{\omega}{k} = \frac{2 \pi \nu}{2 \pi / \lambda} = \nu \lambda$.
The particle velocity $v_p$ is the derivative of displacement with respect to time: $v_p = \frac{\partial y}{\partial t} = y_0 (2 \pi \nu) \cos 2 \pi \left( \nu t - \frac{x}{\lambda} \right)$.
The maximum particle velocity is $v_{p, \text{max}} = 2 \pi \nu y_0$.
According to the problem,$v_{p, \text{max}} = 4 v_w$.
Substituting the expressions: $2 \pi \nu y_0 = 4 (\nu \lambda)$.
Solving for $\lambda$: $2 \pi y_0 = 4 \lambda$,which gives $\lambda = \frac{2 \pi y_0}{4} = \frac{\pi y_0}{2}$.

(B) The given equation of the transverse wave is $y=2 \sin (30 t-40 x)$.
Comparing this with the standard wave equation $y=A \sin (\omega t-k x)$,we get the angular frequency $\omega = 30 \ rad/s$ and the wave number $k = 40 \ m^{-1}$.
The velocity of propagation $v$ is given by the formula $v = \frac{\omega}{k}$.
Substituting the values,we get $v = \frac{30}{40} = 0.75 \ ms^{-1}$.

(D) The standard equation of a travelling wave is given by $y(x, t) = A \sin (kx - \omega t)$.
Comparing this with the given equation $y(x, t) = 0.5 \sin (70.1 x - 10 \pi t)$,we get the angular frequency $\omega = 10 \pi \text{ rad/s}$.
The relationship between angular frequency $\omega$ and frequency $f$ is given by $\omega = 2 \pi f$.
Substituting the value of $\omega$,we have $10 \pi = 2 \pi f$.
Solving for $f$,we get $f = \frac{10 \pi}{2 \pi} = 5 \text{ Hz}$.
Therefore,the frequency of the wave is $5 \text{ Hz}$.

(A) The frequency of the sound wave is $f = 500 \text{ Hz}$.
The time taken for the wave to travel from $X$ to $Y$ is $t = 2 \text{ s}$.
The number of waves $N$ produced by a source in time $t$ is given by the formula $N = f \times t$.
Substituting the given values: $N = 500 \text{ Hz} \times 2 \text{ s} = 1000$.
Thus,there are $1000$ waves between points $X$ and $Y$.

(B) The speed of a wave $v$ is given by $v = f \lambda$. Since the medium is not changed,the speed $v$ remains constant.
Therefore,$f_1 \lambda_1 = f_2 \lambda_2$,which implies $f \propto \frac{1}{\lambda}$.
Given that the frequency is increased by $25 \%$,the new frequency $f_2 = f_1 + 0.25 f_1 = 1.25 f_1 = \frac{5}{4} f_1$.
Using the inverse relationship,$\frac{\lambda_2}{\lambda_1} = \frac{f_1}{f_2} = \frac{f_1}{1.25 f_1} = \frac{1}{1.25} = \frac{4}{5} = 0.8$.
This means $\lambda_2 = 0.8 \lambda_1$.
The percentage change in wavelength is $\frac{\lambda_2 - \lambda_1}{\lambda_1} \times 100 = \frac{0.8 \lambda_1 - \lambda_1}{\lambda_1} \times 100 = -0.2 \times 100 = -20 \%$.
The negative sign indicates a $20 \%$ decrease in wavelength.

(B) Given: Frequency $f = 100 \ Hz$,speed $v = 100 \ cm/s$,and path difference $\Delta x = 2.75 \ cm$.
First,we calculate the wavelength $\lambda$ using the relation $v = f \lambda$:
$\lambda = \frac{v}{f} = \frac{100 \ cm/s}{100 \ Hz} = 1 \ cm$.
The phase difference $\Delta \phi$ is related to the path difference $\Delta x$ by the formula:
$\Delta \phi = \frac{2 \pi}{\lambda} \Delta x$.
Substituting the values:
$\Delta \phi = \frac{2 \pi}{1 \ cm} \times 2.75 \ cm = 5.5 \pi$.
Converting $5.5 \pi$ to a fraction:
$5.5 \pi = \frac{11}{2} \pi$ radians.

(B) The given wave equation is $y=0.03 \sin (\pi[2 t-0.01 x])$.
Expanding the equation,we get $y=0.03 \sin (2\pi t - 0.01\pi x)$.
Comparing this with the standard wave equation $y=A \sin (\omega t - kx)$,we identify the wave number $k = 0.01\pi \ rad/m$.
The path difference between the two particles is given as $\Delta x = 25 \ m$.
The relationship between phase difference $\Delta \phi$ and path difference $\Delta x$ is $\Delta \phi = k \cdot \Delta x$.
Substituting the values,$\Delta \phi = (0.01\pi) \times 25 = 0.25\pi$.
Converting $0.25\pi$ to a fraction,we get $\Delta \phi = \frac{1}{4}\pi = \frac{\pi}{4} \ rad$.