Wave Equation and Characteristics of Waves Questions in English

(B) Given equations are:
$y_1 = a \sin(\omega t - kx)$
$y_2 = b \cos(\omega t - kx + \frac{\pi}{3})$
To find the phase difference,we convert the cosine function into a sine function using the identity $\cos(\theta) = \sin(\theta + \frac{\pi}{2})$.
$y_2 = b \sin(\omega t - kx + \frac{\pi}{3} + \frac{\pi}{2})$
$y_2 = b \sin(\omega t - kx + \frac{2\pi + 3\pi}{6})$
$y_2 = b \sin(\omega t - kx + \frac{5\pi}{6})$
The phase of $y_1$ is $\phi_1 = \omega t - kx$.
The phase of $y_2$ is $\phi_2 = \omega t - kx + \frac{5\pi}{6}$.
The phase difference $\Delta\phi = \phi_2 - \phi_1 = \frac{5\pi}{6}$.

(A) Given: Frequency $f = 200 Hz$,Speed $v = 340 m s^{-1}$,Distance $\Delta x = 85 cm = 0.85 m$.
First,calculate the wavelength $\lambda$ using the formula $\lambda = \frac{v}{f}$.
$\lambda = \frac{340}{200} = 1.7 m$.
The phase difference $\Delta \phi$ is related to the path difference $\Delta x$ by the formula $\Delta \phi = \frac{2 \pi}{\lambda} \times \Delta x$.
Substituting the values: $\Delta \phi = \frac{2 \pi}{1.7} \times 0.85$.
$\Delta \phi = \frac{2 \pi}{1.7} \times \frac{1.7}{2} = \pi$ radians.

(D) The given wave equation is $y = (0.02 \ m) \sin (5 \pi x - 20 t)$.
Comparing this with the standard wave equation $y = A \sin (kx - \omega t)$,we get the wave number $k = 5 \pi \ m^{-1}$.
The wavelength $\lambda$ is given by $k = \frac{2 \pi}{\lambda}$,so $\lambda = \frac{2 \pi}{k} = \frac{2 \pi}{5 \pi} = 0.4 \ m$.
Particles in a wave have the same speed if they are separated by a distance equal to half the wavelength $(\frac{\lambda}{2})$ or an integer multiple of the wavelength. The minimum distance between two particles having the same speed is $\frac{\lambda}{2}$.
Therefore,the minimum distance $= \frac{0.4 \ m}{2} = 0.2 \ m$.

(A) The distance between two successive minima is equal to the wavelength $\lambda$. Therefore,$\lambda = 2.7 \ m$.
Frequency $f$ is defined as the number of waves passing a point per unit time.
Given that $5$ crests pass in $15.0 \ s$,the frequency is $f = \frac{5}{15} = \frac{1}{3} \ s^{-1}$.
The speed of the wave $v$ is given by the relation $v = f \lambda$.
Substituting the values,$v = \frac{1}{3} \times 2.7 = 0.9 \ m \ s^{-1}$.

(A) The given wave function is $y(x, t) = e^{-(\sqrt{a}x + \sqrt{b}t)^2}$.
For a wave to travel, the argument of the function must be of the form $(x \pm vt)$.
We can rewrite the exponent as $-(\sqrt{a}(x + \sqrt{\frac{b}{a}}t))^2$.
This is in the form $f(x + vt)$, where $v = \sqrt{\frac{b}{a}}$.
A function of the form $f(x + vt)$ represents a wave traveling in the negative $x$-direction with speed $v = \sqrt{\frac{b}{a}}$.

(B) The general equation of a progressive wave is $s(x, t) = A \sin(kx - \omega t)$.
$1$. Amplitude $(A)$: The particle moves $10 \text{ cm}$ between two extreme points,which is the total path length $(2A)$. Thus,$2A = 10 \text{ cm} = 0.10 \text{ m}$,so $A = 0.05 \text{ m}$.
$2$. Angular frequency $(\omega)$: $\omega = 2\pi f = 2 \times \pi \times 210 = 420\pi \approx 1319.47 \approx 1320 \text{ rad/s}$.
$3$. Wave number $(k)$: $v = \frac{\omega}{k} \implies k = \frac{\omega}{v} = \frac{420\pi}{330} = \frac{42\pi}{33} = \frac{14\pi}{11} \approx 4 \text{ rad/m}$.
$4$. Substituting these values into the wave equation: $s(x, t) = 0.05 \sin(4x - 1320t) \text{ m}$.

(C) The given equation of the wave is $y = 0.02 \sin(\pi x - 8 \pi t)$.
Comparing this with the standard wave equation $y = A \sin(kx - \omega t)$:
Here,the angular wave number $k = \pi \ m^{-1}$ and the angular frequency $\omega = 8 \pi \ rad/s$.
The velocity of the wave $v$ is given by the formula $v = \frac{\omega}{k}$.
Substituting the values,we get $v = \frac{8 \pi}{\pi} = 8 \ m/s$.

(A) The given wave equation is $y = 5 \times 10^{-3} \sin \left(12.5 \pi x - \frac{\pi}{2} t\right)$.
Comparing this with the standard wave equation $y = A \sin(kx - \omega t)$,we get:
Wave number $k = 12.5 \pi$ and angular frequency $\omega = \frac{\pi}{2}$.
We know that $k = \frac{2 \pi}{\lambda}$,so $\lambda = \frac{2 \pi}{k} = \frac{2 \pi}{12.5 \pi} = \frac{2}{12.5} = 0.16 \ m$.
We also know that $\omega = \frac{2 \pi}{T}$,so $T = \frac{2 \pi}{\omega} = \frac{2 \pi}{\pi / 2} = 4 \ s$.
Therefore,the wavelength is $0.16 \ m$ and the time period is $4 \ s$.

(B) The general equation of a progressive wave is $y(x, t) = A \sin(kx - \omega t)$.
The wave speed $v$ is given by the ratio of the angular frequency $\omega$ to the wave number $k$:
$v = \frac{\omega}{k} = \frac{\text{coefficient of } t}{\text{coefficient of } x}$.
Calculating the speed for each option:
$(a)$ $v = \frac{2}{2} = 1.0 \text{ m/s}$
$(b)$ $v = \frac{3}{2} = 1.5 \text{ m/s}$
$(c)$ $v = \frac{2}{3} \approx 0.67 \text{ m/s}$
$(d)$ $v = \frac{2}{5} = 0.4 \text{ m/s}$
Comparing these values,the wave speed is largest for the equation in option $(b)$.

(C) The maximum velocity of a particle in a wave is given by $v_{\max} = A \omega$,where $A$ is the amplitude and $\omega$ is the angular frequency.
Given that the maximum particle velocity is equal to the wave velocity $(v)$,we have $v_{\max} = v$.
Substituting the expressions,we get $A \omega = v$.
Since $\omega = 2 \pi f$ and the wave velocity $v = f \lambda$,we can write $\omega = 2 \pi (v / \lambda)$.
Substituting this into the equation: $A \times (2 \pi v / \lambda) = v$.
Dividing both sides by $v$,we get $A \times (2 \pi / \lambda) = 1$.
Therefore,the amplitude $A = \frac{\lambda}{2 \pi}$.

(A) The standard equation of a progressive wave is given by $y = A \sin(\omega t - kx + \phi)$.
Comparing this with the given equation $y = 4 \sin(4 \pi t - 0.04 x + \pi / 3)$,we identify the angular frequency $\omega = 4 \pi \ rad/s$ and the wave number $k = 0.04 \ rad/m$.
The velocity of the wave $(v)$ is given by the ratio of angular frequency to the wave number: $v = \frac{\omega}{k}$.
Substituting the values: $v = \frac{4 \pi}{0.04} = \frac{400 \pi}{4} = 100 \pi \ m/s$.

(C) The given wave equation is $x = x_0 \sin 2 \pi (nt - x/\lambda)$.
Comparing this with the standard form $x = A \sin (\omega t - kx)$,we have amplitude $A = x_0$,angular frequency $\omega = 2 \pi n$,and wave number $k = 2 \pi / \lambda$.
The maximum particle velocity $V_p$ is given by $V_p = A \omega = x_0 (2 \pi n) = 2 \pi n x_0$.
The wave velocity $V_w$ is given by $V_w = \frac{\omega}{k} = \frac{2 \pi n}{2 \pi / \lambda} = n \lambda$.
According to the problem,$V_p = 4 V_w$.
Substituting the expressions,we get $2 \pi n x_0 = 4 (n \lambda)$.
Dividing both sides by $n$,we get $2 \pi x_0 = 4 \lambda$.
Therefore,$\lambda = \frac{2 \pi x_0}{4} = \frac{\pi x_0}{2}$.

(C) The frequency of the wave is $f = 500 Hz$ and the velocity is $v = 300 ms^{-1}$.
First,we calculate the wavelength $\lambda$ using the relation $v = f \lambda$:
$\lambda = \frac{v}{f} = \frac{300}{500} = 0.6 m = 60 cm$.
The phase difference $\Delta \phi$ is given as $60^{\circ}$,which is $\frac{\pi}{3}$ radians.
The relationship between phase difference $\Delta \phi$ and path difference $\Delta x$ is given by $\Delta \phi = \frac{2 \pi}{\lambda} \Delta x$.
Substituting the values: $\frac{\pi}{3} = \frac{2 \pi}{60 cm} \Delta x$.
Solving for $\Delta x$: $\Delta x = \frac{\pi}{3} \times \frac{60 cm}{2 \pi} = \frac{60}{6} cm = 10 cm$.

(A) The standard equation of a plane progressive wave is given by $y = A \cos(2\pi(\nu t - \frac{x}{\lambda}))$,where $\nu$ is the frequency.
Given equation: $y = 2 \cos 6.284(330t - x)$.
Since $6.284 \approx 2\pi$,we can rewrite the equation as $y = 2 \cos 2\pi(330t - x)$.
Comparing this with the standard form,the frequency $\nu = 330 \text{ Hz}$.
The time period $T$ is the reciprocal of the frequency,$T = \frac{1}{\nu}$.
Therefore,$T = \frac{1}{330} \text{ s}$.

(A) The standard equation of a transverse wave is given by $y = A \sin(\omega t + kx + \phi)$.
Comparing this with the given equation $y = 3 \sin(36t + 0.018x + \pi/4)$,we identify the wave number $k = 0.018 \ cm^{-1}$.
The distance between two successive crests is equal to the wavelength $\lambda$.
The relationship between wavelength and wave number is $\lambda = 2\pi / k$.
Substituting the values: $\lambda = 2 \times 3.14 / 0.018 = 6.28 / 0.018$.
Calculating the value: $\lambda \approx 348.88 \ cm$.
Rounding to the nearest integer,we get $\lambda = 349 \ cm$.

(C) The given wave equation is $y(x, t) = 2.0 \cos(2\pi(10t - 0.0080x + 0.35))$.
Comparing this with the standard wave equation $y = A \cos(2\pi ft - kx + \phi)$, we identify the wave number $k$.
The term inside the cosine function is $2\pi(10t - 0.0080x + 0.35) = 20\pi t - 0.016\pi x + 0.7\pi$.
Thus, the wave number $k = 0.016\pi \text{ cm}^{-1}$.
The phase difference $\Delta\phi$ between two points separated by a distance $\Delta x$ is given by $\Delta\phi = k \Delta x$.
Given $\Delta x = 0.5 \text{ m} = 50 \text{ cm}$.
Substituting the values, $\Delta\phi = (0.016\pi \text{ cm}^{-1}) \times 50 \text{ cm} = 0.8\pi \text{ rad}$.
Therefore, the correct option is $C$.