Welcome to Lecture 121 (Master Index Entry: L115) of our comprehensive JEE Physics Master Series! Advancing deeper into continuous wave kinematics in Unit 10: Oscillations and Waves, we present the master mathematical lecture of wave theory: The Displacement Relation for a Progressive Wave — Mathematical Wave Functions, Phase Velocity, Slope-Velocity Relations & Energy Transport. While an oscillation tracks the displacement of a single mass element over time, a progressive wave represents a two-variable field $y(x, t)$ linking space and time. In this comprehensive tutorial, we derive the plane progressive harmonic wave equation from first principles, explore its diverse mathematical formats, evaluate the sign conventions governing propagation direction, prove the fundamental particle velocity-slope relation $v_p = -v(\partial y/\partial x)$, demonstrate why kinetic and potential energy densities are perfectly in phase, and master high-yield JEE Advanced problem patterns.
Detailed Concept Card: Mathematical Formulations of Progressive Waves
1. The Physical Derivation of the Displacement Relation
Consider an elastic medium extending along the $x$-axis. Suppose a vibration source located at the origin ($x = 0$) executes simple harmonic motion along the $y$-axis with amplitude $A$ and angular frequency $\omega$:
$$\mathbf{y(0, t) = A \sin(\omega t + \phi_0)}$$
If the disturbance propagates along the $+x$-direction with a constant phase velocity $v$, the state of motion of a particle at coordinate $x$ at time $t$ is identical to the state of motion that the source particle at $x = 0$ possessed at an earlier instant $t’ = t – \frac{x}{v}$:
$$y(x, t) = y\left(0, t – \frac{x}{v}\right) = A \sin\left[\omega\left(t – \frac{x}{v}\right) + \phi_0\right]$$
Substituting the propagation constant (angular wave number) $k = \frac{\omega}{v} = \frac{2\pi}{\lambda}$:
$$\mathbf{y(x, t) = A \sin(\omega t – kx + \phi_0)}$$
2. Equivalent Mathematical Formats of a Progressive Wave
| Mathematical Representation | Explicit Variable Dependency | Key Analytical Feature |
|---|---|---|
| $$\mathbf{y(x, t) = A \sin(\omega t – kx)}$$ | Angular frequency $\omega$ and wave number $k$ | Standard textbook form; $\omega = 2\pi f, k = 2\pi/\lambda$. |
| $$\mathbf{y(x, t) = A \sin\left[2\pi\left(\frac{t}{T} – \frac{x}{\lambda}\right)\right]}$$ | Time period $T$ and wavelength $\lambda$ | Directly reveals spatial and temporal periodicities. |
| $$\mathbf{y(x, t) = A \sin\left[\frac{2\pi}{\lambda}(vt – x)\right]}$$ | Wavelength $\lambda$ and wave velocity $v$ | Emphasizes the moving frame coordinate $(vt – x)$. |
| $$\mathbf{y(x, t) = A \sin[k(vt – x)]}$$ | Wave number $k$ and wave velocity $v$ | Compact spatial-propagation form. |
3. Propagation Direction Sign Conventions
- Wave Travelling in Positive $x$-Direction ($+x$): The phase argument has opposite signs for $t$ and $x$: $$y(x, t) = A \sin(\omega t – kx) \quad \text{or} \quad y(x, t) = A \sin(kx – \omega t)$$
- Wave Travelling in Negative $x$-Direction ($-x$): The phase argument has identical signs for $t$ and $x$: $$y(x, t) = A \sin(\omega t + kx) \quad \text{or} \quad y(x, t) = A \sin(kx + \omega t)$$
- Phase Velocity Determination: For any constant phase $\Phi(x, t) = \text{constant}$: $$d\Phi = \omega \, dt \mp k \, dx = 0 \implies \mathbf{v = \frac{dx}{dt} = \pm \frac{\omega}{k}}$$
4. Particle Kinematics vs Wave Propagation
Differentiating the displacement relation $y(x, t) = A \sin(\omega t – kx)$:
- Transverse Particle Velocity ($v_p$): $$\mathbf{v_p = \frac{\partial y}{\partial t} = \omega A \cos(\omega t – kx)}$$ $$\mathbf{v_{p,\text{max}} = \omega A = 2\pi f A}$$
- Transverse Particle Acceleration ($a_p$): $$\mathbf{a_p = \frac{\partial^2 y}{\partial t^2} = -\omega^2 A \sin(\omega t – kx) = -\omega^2 y}$$ $$\mathbf{a_{p,\text{max}} = \omega^2 A = 4\pi^2 f^2 A}$$
- The Fundamental Velocity-Slope Link:
Taking the spatial derivative $\frac{\partial y}{\partial x} = -k A \cos(\omega t – kx)$:
$$\mathbf{v_p = -v \left(\frac{\partial y}{\partial x}\right)}$$
$$\mathbf{\text{Particle Velocity} = -(\text{Wave Velocity}) \times (\text{Spatial Slope})}$$
- For a forward wave ($v > 0$): where the wave profile has a negative slope ($\frac{\partial y}{\partial x} < 0$), the particle moves upwards ($v_p > 0$); where the slope is positive, the particle moves downwards ($v_p < 0$).
- For a backward wave ($v < 0$): the particle velocity shares the exact same sign as the spatial slope!
5. Energy Density & Power Transmission in a Progressive Wave
For a transverse wave on a string of linear density $\mu$ under tension $T$:
- Kinetic Energy Density: $$\frac{dK}{dx} = \frac{1}{2} \mu v_p^2 = \frac{1}{2} \mu \omega^2 A^2 \cos^2(\omega t – kx)$$
- Potential Energy Density: $$\frac{dU}{dx} = \frac{1}{2} T \left(\frac{\partial y}{\partial x}\right)^2 = \frac{1}{2} (\mu v^2) (k^2 A^2) \cos^2(\omega t – kx) = \frac{1}{2} \mu \omega^2 A^2 \cos^2(\omega t – kx)$$
- Crucial Theorem: Unlike simple harmonic motion of a localized particle where kinetic and potential energies are $180^\circ$ out of phase, in a progressive wave, kinetic energy density and potential energy density are strictly in phase and equal at all coordinates and times! $$\mathbf{\frac{dK}{dx} = \frac{dU}{dx}}$$ Both reach their maximum values simultaneously when a particle crosses the mean position ($y = 0$, where speed is maximum and string slope is steepest).
- Average Power Transmitted ($P_{\text{avg}}$): $$\mathbf{P_{\text{avg}} = \frac{1}{2} \mu v \omega^2 A^2 = 2\pi^2 \mu v f^2 A^2}$$
Four Solved Examples (Graded Basic to Advanced)
Example 1: Basic Application — Extracting Wave Characteristics from Standard Displacement Form
Problem Statement: A transverse progressive wave on a string is represented by: $$y(x, t) = 0.04 \sin(20\pi t – 0.5\pi x + 0.2\pi)$$ where $x$ and $y$ are in meters and $t$ is in seconds.
(a) Determine the amplitude $A$, angular frequency $\omega$, wave number $k$, frequency $f$, wavelength $\lambda$, and wave speed $v$.
(b) Identify the direction of propagation and the initial phase angle $\phi_0$.
(c) Calculate the maximum transverse particle speed $v_{p,\text{max}}$ and maximum particle acceleration $a_{p,\text{max}}$.
(d) Find the transverse velocity of a particle located at $x = 2.0\text{ m}$ at time $t = 0.10\text{ s}$.
Solution:
(a) Wave Parameters:
Comparing with $y(x, t) = A \sin(\omega t – kx + \phi_0)$:
$$\mathbf{A = 0.04\text{ m} = 4.0\text{ cm}}$$
$$\mathbf{\omega = 20\pi\text{ rad/s}}$$
$$\mathbf{k = 0.5\pi\text{ rad/m}}$$
$$f = \frac{\omega}{2\pi} = \frac{20\pi}{2\pi} = \mathbf{10\text{ Hz}}$$
$$\lambda = \frac{2\pi}{k} = \frac{2\pi}{0.5\pi} = \mathbf{4.0\text{ m}}$$
$$v = \frac{\omega}{k} = \frac{20\pi}{0.5\pi} = \mathbf{40\text{ m/s}}$$
(b) Direction and Initial Phase:
Because the signs of $t$ and $x$ are opposite, the wave travels in the positive $x$-direction ($+x$).
Initial phase: $\mathbf{\phi_0 = 0.2\pi\text{ rad} = 36^\circ}$.
(c) Maximum Particle Kinematics:
$$v_{p,\text{max}} = \omega A = (20\pi\text{ rad/s})(0.04\text{ m}) = \mathbf{0.80\pi\text{ m/s} \approx 2.513\text{ m/s}}$$
$$a_{p,\text{max}} = \omega^2 A = (20\pi)^2 (0.04) = 400\pi^2 (0.04) = \mathbf{16\pi^2\text{ m/s}^2 \approx 157.9\text{ m/s}^2}$$
(d) Particle Velocity at $x = 2.0\text{ m}, t = 0.10\text{ s}$:
$$v_p = \frac{\partial y}{\partial t} = (0.04)(20\pi) \cos(20\pi t – 0.5\pi x + 0.2\pi)$$
Evaluating the argument $\Phi$:
$$\Phi = 20\pi(0.10) – 0.5\pi(2.0) + 0.2\pi = 2.0\pi – 1.0\pi + 0.2\pi = 1.2\pi\text{ rad} = 216^\circ$$
$$\cos(1.2\pi) = \cos(\pi + 0.2\pi) = -\cos(0.2\pi) = -\cos(36^\circ) \approx -0.8090$$
$$v_p = 0.80\pi(-0.8090) = \mathbf{-0.647\pi\text{ m/s} \approx -2.033\text{ m/s}}$$
Example 2: Intermediate Application — Backward Propagating Wave & Slope-Velocity Dynamics
Problem Statement: A transverse harmonic wave propagates along a horizontal string in the negative $x$-direction with speed $v = 8.0\text{ m/s}$. At $t = 0$, the profile of the string is given by $y(x, 0) = A \sin\left(\frac{\pi x}{4}\right)$, where $x$ and $y$ are in meters and $A = 0.05\text{ m}$.
(a) Write the complete wave equation $y(x, t)$ for all $t$.
(b) Calculate the slope of the string and the transverse velocity of the particle at $x = 1.0\text{ m}$ at time $t = 0$.
(c) Demonstrate that for a wave travelling in the $-x$ direction, particle velocity and wave slope have identical signs.
Solution:
(a) Complete Wave Equation:
For propagation in the $-x$ direction with speed $v = 8.0\text{ m/s}$, replace $x$ by $(x + vt) = (x + 8.0t)$:
$$\mathbf{y(x, t) = 0.05 \sin\left[\frac{\pi}{4}(x + 8.0t)\right] = 0.05 \sin\left(\frac{\pi x}{4} + 2\pi t\right)\text{ m}}$$
(b) Slope and Velocity at $x = 1.0\text{ m}, t = 0$:
Spatial slope:
$$\frac{\partial y}{\partial x} = 0.05\left(\frac{\pi}{4}\right) \cos\left(\frac{\pi x}{4} + 2\pi t\right)$$
At $x = 1.0\text{ m}, t = 0$:
$$\left.\frac{\partial y}{\partial x}\right|_{x=1, t=0} = \frac{0.05\pi}{4} \cos\left(\frac{\pi}{4}\right) = \frac{0.05\pi}{4\sqrt{2}} \approx \mathbf{+0.0278}$$
Transverse particle velocity:
$$v_p = \frac{\partial y}{\partial t} = 0.05(2\pi) \cos\left(\frac{\pi x}{4} + 2\pi t\right)$$
At $x = 1.0\text{ m}, t = 0$:
$$v_p = 0.10\pi \cos\left(\frac{\pi}{4}\right) = \frac{0.10\pi}{\sqrt{2}} \approx \mathbf{+0.222\text{ m/s}}$$
(c) Sign Alignment Proof:
Using the universal relation $v_p = -v_{\text{wave}}\left(\frac{\partial y}{\partial x}\right)$:
Here $v_{\text{wave}} = -8.0\text{ m/s}$ (since it moves along $-x$):
$$v_p = -(-8.0) \left(\frac{\partial y}{\partial x}\right) = +8.0 \left(\frac{\partial y}{\partial x}\right)$$
Evaluating: $8.0 \times (+0.02777) \approx +0.222\text{ m/s}$. Both the slope and the particle velocity are positive simultaneously.
Example 3: Advanced Application — Energy Density and Power Flow on a Stretched Wire
Problem Statement: A uniform steel wire of length $L = 10.0\text{ m}$, cross-sectional area $S = 1.0\text{ mm}^2 = 1.0 \times 10^{-6}\text{ m}^2$, and density $\rho = 7800\text{ kg/m}^3$ is held under tension $T = 312\text{ N}$. A sinusoidal wave of frequency $f = 120\text{ Hz}$ and amplitude $A = 2.0\text{ mm} = 2.0 \times 10^{-3}\text{ m}$ propagates along the wire.
(a) Determine the linear mass density $\mu$ and wave speed $v$.
(b) Calculate the angular frequency $\omega$ and wavelength $\lambda$.
(c) Determine the time-averaged power $P_{\text{avg}}$ transmitted along the wire.
(d) Calculate the total mechanical energy stored in a $1.0\text{ m}$ length of the wire.
Solution:
(a) Linear Density & Speed:
$$\mu = \rho S = (7800\text{ kg/m}^3)(1.0 \times 10^{-6}\text{ m}^2) = \mathbf{7.8 \times 10^{-3}\text{ kg/m} = 0.0078\text{ kg/m}}$$
$$v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{312\text{ N}}{0.0078\text{ kg/m}}} = \sqrt{40000} = \mathbf{200\text{ m/s}}$$
(b) Frequency & Wavelength:
$$\omega = 2\pi f = 2\pi(120\text{ Hz}) = \mathbf{240\pi\text{ rad/s} \approx 753.98\text{ rad/s}}$$
$$\lambda = \frac{v}{f} = \frac{200\text{ m/s}}{120\text{ Hz}} = \mathbf{\frac{5}{3}\text{ m} \approx 1.667\text{ m}}$$
(c) Average Power Transmitted:
$$P_{\text{avg}} = \frac{1}{2} \mu v \omega^2 A^2 = \frac{1}{2}(0.0078\text{ kg/m})(200\text{ m/s})(240\pi\text{ rad/s})^2 (2.0 \times 10^{-3}\text{ m})^2$$
$$P_{\text{avg}} = (0.78)(57600\pi^2)(4.0 \times 10^{-6}) = (0.78)(0.2304\pi^2) = 0.1797\pi^2\text{ W} \approx \mathbf{1.774\text{ W}}$$
(d) Energy in $1.0\text{ m}$ Length:
Linear energy density:
$$\langle u \rangle = \frac{P_{\text{avg}}}{v} = \frac{1.774\text{ W}}{200\text{ m/s}} = \mathbf{8.87 \times 10^{-3}\text{ J/m} = 8.87\text{ mJ/m}}$$
For a length $\Delta x = 1.0\text{ m}$, the energy content is $\mathbf{8.87\text{ mJ}}$.
Example 4: JEE Advanced Scenario — Analytical Kinematics & Kinetic Energy of a Lorentzian Pulse
Problem Statement: A localized non-sinusoidal wave pulse on a string ($\mu = 0.050\text{ kg/m}$) propagates with speed $v = 20\text{ m/s}$ according to the displacement relation: $$y(x, t) = \frac{b^3}{b^2 + (x – vt)^2}$$ where $b = 0.10\text{ m}$.
(a) Determine the maximum transverse displacement and the location of the pulse peak at time $t$.
(b) Derive the expression for particle velocity $v_p(x, t)$ and determine its maximum magnitude $v_{p,\text{max}}$.
(c) Evaluate the total kinetic energy $K_{\text{total}}$ of the entire pulse on the string.
Solution:
(a) Maximum Displacement:
$y(x, t)$ is maximized when the denominator is minimized, which occurs when $(x – vt) = 0 \implies \mathbf{x = vt}$.
$$\mathbf{y_{\text{max}} = \frac{b^3}{b^2 + 0} = b = 0.10\text{ m} = 10\text{ cm}}$$
(b) Particle Velocity and Maximum Speed:
Let $u = x – vt$.
$$v_p = \frac{\partial y}{\partial t} = \frac{-b^3}{(b^2 + u^2)^2} \cdot (-2v u) = \frac{2b^3 v u}{(b^2 + u^2)^2}$$
Setting $\frac{d v_p}{du} = 0$:
$$(b^2 + u^2)^2(1) – u[2(b^2 + u^2)(2u)] = 0 \implies (b^2 + u^2) – 4u^2 = 0 \implies 3u^2 = b^2 \implies u = \pm \frac{b}{\sqrt{3}}$$
Substituting $u = \frac{b}{\sqrt{3}}$ into $v_p$:
$$b^2 + u^2 = \frac{4}{3}b^2 \implies (b^2 + u^2)^2 = \frac{16}{9}b^4$$
$$v_{p,\text{max}} = \frac{2b^3 v (b/\sqrt{3})}{\frac{16}{9}b^4} = 2v \left(\frac{9}{16\sqrt{3}}\right) = \mathbf{\frac{3\sqrt{3}}{8} v}$$
With $v = 20\text{ m/s}$:
$$\mathbf{v_{p,\text{max}} = \frac{3\sqrt{3}}{8}(20) = \frac{15\sqrt{3}}{2}\text{ m/s} \approx 12.99\text{ m/s} \approx 13.0\text{ m/s}}$$
(c) Total Kinetic Energy of the Pulse:
$$K_{\text{total}} = \int_{-\infty}^\infty \frac{1}{2} \mu v_p^2 \, dx = \frac{1}{2} \mu (2b^3 v)^2 \int_{-\infty}^\infty \frac{u^2}{(b^2 + u^2)^4} \, du = 2\mu b^6 v^2 \int_{-\infty}^\infty \frac{u^2}{(b^2 + u^2)^4} \, du$$
Using the trigonometric substitution $u = b \tan\theta, du = b \sec^2\theta \, d\theta$:
$$\int_{-\infty}^\infty \frac{u^2}{(b^2 + u^2)^4} \, du = \frac{2}{b^5} \int_0^{\pi/2} \sin^2\theta \cos^4\theta \, d\theta = \frac{2}{b^5} \left(\frac{\pi}{32}\right) = \frac{\pi}{16 b^5}$$
Substituting this back:
$$\mathbf{K_{\text{total}} = 2\mu b^6 v^2 \left(\frac{\pi}{16 b^5}\right) = \frac{\pi}{8} \mu v^2 b}$$
Evaluating numerically with $\mu = 0.050\text{ kg/m}, v = 20\text{ m/s}, b = 0.10\text{ m}$:
$$\mathbf{K_{\text{total}} = \frac{\pi}{8}(0.050)(20)^2(0.10) = \frac{\pi}{8}(0.050 \times 400 \times 0.10) = \frac{\pi}{8}(2.0) = \frac{\pi}{4}\text{ J} \approx 0.785\text{ J}}$$
Practice Worksheet: 10 Target JEE Problems
Master wave equations, phase shifts, and power transmission relations. Attempt all 10 problems before consulting the detailed solutions at the end.
Section I: Single Correct Choice Questions (Q1–Q5)
Q1. A progressive harmonic wave is represented by $y = A \sin(100\pi t – 2\pi x)$ in SI units. The distance between two points on the wave that differ in phase by $\pi/3$ radians is:
(A) $\frac{1}{2}\text{ m}$
(B) $\frac{1}{3}\text{ m}$
(C) $\frac{1}{6}\text{ m}$
(D) $\frac{1}{12}\text{ m}$
Q2. Which of the following displacement equations represents a wave propagating strictly along the negative $x$-direction?
(A) $y = A \sin(\omega t – kx)$
(B) $y = A \cos(kx – \omega t)$
(C) $y = A \sin(kx + \omega t)$
(D) $y = A \sin(kx) \cos(\omega t)$
Q3. A transverse wave of amplitude $A$ and angular frequency $\omega$ travels along a string with wave number $k$. The ratio of the maximum particle acceleration to the wave speed is equal to:
(A) $\omega A$
(B) $\omega k A$
(C) $k^2 A$
(D) $\omega^2 / k$
Q4. In a simple harmonic progressive wave on a stretched string, the kinetic energy density and the elastic potential energy density of any string element:
(A) Are $180^\circ$ out of phase
(B) Are $90^\circ$ out of phase
(C) Are in phase and equal at all coordinates and times
(D) Sum to a constant independent of space and time
Q5. The average power transmitted along a string carrying a progressive wave $y = A \sin(\omega t – kx)$ is directly proportional to:
(A) $A \omega$
(B) $A^2 \omega$
(C) $A^2 \omega^2$
(D) $A \omega^2$
Section II: One or More Than One Correct Questions (Q6–Q8)
Q6. For the progressive harmonic wave $y(x, t) = A \sin(\omega t – kx)$ propagating along a string:
(A) The phase velocity of the wave is $v = \omega/k$.
(B) The particle velocity is related to the spatial slope by $v_p = -v \left(\frac{\partial y}{\partial x}\right)$.
(C) At the crest of the wave, the particle velocity is zero and its acceleration is negative maximum.
(D) As a particle passes through its equilibrium position ($y = 0$), its speed is maximum.
Q7. A stretched string under tension $T$ transmits a transverse wave. Which of the following independent modifications will quadruple the average power transmitted along the string?
(A) Doubling the wave amplitude $A$ while holding frequency and tension constant.
(B) Doubling the wave frequency $f$ while holding amplitude and tension constant.
(C) Quadrupling the string tension $T$ while holding amplitude and frequency constant.
(D) Increasing the string tension 16 times while holding amplitude and frequency constant.
Q8. Consider the non-sinusoidal wave pulse $y(x, t) = \frac{0.05}{2 + (x – 4t)^2}$ (in SI units). Which of the following statements is/are correct?
(A) The wave pulse propagates in the positive $x$-direction with speed $4.0\text{ m/s}$.
(B) The maximum transverse displacement of any particle on the string is $0.025\text{ m}$.
(C) The pulse satisfies the partial differential wave equation $\frac{\partial^2 y}{\partial x^2} = \frac{1}{16} \frac{\partial^2 y}{\partial t^2}$.
(D) At $x = 4t$, the particle velocity is zero.
Section III: Numerical / Integer Type Questions (Q9–Q10)
Q9. A transverse harmonic wave propagating on a string has amplitude $A = 0.05\text{ m}$, frequency $f = 50\text{ Hz}$, and wave speed $v = 100\text{ m/s}$. The maximum particle speed is expressible as $v_{p,\text{max}} = n\pi\text{ m/s}$. Determine the integer value of $n$.
Q10. Two particles on a string carrying a travelling progressive wave $y = A \sin(\omega t – kx)$ are separated by a spatial distance $\Delta x = 1.25\text{ m}$. If the wavelength of the wave is $\lambda = 0.50\text{ m}$, the phase difference between the two particles is $\Delta\Phi = m\pi\text{ rad}$. Calculate the integer value of $m$.
Comprehensive Solutions & Answer Key
Answer Key
| Question No. | Correct Answer | Key Concept / Formula Tested |
|---|---|---|
| Q1 | (C) | $\Delta\Phi = (2\pi/\lambda)\Delta x \implies \Delta x = \lambda/6 = 1/6\text{ m}$ |
| Q2 | (C) | Identical signs of $kx$ and $\omega t$ indicate propagation along $-x$ |
| Q3 | (B) | $a_{p,\text{max}} = \omega^2 A, v = \omega/k \implies a_{p,\text{max}}/v = \omega k A$ |
| Q4 | (C) | $dK/dx = dU/dx = \frac{1}{2}\mu\omega^2 A^2 \cos^2(\omega t – kx)$ |
| Q5 | (C) | $P_{\text{avg}} = \frac{1}{2}\mu v \omega^2 A^2 \propto A^2 \omega^2$ |
| Q6 | (A, B, C, D) | Harmonic wave kinematics: phase speed, slope relation, peak acceleration |
| Q7 | (A, B, D) | $P \propto A^2 \omega^2 \sqrt{T}$; doubling $A$ or $\omega$, or $16\times T$ quadruples power |
| Q8 | (A, B, C, D) | Pulse characteristics: $v = 4\text{ m/s}, y_{\text{max}} = 0.05/2 = 0.025\text{ m}$, wave PDE |
| Q9 | 5 | $v_{p,\text{max}} = \omega A = (2\pi \times 50)(0.05) = 5\pi\text{ m/s} \implies n = 5$ |
| Q10 | 5 | $\Delta\Phi = (2\pi/0.50)(1.25) = 5\pi\text{ rad} \implies m = 5$ |
Detailed Step-by-Step Solutions
Solution to Q1:
Comparing $y = A \sin(100\pi t – 2\pi x)$ with $y = A \sin(\omega t – kx)$:
$$k = 2\pi\text{ rad/m} \implies \lambda = \frac{2\pi}{k} = \frac{2\pi}{2\pi} = 1.0\text{ m}$$
The phase difference is related to path difference by:
$$\Delta\Phi = \frac{2\pi}{\lambda} \Delta x \implies \frac{\pi}{3} = \frac{2\pi}{1.0} \Delta x$$
$$\mathbf{\Delta x = \frac{1}{6}\text{ m}}$$
Hence, the correct choice is (C).
Solution to Q2:
For a wave function $y = f(kx \pm \omega t)$, the propagation velocity is $v = -\frac{\text{coeff of } t}{\text{coeff of } x}$.
– For $y = A \sin(kx + \omega t)$, $v = -\frac{\omega}{k} < 0$, which represents propagation in the negative $x$-direction.
– Expressions (A) and (B) propagate in the positive $x$-direction.
– Expression (D) is a standing wave with separable space and time functions.
Hence, the correct choice is (C).
Solution to Q3:
The maximum particle acceleration is:
$$a_{p,\text{max}} = \omega^2 A$$
The wave propagation speed is:
$$v = \frac{\omega}{k}$$
Dividing the two quantities:
$$\mathbf{\frac{a_{p,\text{max}}}{v} = \frac{\omega^2 A}{\omega / k} = \omega k A}$$
Hence, the correct choice is (B).
Solution to Q4:
For a progressive wave on a string, the kinetic energy per unit length is $\frac{dK}{dx} = \frac{1}{2}\mu v_p^2 = \frac{1}{2}\mu \omega^2 A^2 \cos^2(\omega t – kx)$.
The elastic potential energy per unit length is $\frac{dU}{dx} = \frac{1}{2}T \left(\frac{\partial y}{\partial x}\right)^2 = \frac{1}{2}(\mu v^2)(k^2 A^2) \cos^2(\omega t – kx) = \frac{1}{2}\mu \omega^2 A^2 \cos^2(\omega t – kx)$.
Both energy densities vary with the exact same spatial and temporal phase $\cos^2(\omega t – kx)$ and have identical peak values. They are in phase and equal at all coordinates and times.
Hence, the correct choice is (C).
Solution to Q5:
The time-averaged power transmitted by a progressive wave is:
$$P_{\text{avg}} = \frac{1}{2} \mu v \omega^2 A^2$$
Therefore:
$$\mathbf{P_{\text{avg}} \propto A^2 \omega^2}$$
Hence, the correct choice is (C).
Solution to Q6:
Let us evaluate each statement:
(A) The phase speed is $v = \omega/k$. (A) is correct.
(B) $v_p = \frac{\partial y}{\partial t} = \omega A \cos(\omega t – kx)$, while $\frac{\partial y}{\partial x} = -kA \cos(\omega t – kx)$. Thus $-v(\frac{\partial y}{\partial x}) = -(\frac{\omega}{k})[-kA \cos(\omega t – kx)] = \omega A \cos(\omega t – kx) = v_p$. (B) is correct.
(C) At a crest, $y = +A \implies \sin(\omega t – kx) = +1 \implies \cos(\omega t – kx) = 0 \implies v_p = 0$. The acceleration is $a_p = -\omega^2 y = -\omega^2 A$ (negative maximum). (C) is correct.
(D) At the equilibrium crossing ($y = 0$), $\cos(\omega t – kx) = \pm 1$, so particle speed is maximum ($|v_p| = \omega A$). (D) is correct.
Hence, all choices (A, B, C, D) are correct.
Solution to Q7:
The average power is $P = \frac{1}{2}\mu v \omega^2 A^2 = \frac{1}{2}\mu \sqrt{\frac{T}{\mu}} \omega^2 A^2 = \frac{1}{2}\sqrt{\mu T} \omega^2 A^2$.
– If $A \to 2A$: $P \propto A^2 \implies P’ = 4P$. (A is correct).
– If $f \to 2f \implies \omega \to 2\omega$: $P \propto \omega^2 \implies P’ = 4P$. (B is correct).
– If $T \to 4T$: $P \propto \sqrt{T} \implies P’ = 2P$. (C is incorrect).
– If $T \to 16T$: $P \propto \sqrt{16} = 4P$. (D is correct).
Hence, correct choices are (A, B, D).
Solution to Q8:
Let us evaluate each statement:
(A) The argument is $(x – 4t) \implies x – vt$ with $v = 4.0\text{ m/s}$ along $+x$. (A) is correct.
(B) $y$ is maximum when denominator is minimum: $(x – 4t) = 0 \implies y_{\text{max}} = 0.05/2 = 0.025\text{ m}$. (B) is correct.
(C) Any function of $(x – vt)$ satisfies the wave equation with $v = 4\text{ m/s} \implies v^2 = 16 \implies \frac{\partial^2 y}{\partial x^2} = \frac{1}{16}\frac{\partial^2 y}{\partial t^2}$. (C) is correct.
(D) $v_p = \frac{\partial y}{\partial t} = \frac{-0.05}{[2+(x-4t)^2]^2}[-8(x-4t)]$. At $x = 4t$, $v_p = 0$. (D) is correct.
Hence, all choices (A, B, C, D) are correct.
Solution to Q9:
The maximum particle speed is:
$$v_{p,\text{max}} = \omega A = (2\pi f) A$$
Substituting $f = 50\text{ Hz}$ and $A = 0.05\text{ m}$:
$$v_{p,\text{max}} = [2\pi(50\text{ s}^{-1})](0.05\text{ m}) = (100\pi)(0.05) = \mathbf{5\pi\text{ m/s}}$$
Comparing with $n\pi\text{ m/s}$:
$$\mathbf{n = 5}$$
Answer: 5
Solution to Q10:
The phase difference corresponding to spatial separation $\Delta x = 1.25\text{ m}$ for wavelength $\lambda = 0.50\text{ m}$ is:
$$\Delta\Phi = \frac{2\pi}{\lambda} \Delta x = \frac{2\pi}{0.50\text{ m}} (1.25\text{ m}) = 2\pi (2.5) = \mathbf{5\pi\text{ rad}}$$
Comparing with $m\pi\text{ rad}$:
$$\mathbf{m = 5}$$
Answer: 5