Welcome to Lecture 108 of our comprehensive JEE Physics Master Series. Having mastered simple harmonic oscillators, we now investigate the collective propagation of oscillatory disturbances across space and time: Wave Motion. In this lecture, we formulate the classical one-dimensional wave equation from first principles, dissect the kinematics of progressive transverse waves on a stretched string, derive the wave propagation velocity from Newton’s second law, analyze the rate of energy transport and transmitted power, and solve 4 graded benchmark examples followed by a 10-question practice worksheet (DPP) with complete solutions.
1. Kinematic Description of Progressive Waves
A wave is a self-sustaining disturbance that propagates through a medium or space, transporting energy and momentum without the continuous transport of bulk matter.
If a disturbance at origin $x = 0$ is described as a function of time $f(t)$, and it travels undistorted with a constant wave speed $v$ along the positive $x$-axis, the disturbance at position $x$ at time $t$ corresponds to the state of the origin at the earlier time $t’ = t – \frac{x}{v}$:
$$y(x, t) = f\left(t – \frac{x}{v}\right) \quad \text{or} \quad y(x, t) = g(x – v t)$$
Similarly, for a wave propagating in the negative $x$-direction:
$$y(x, t) = g(x + v t)$$
2. The Classical 1D Differential Wave Equation
Differentiating $y(x, t) = g(x \mp v t)$ twice with respect to position $x$ and time $t$:
$$\frac{\partial y}{\partial x} = g'(u), \quad \frac{\partial^2 y}{\partial x^2} = g”(u)$$
$$\frac{\partial y}{\partial t} = \mp v g'(u), \quad \frac{\partial^2 y}{\partial t^2} = v^2 g”(u)$$
Eliminating $g”(u)$ yields the fundamental linear differential wave equation:
$$\frac{\partial^2 y}{\partial x^2} = \frac{1}{v^2} \frac{\partial^2 y}{\partial t^2}$$
Criterion for a Valid Wave Function: Any physical mathematical function $y(x, t)$ represents a progressive wave if and only if: (1) it satisfies the wave equation $\frac{\partial^2 y}{\partial x^2} = \frac{1}{v^2}\frac{\partial^2 y}{\partial t^2}$, and (2) it remains finite and bounded for all real values of $x \in (-\infty, \infty)$ and $t \ge 0$.
3. Transverse Waves on a Stretched String
Consider a flexible, uniform string under longitudinal tension $T$ with linear mass density $\mu$ ($\text{kg/m}$). Consider an infinitesimal curved element of the string of unstretched length $\Delta x$ and mass $\Delta m = \mu \Delta x$.
For small transverse deflections ($\theta \ll 1\text{ rad}$, where $\sin\theta \approx \tan\theta = \frac{\partial y}{\partial x}$):
1. Net Transverse Force:
$$F_y = T \sin(\theta + \Delta \theta) – T \sin\theta \approx T \left[ \left(\frac{\partial y}{\partial x}\right)_{x + \Delta x} – \left(\frac{\partial y}{\partial x}\right)_x \right] = T \frac{\partial^2 y}{\partial x^2} \Delta x$$
2. Newton’s Second Law:
$$F_y = (\Delta m) a_y = (\mu \Delta x) \frac{\partial^2 y}{\partial t^2}$$
Equating the two expressions:
$$T \frac{\partial^2 y}{\partial x^2} \Delta x = \mu \Delta x \frac{\partial^2 y}{\partial t^2} \implies \frac{\partial^2 y}{\partial x^2} = \frac{\mu}{T} \frac{\partial^2 y}{\partial t^2}$$
Comparing directly with the standard wave equation $\frac{\partial^2 y}{\partial x^2} = \frac{1}{v^2}\frac{\partial^2 y}{\partial t^2}$, the propagation speed of transverse waves is:
$$v = \sqrt{\frac{T}{\mu}}$$
4. Particle Kinematics vs. Wave Kinematics
For a sinusoidal progressive harmonic wave moving in the $+x$ direction:
$$y(x, t) = A \sin(k x – \omega t + \phi)$$
where $k = \frac{2\pi}{\lambda}$ is the angular wave number, $\omega = 2\pi f$ is the angular frequency, and $v = \frac{\omega}{k} = f \lambda$.
1. Particle Velocity ($v_p$): The instantaneous vertical velocity of a medium particle at fixed coordinate $x$:
$$v_p = \frac{\partial y}{\partial t} = -A \omega \cos(k x – \omega t + \phi)$$
2. Wave Slope ($\frac{\partial y}{\partial x}$): Spatial rate of deformation:
$$\frac{\partial y}{\partial x} = A k \cos(k x – \omega t + \phi)$$
3. Fundamental Velocity-Slope Relation: Dividing particle velocity by wave slope:
$$v_p = -\left(\frac{\omega}{k}\right) \frac{\partial y}{\partial x} = -v_{\text{wave}} \times (\text{Slope of waveform at that point})$$
5. Energy Transport & Transmitted Power
For an element of length $dx$ and mass $dm = \mu dx$ on the vibrating string:
1. Kinetic Energy per unit length ($u_K$):
$$u_K = \frac{dK}{dx} = \frac{1}{2} \mu v_p^2 = \frac{1}{2} \mu \omega^2 A^2 \cos^2(k x – \omega t)$$
2. Potential Energy per unit length ($u_U$): Due to the elastic stretching of the string against tension $T$:
$$u_U = \frac{dU}{dx} = \frac{1}{2} T \left(\frac{\partial y}{\partial x}\right)^2 = \frac{1}{2} T k^2 A^2 \cos^2(k x – \omega t)$$
Since $T k^2 = (\mu v^2)\left(\frac{\omega}{v}\right)^2 = \mu \omega^2$, we have $u_K = u_U$. The potential energy and kinetic energy are equal at every point in a progressive wave and oscillate strictly in phase.
3. Average Power Transmitted ($\langle P \rangle$):
$$\langle P \rangle = \frac{1}{2} \mu \omega^2 A^2 v = 2 \pi^2 \mu f^2 A^2 v$$
6. Solved Examples
Example 1 (Foundational – JEE Main):
A steel wire of length $2.0\text{ m}$ and mass $40\text{ g}$ is under a tension of $80\text{ N}$. Find: (a) the linear mass density $\mu$, (b) the speed of transverse waves on the wire, and (c) the time taken by a transverse wave pulse to travel from one end of the wire to the other.
Solution:
(a) Linear mass density:
$$\mu = \frac{m}{L} = \frac{40 \times 10^{-3}\text{ kg}}{2.0\text{ m}} = 0.02\text{ kg/m} = \mathbf{2.0 \times 10^{-2}\text{ kg/m}}$$
(b) Transverse wave velocity:
$$v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{80\text{ N}}{0.02\text{ kg/m}}} = \sqrt{4000} \approx \mathbf{63.25\text{ m/s}}$$
(c) Transit time across length $L$:
$$t = \frac{L}{v} = \frac{2.0}{\sqrt{4000}} = \frac{2.0}{63.245} \approx \mathbf{0.0316\text{ s} = 31.6\text{ ms}}$$
Example 2 (Particle Velocity vs. Wave Slope – JEE Main/Advanced):
A sinusoidal transverse wave is propagating along the $+x$ direction on a string with speed $v = 20\text{ m/s}$. The equation of the wave is $y(x, t) = 0.05 \sin(4\pi x – 80\pi t)$, where $x$ and $y$ are in meters and $t$ is in seconds. At $t = 0$ and $x = 0.125\text{ m}$, find: (a) the transverse displacement, (b) the slope of the string, and (c) the particle velocity.
Solution:
Given: $A = 0.05\text{ m}$, $k = 4\pi\text{ rad/m}$, $\omega = 80\pi\text{ rad/s}$.
At $t = 0, x = 0.125\text{ m}$, the phase is $\theta = k x – \omega t = 4\pi(0.125) – 0 = \frac{\pi}{2}\text{ rad}$.
(a) Displacement:
$$y = 0.05 \sin\left(\frac{\pi}{2}\right) = \mathbf{0.05\text{ m} = 5.0\text{ cm}}$$
(b) Slope of the string:
$$\frac{\partial y}{\partial x} = A k \cos(k x – \omega t) = (0.05)(4\pi) \cos\left(\frac{\pi}{2}\right) = \mathbf{0}$$
(c) Particle velocity:
$$v_p = \frac{\partial y}{\partial t} = -A \omega \cos(k x – \omega t) = -(0.05)(80\pi) \cos\left(\frac{\pi}{2}\right) = \mathbf{0}$$
At maximum displacement ($y = A$), the particle reaches a turning point where both slope and velocity vanish.
Example 3 (Vertical Heavy Hanging Rope – JEE Advanced):
A uniform heavy rope of mass $M$ and length $L$ hangs vertically from a rigid ceiling under gravity. A small transverse pulse is initiated at the lower free end. (a) Find the speed of the pulse as a function of height $y$ measured from the free end. (b) Calculate the total time taken by the pulse to reach the top support.
Solution:
(a) At distance $y$ from the bottom, the tension in the rope supports the weight of the rope segment below it:
$$\text{Mass of segment below: } m(y) = \mu y = \left(\frac{M}{L}\right)y$$
Tension at height $y$: $T(y) = m(y) g = \mu g y$.
Local wave speed:
$$v(y) = \sqrt{\frac{T(y)}{\mu}} = \sqrt{\frac{\mu g y}{\mu}} = \mathbf{\sqrt{g y}}$$
(b) Transit time from $y = 0$ to $y = L$:
Since $v(y) = \frac{dy}{dt} = \sqrt{g y}$:
$$dt = \frac{dy}{\sqrt{g y}} = \frac{1}{\sqrt{g}} y^{-1/2} dy$$
Integrating from $y = 0$ to $y = L$:
$$t_{\text{total}} = \frac{1}{\sqrt{g}} \int_0^L y^{-1/2} dy = \frac{1}{\sqrt{g}} \left[ 2\sqrt{y} \right]_0^L = \mathbf{2\sqrt{\frac{L}{g}}}$$
Alternative acceleration perspective: $a = \frac{dv}{dt} = \frac{dv}{dy}\frac{dy}{dt} = \left(\frac{\sqrt{g}}{2\sqrt{y}}\right)\sqrt{gy} = \frac{g}{2} = \text{constant}$. Using kinematics $L = \frac{1}{2} a t^2 \implies L = \frac{1}{2}\left(\frac{g}{2}\right)t^2 \implies t = 2\sqrt{\frac{L}{g}}$.
Example 4 (Power Transmission and Energy Density – JEE Advanced):
A long string with linear mass density $\mu = 0.05\text{ kg/m}$ is held under tension $T = 45\text{ N}$. A wave generator at one end oscillates transversely at frequency $f = 60\text{ Hz}$ with amplitude $A = 2.0\text{ cm}$. Calculate: (a) the wave velocity, (b) the total energy per unit length of the string, and (c) the average mechanical power transmitted along the string.
Solution:
(a) Wave velocity:
$$v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{45}{0.05}} = \sqrt{900} = \mathbf{30\text{ m/s}}$$
(b) Angular frequency: $\omega = 2\pi f = 2\pi(60) = 120\pi\text{ rad/s}$.
Average total energy density (energy per unit length):
$$\langle u_{\text{total}} \rangle = \langle u_K \rangle + \langle u_U \rangle = \frac{1}{2}\mu \omega^2 A^2$$
$$\langle u_{\text{total}} \rangle = \frac{1}{2}(0.05)(120\pi)^2 (0.02)^2 = 0.025 \times (14400 \pi^2) \times (0.0004) = 0.144 \pi^2 \approx \mathbf{1.421\text{ J/m}}$$
(c) Average transmitted power:
$$\langle P \rangle = \langle u_{\text{total}} \rangle \cdot v = (1.421\text{ J/m})(30\text{ m/s}) \approx \mathbf{42.64\text{ W}}$$
$$\text{Direct formula: } \langle P \rangle = \frac{1}{2} \mu \omega^2 A^2 v = \frac{1}{2}(0.05)(14400\pi^2)(0.0004)(30) = 4.32 \pi^2 \approx 4.32 \times 9.8696 \approx \mathbf{42.64\text{ W}}$$
7. 10-Question Practice Worksheet (DPP)
Q1 (Single Choice): Which of the following functions represents a traveling progressive wave?
(A) $y = (x – vt)^2$
(B) $y = \ln(x + vt)$
(C) $y = \frac{1}{1 + (x – vt)^2}$
(D) $y = e^{-(x^2 + v^2 t^2)}$
Q2 (Single Choice): A transverse wave travels along the positive $x$-axis on a taut string. At an instant, a segment of the string has a positive slope ($\frac{\partial y}{\partial x} > 0$). The particle velocity $v_p$ of this segment is:
(A) Upward (positive $y$)
(B) Downward (negative $y$)
(C) Zero
(D) Directed along positive $x$
Q3 (Single Choice): The tension in a stretched string is increased by $69\%$. Keeping the linear density constant, the percentage increase in transverse wave velocity is:
(A) $30\%$
(B) $34.5\%$
(C) $69\%$
(D) $13\%$
Q4 (Single Choice): A wave equation is given by $y = 0.02 \sin(0.5 x – 10 t)$, with SI units. The maximum particle acceleration is:
(A) $0.2\text{ m/s}^2$
(B) $2.0\text{ m/s}^2$
(C) $5.0\text{ m/s}^2$
(D) $20\text{ m/s}^2$
Q5 (Single Choice): If the amplitude of a progressive wave on a string is doubled and its frequency is halved, the average power transmitted by the wave changes by a factor of:
(A) $1$ (unchanged)
(B) $2$
(C) $4$
(D) $1/2$
Q6 (Single Choice): A heavy uniform rope of length $L$ hangs freely from the ceiling. The ratio of wave speed at distance $L/4$ from the bottom to that at the top suspension is:
(A) $1 : 4$
(B) $1 : 2$
(C) $2 : 1$
(D) $1 : \sqrt{2}$
Q7 (One or More than One Correct): Which of the following statements regarding a transverse harmonic progressive wave are strictly TRUE?
(A) At the crests and troughs, the particle velocity is zero while the wave slope is zero.
(B) At the nodes of displacement ($y = 0$), the particle velocity has maximum magnitude.
(C) Kinetic energy density and potential energy density are identical in magnitude at every point along the string.
(D) The wave velocity is equal to the maximum particle velocity.
Q8 (One or More than One Correct): Two strings $A$ and $B$ of the same material are subjected to the same tension. The radius of string $A$ is double that of string $B$. Transverse waves are established in both strings. Then:
(A) The linear mass density of string $A$ is 4 times that of string $B$.
(B) The wave velocity in string $A$ is half that in string $B$.
(C) If both carry waves of the same frequency and amplitude, the power transmitted in $A$ is double that in $B$.
(D) If both carry waves of the same wavelength, the frequency in $A$ is double that in $B$.
Q9 (Numerical Value): A transverse harmonic wave of amplitude $0.04\text{ m}$ and frequency $50\text{ Hz}$ travels along a string under tension $100\text{ N}$ having linear mass density $0.01\text{ kg/m}$. The average power transmitted along the string in Watts (rounded to the nearest integer) is ________. (Take $\pi = 3.1416$)
Q10 (Numerical Value): A uniform rope of mass $0.9\text{ kg}$ and length $10\text{ m}$ hangs vertically from a ceiling. A small transverse pulse generated at the lower end travels upward. Taking $g = 10\text{ m/s}^2$, the time taken in seconds for the pulse to reach the top ceiling is ________.
8. Answer Key & Detailed Step-by-Step Solutions
Answer Key:
Q1: (C) | Q2: (B) | Q3: (A) | Q4: (B) | Q5: (A) | Q6: (B) | Q7: (A, B, C) | Q8: (A, B, C) | Q9: 79 | Q10: 2
Detailed Solutions
Solution 1:
A valid wave function must remain finite and bounded for all coordinates $x$ and $t$. For (A) and (B), as $x \to \infty$, the function diverges to infinity. Function (D) is not in the form $f(x \pm vt)$. Function (C) has the valid traveling form $g(x – vt)$ and remains strictly bounded ($0 < y \le 1$). Hence, (C) is correct.
Solution 2:
The fundamental relation between particle velocity and waveform slope is:
$$v_p = -v_{\text{wave}} \left(\frac{\partial y}{\partial x}\right)$$
Given $v_{\text{wave}} > 0$ (moving along $+x$) and $\frac{\partial y}{\partial x} > 0$, we have $v_p < 0$. The particle moves downward. Hence, (B) is correct.
Solution 3:
Wave velocity $v = \sqrt{T/\mu} \propto \sqrt{T}$.
When tension increases by $69\%$, $T’ = 1.69 T$.
New velocity: $v’ = \sqrt{1.69 T / \mu} = 1.30 v$.
Percentage increase: $\frac{v’ – v}{v} \times 100\% = (1.30 – 1.00) \times 100\% = \mathbf{30\%}$. Hence, (A) is correct.
Solution 4:
From $y = A \sin(kx – \omega t)$, we have $A = 0.02\text{ m}$ and $\omega = 10\text{ rad/s}$.
Maximum particle acceleration:
$$a_{\text{max}} = \omega^2 A = (10)^2 (0.02) = 100 \times 0.02 = \mathbf{2.0\text{ m/s}^2}$$
Hence, (B) is correct.
Solution 5:
Average power transmitted is $\langle P \rangle = 2\pi^2 \mu f^2 A^2 v \propto f^2 A^2$.
If $A’ = 2A$ and $f’ = f/2$:
$$(f’)^2 (A’)^2 = \left(\frac{f}{2}\right)^2 (2A)^2 = \frac{f^2}{4} (4A^2) = f^2 A^2$$
The power is unchanged (factor of 1). Hence, (A) is correct.
Solution 6:
Wave velocity at distance $y$ from the bottom of a hanging rope is $v(y) = \sqrt{gy}$.
$$\frac{v(L/4)}{v(L)} = \frac{\sqrt{g(L/4)}}{\sqrt{gL}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$$
Hence, (B) is correct.
Solution 7:
(A) True: At turning points (crests/troughs), $v_p = 0$ and the local tangent is horizontal ($\partial y/\partial x = 0$).
(B) True: At $y = 0$, $\cos(kx – \omega t) = \pm 1$, so $|v_p| = A\omega = v_{p,\text{max}}$.
(C) True: In progressive waves, $u_K = u_U = \frac{1}{2}\mu \omega^2 A^2 \cos^2(kx – \omega t)$ everywhere.
(D) False: Wave velocity $v = \omega/k$ is an intrinsic property of the medium, while $v_{p,\text{max}} = \omega A = k A v$, which equals $v$ only if $kA = 1$.
Hence, (A, B, C) are correct.
Solution 8:
(A) True: $\mu = \rho (\pi r^2) \propto r^2$. Since $r_A = 2r_B$, $\mu_A = 4\mu_B$.
(B) True: $v = \sqrt{T/\mu} \propto 1/\sqrt{\mu} \implies v_A = v_B/\sqrt{4} = v_B/2$.
(C) True: Power $P = \frac{1}{2}\mu \omega^2 A^2 v$. Since $\mu_A v_A = (4\mu_B)(v_B/2) = 2\mu_B v_B$, $P_A = 2P_B$.
(D) False: Frequency $f = v/\lambda \propto v$, so $f_A = f_B/2$.
Hence, (A, B, C) are correct.
Solution 9:
Wave velocity: $v = \sqrt{T/\mu} = \sqrt{100 / 0.01} = \sqrt{10000} = 100\text{ m/s}$.
Angular frequency: $\omega = 2\pi(50) = 100\pi\text{ rad/s}$.
Average power:
$$\langle P \rangle = \frac{1}{2}\mu \omega^2 A^2 v = \frac{1}{2}(0.01)(100\pi)^2 (0.04)^2 (100)$$
$$\langle P \rangle = \frac{1}{2}(0.01)(10000\pi^2)(0.0016)(100) = \frac{1}{2}(100\pi^2)(0.16) = 8\pi^2\text{ W}$$
Using $\pi^2 \approx 9.8696$: $\langle P \rangle \approx 8 \times 9.8696 = 78.96\text{ W} \approx \mathbf{79\text{ W}}$.
Answer is 79.
Solution 10:
Transit time for a transverse pulse along a heavy vertical rope is:
$$t = 2\sqrt{\frac{L}{g}}$$
Substituting $L = 10\text{ m}$ and $g = 10\text{ m/s}^2$:
$$t = 2\sqrt{\frac{10}{10}} = 2\sqrt{1} = \mathbf{2\text{ seconds}}$$
Answer is 2.