Welcome to Lecture 109 of our comprehensive JEE Physics Master Series. Having formulated the propagation of traveling waves, we now analyze what happens when multiple waves occupy the same region of space or encounter material discontinuities: the Principle of Superposition, Boundary Reflections and Transmissions, and the formation of Standing (Stationary) Waves on a Stretched String. We derive the amplitude reflection and transmission coefficients from boundary continuity conditions, formulate the spatial and temporal nodes and antinodes of standing waves, evaluate normal vibration modes (harmonics and overtones) of fixed-fixed and fixed-free strings, and solve 4 graded benchmark examples followed by a 10-question practice worksheet (DPP) with complete step-by-step solutions.
1. Principle of Linear Superposition
Because the governing 1D differential wave equation $\frac{\partial^2 y}{\partial x^2} = \frac{1}{v^2}\frac{\partial^2 y}{\partial t^2}$ is strictly linear, any linear combination of individual solutions is also an exact solution. When two or more waves travel simultaneously through an elastic medium, the resultant displacement $y(x, t)$ of any particle at position $x$ and time $t$ is the algebraic vector sum of the individual displacements:
$$y(x, t) = y_1(x, t) + y_2(x, t) + \dots + y_n(x, t)$$
Superposition holds whenever the strain in the medium remains within the linear elastic Hookean regime (small amplitudes). At extreme amplitudes (non-linear regime), shock waves form and the principle ceases to apply.
2. Boundary Reflections and Transmissions
Consider a junction at $x = 0$ between two semi-infinite strings under the same tension $T$: String 1 has linear density $\mu_1$ (wave speed $v_1 = \sqrt{T/\mu_1}$) for $x < 0$, and String 2 has linear density $\mu_2$ (wave speed $v_2 = \sqrt{T/\mu_2}$) for $x > 0$.
Let an incident harmonic wave travel along $+x$: $y_i(x, t) = A_i \sin(k_1 x – \omega t)$. Upon striking the boundary, a reflected wave $y_r(x, t) = A_r \sin(-k_1 x – \omega t)$ travels backward into String 1, and a transmitted wave $y_t(x, t) = A_t \sin(k_2 x – \omega t)$ propagates forward into String 2.
Two essential boundary conditions must hold at $x = 0$ at all times:
1. Displacement Continuity: The string cannot break at the joint:
$$y_i(0, t) + y_r(0, t) = y_t(0, t) \implies A_i + A_r = A_t$$
2. Slope Continuity (Transverse Force Balance): A massless junction point cannot support a finite net transverse force ($T \left.\frac{\partial y_1}{\partial x}\right|_{x=0} = T \left.\frac{\partial y_2}{\partial x}\right|_{x=0}$):
$$k_1 A_i – k_1 A_r = k_2 A_t$$
Solving these two simultaneous equations using $k = \omega / v$ yields the Reflection and Transmission Amplitude Coefficients:
$$A_r = \left(\frac{v_2 – v_1}{v_1 + v_2}\right) A_i = \left(\frac{k_1 – k_2}{k_1 + k_2}\right) A_i = \left(\frac{\sqrt{\mu_1} – \sqrt{\mu_2}}{\sqrt{\mu_1} + \sqrt{\mu_2}}\right) A_i$$
$$A_t = \left(\frac{2 v_2}{v_1 + v_2}\right) A_i = \left(\frac{2 k_1}{k_1 + k_2}\right) A_i = \left(\frac{2\sqrt{\mu_1}}{\sqrt{\mu_1} + \sqrt{\mu_2}}\right) A_i$$
Phase Change Rules:
• Denser Medium Reflection ($v_2 < v_1$, i.e., $\mu_2 > \mu_1$): $A_r < 0$. The reflected wave undergoes an abrupt phase reversal of $\pi\text{ rad}$ ($180^\circ$). For an infinitely rigid wall ($v_2 = 0$), $A_r = -A_i$ and $A_t = 0$.
• Rarer Medium Reflection ($v_2 > v_1$, i.e., $\mu_2 < \mu_1$): $A_r > 0$. The reflected wave is reflected in phase ($\Delta\phi = 0$). For a completely free end ($v_2 \to \infty$), $A_r = +A_i$.
• Transmitted Wave: Always in phase with the incident wave ($\Delta\phi = 0$), because $A_t > 0$ for all real physical parameters.
3. Mathematical Formulation of Standing Waves
When two identical harmonic waves of equal amplitude $A$ and equal frequency $\omega$ travel along a string in opposite directions, their superposition produces a stationary wave pattern:
$$y_1(x, t) = A \sin(k x – \omega t), \quad y_2(x, t) = A \sin(k x + \omega t)$$
Using the trigonometric identity $\sin(A – B) + \sin(A + B) = 2\sin A \cos B$:
$$y(x, t) = [2 A \sin(k x)] \cos(\omega t) = A(x) \cos(\omega t)$$
where $A(x) = 2 A \sin(k x)$ is the position-dependent amplitude envelope.
1. Nodes ($A(x) = 0$): Points of permanently zero displacement:
$$\sin(k x) = 0 \implies k x = n \pi \implies x = \frac{n \lambda}{2} \quad (n = 0, 1, 2, \dots)$$
Distance between two consecutive nodes is $\Delta x_{\text{node}} = \frac{\lambda}{2}$.
2. Antinodes ($|A(x)| = 2A$): Points of maximum oscillatory amplitude:
$$|\sin(k x)| = 1 \implies k x = \left(n + \frac{1}{2}\right)\pi \implies x = \left(n + \frac{1}{2}\right)\frac{\lambda}{2} \quad (n = 0, 1, 2, \dots)$$
Distance between a node and its adjacent antinode is $\frac{\lambda}{4}$.
3. Energetics: In a standing wave, zero net mechanical energy is transported past any node ($\langle P \rangle = 0$). Energy remains localized between adjacent nodes, cyclically interchanging between pure kinetic energy (when string passes through $y = 0$) and pure elastic potential energy (at maximum deflection $t = 0, T/2$).
4. Normal Modes of Vibration on a Stretched String
Case 1: String Clamped at Both Ends ($x = 0$ and $x = L$):
Boundary conditions require nodes at both boundaries: $y(0, t) = 0$ and $y(L, t) = 0$.
This implies $\sin(k L) = 0 \implies k L = n \pi \implies \frac{2\pi}{\lambda_n} L = n \pi$:
$$\lambda_n = \frac{2L}{n}, \quad f_n = \frac{v}{\lambda_n} = n \left(\frac{v}{2L}\right) = n f_1 \quad (n = 1, 2, 3, \dots)$$
All integer harmonics (both odd and even) are naturally excited:
- $n = 1$: Fundamental mode / 1st Harmonic ($f_1 = \frac{v}{2L}$)
- $n = 2$: 1st Overtone / 2nd Harmonic ($f_2 = 2 f_1$)
- $n = 3$: 2nd Overtone / 3rd Harmonic ($f_3 = 3 f_1$)
- Number of nodes = $n + 1$; Number of antinodes = $n$.
Case 2: String Fixed at One End ($x = 0$) and Free at the Other ($x = L$):
Node at $x = 0$ and antinode at $x = L$. The boundary condition requires $\cos(k L) = 0 \implies k L = (2n – 1)\frac{\pi}{2}$:
$$\lambda_n = \frac{4L}{2n – 1}, \quad f_n = (2n – 1)\left(\frac{v}{4L}\right) = (2n – 1) f_1 \quad (n = 1, 2, 3, \dots)$$
Only odd harmonics are present ($f_1, 3f_1, 5f_1, \dots$).
5. Solved Examples
Example 1 (Foundational – JEE Main):
A traveling wave of amplitude $A_i = 4.0\text{ mm}$ propagates on a string of linear mass density $\mu_1 = 0.04\text{ kg/m}$ under tension $T = 36\text{ N}$. It strikes a boundary connected to a second string of linear mass density $\mu_2 = 0.16\text{ kg/m}$ under the same tension. Find: (a) the amplitude of the reflected wave $A_r$, (b) the amplitude of the transmitted wave $A_t$, and (c) the phase shift of the reflected wave.
Solution:
Wave speeds in the two media:
$$v_1 = \sqrt{\frac{T}{\mu_1}} = \sqrt{\frac{36}{0.04}} = \sqrt{900} = 30\text{ m/s}$$
$$v_2 = \sqrt{\frac{T}{\mu_2}} = \sqrt{\frac{36}{0.16}} = \sqrt{225} = 15\text{ m/s}$$
(a) Reflected amplitude:
$$A_r = \left(\frac{v_2 – v_1}{v_1 + v_2}\right) A_i = \left(\frac{15 – 30}{30 + 15}\right)(4.0\text{ mm}) = \left(-\frac{15}{45}\right)(4.0) = -\frac{1}{3}(4.0) = \mathbf{-1.33\text{ mm}}$$
Magnitude is $1.33\text{ mm}$.
(b) Transmitted amplitude:
$$A_t = \left(\frac{2 v_2}{v_1 + v_2}\right) A_i = \left(\frac{2 \times 15}{30 + 15}\right)(4.0\text{ mm}) = \left(\frac{30}{45}\right)(4.0) = \frac{2}{3}(4.0) = \mathbf{+2.67\text{ mm}}$$
(c) Since $A_r < 0$, the wave undergoes a phase shift of $\mathbf{\pi\text{ rad} = 180^\circ}$ upon reflection from the denser string.
Example 2 (Standing Wave Node-Antinode Analysis – JEE Main/Advanced):
A standing wave on a string clamped at both ends is represented by $y(x, t) = 0.06 \sin(0.5\pi x) \cos(40\pi t)$, where $x$ and $y$ are in meters and $t$ is in seconds. (a) Determine the amplitude and velocity of the two constituent traveling waves. (b) Find the distance between two consecutive nodes. (c) Calculate the transverse velocity of a string particle at $x = 1.5\text{ m}$ at time $t = 0.125\text{ s}$.
Solution:
(a) Comparing with $y(x, t) = 2 A \sin(k x) \cos(\omega t)$:
$$2A = 0.06\text{ m} \implies A = \mathbf{0.03\text{ m} = 3\text{ cm}}$$
Wave parameters: $k = 0.5\pi\text{ rad/m}$, $\omega = 40\pi\text{ rad/s}$.
Wave velocity: $v = \frac{\omega}{k} = \frac{40\pi}{0.5\pi} = \mathbf{80\text{ m/s}}$.
(b) Wavelength $\lambda = \frac{2\pi}{k} = \frac{2\pi}{0.5\pi} = 4.0\text{ m}$.
Distance between consecutive nodes: $\Delta x = \frac{\lambda}{2} = \frac{4.0}{2} = \mathbf{2.0\text{ m}}$.
(c) Particle velocity: $v_p = \frac{\partial y}{\partial t} = -0.06 \times (40\pi) \sin(0.5\pi x) \sin(40\pi t) = -2.4\pi \sin(0.5\pi x) \sin(40\pi t)$.
Substitute $x = 1.5\text{ m}$ and $t = 0.125\text{ s} = 1/8\text{ s}$:
$$\sin(0.5\pi \times 1.5) = \sin\left(\frac{3\pi}{4}\right) = \frac{1}{\sqrt{2}}$$
$$\sin(40\pi \times 0.125) = \sin(5\pi) = 0$$
$$v_p = -2.4\pi \left(\frac{1}{\sqrt{2}}\right)(0) = \mathbf{0\text{ m/s}}$$
At $t = 1/8\text{ s}$, the string is at maximum deflection, so all particles momentarily stop.
Example 3 (Sonometer Wire Frequency Tuning – JEE Advanced):
A sonometer wire of length $L = 1.0\text{ m}$ and mass $m = 5.0\text{ g}$ is clamped between two bridges and stretched by a suspended mass $M = 9.0\text{ kg}$. (Take $g = 9.8\text{ m/s}^2$). (a) Find the fundamental frequency of transverse vibration. (b) If the length between bridges is reduced to $0.70\text{ m}$, what hanging mass $M’$ must be suspended to maintain the exact same fundamental frequency?
Solution:
(a) Linear mass density: $\mu = \frac{5.0 \times 10^{-3}\text{ kg}}{1.0\text{ m}} = 5.0 \times 10^{-3}\text{ kg/m}$.
Tension: $T = M g = 9.0 \times 9.8 = 88.2\text{ N}$.
Wave speed: $v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{88.2}{0.005}} = \sqrt{17640} \approx 132.815\text{ m/s}$.
Fundamental frequency:
$$f_1 = \frac{v}{2L} = \frac{132.815}{2(1.0)} \approx \mathbf{66.41\text{ Hz}}$$
(b) Fundamental frequency formula: $f = \frac{1}{2L}\sqrt{\frac{M g}{\mu}}$.
For $f$ to remain constant when length changes from $L_1$ to $L_2$:
$$\frac{\sqrt{M_1}}{L_1} = \frac{\sqrt{M_2}}{L_2} \implies \sqrt{M_2} = \sqrt{M_1} \left(\frac{L_2}{L_1}\right) = \sqrt{9.0} \left(\frac{0.70}{1.0}\right) = 3.0 \times 0.70 = 2.1$$
Squaring both sides:
$$M’ = (2.1)^2 = \mathbf{4.41\text{ kg}}$$
Example 4 (Energy Stored in a Standing Wave Loop – JEE Advanced):
A string of linear mass density $\mu$ clamped at $x = 0$ and $x = L$ vibrates in its $n$-th harmonic: $y(x, t) = 2 A \sin(k_n x) \cos(\omega_n t)$. (a) Calculate the total mechanical energy stored in the entire string. (b) Find the energy contained within a single loop between two adjacent nodes.
Solution:
(a) Total mechanical energy is constant and equals the maximum kinetic energy when the string is completely flat ($\cos(\omega_n t) = 0 \implies \dot{y} = \pm 2 A \omega_n \sin(k_n x)$):
$$E_{\text{total}} = \int_0^L \frac{1}{2} (dm) v_p^2 = \frac{1}{2} \mu \int_0^L [2 A \omega_n \sin(k_n x)]^2 dx = 2 \mu \omega_n^2 A^2 \int_0^L \sin^2(k_n x) dx$$
Since $k_n = \frac{n \pi}{L}$, the integral over $n$ full half-wavelengths is:
$$\int_0^L \sin^2\left(\frac{n\pi x}{L}\right) dx = \frac{L}{2}$$
$$E_{\text{total}} = 2 \mu \omega_n^2 A^2 \left(\frac{L}{2}\right) = \mathbf{\mu \omega_n^2 A^2 L} = \mathbf{2 \mu \omega_n^2 A_{\text{anti}}^2 L} \quad (\text{where } A_{\text{anti}} = 2A)$$
In terms of loop antinode amplitude $A_0 = 2A$: $E_{\text{total}} = \frac{1}{4} \mu \omega_n^2 A_0^2 L$.
(b) The string vibrating in the $n$-th harmonic consists of exactly $n$ identical loops. By spatial symmetry, the energy distributed across each loop is:
$$E_{\text{loop}} = \frac{E_{\text{total}}}{n} = \frac{\mu \omega_n^2 A^2 L}{n} = \mathbf{\mu \omega_n^2 A^2 \left(\frac{\lambda_n}{2}\right)}$$
6. 10-Question Practice Worksheet (DPP)
Q1 (Single Choice): When a transverse wave pulse travelling on a string is reflected from a rigid fixed boundary, the reflected pulse suffers a phase shift of:
(A) $0$
(B) $\pi/2$
(C) $\pi$
(D) $2\pi$
Q2 (Single Choice): A string of length $L$ fixed at both ends vibrates in $4$ loops. The wavelength of the standing wave is:
(A) $L/4$
(B) $L/2$
(C) $2L$
(D) $4L$
Q3 (Single Choice): In a standing wave on a string, the distance between any node and the nearest antinode is:
(A) $\lambda$
(B) $\lambda/2$
(C) $\lambda/4$
(D) $\lambda/8$
Q4 (Single Choice): Two identical waves travelling in opposite directions produce a standing wave $y = 10 \cos(\pi x) \sin(20\pi t)$ (with $x, y$ in cm and $t$ in s). The distance between two adjacent nodes is:
(A) $0.5\text{ cm}$
(B) $1.0\text{ cm}$
(C) $2.0\text{ cm}$
(D) $4.0\text{ cm}$
Q5 (Single Choice): The ratio of the fundamental frequency of a string clamped at both ends to that of a string of the same length, tension, and material clamped at one end and free at the other is:
(A) $1 : 2$
(B) $2 : 1$
(C) $1 : 1$
(D) $4 : 1$
Q6 (Single Choice): A string fixed at both ends has consecutive resonant frequencies of $150\text{ Hz}$ and $200\text{ Hz}$. The fundamental frequency of the string is:
(A) $25\text{ Hz}$
(B) $50\text{ Hz}$
(C) $75\text{ Hz}$
(D) $100\text{ Hz}$
Q7 (One or More than One Correct): Which of the following statements regarding standing waves on a stretched string are strictly TRUE?
(A) In a standing wave, all particles between two consecutive nodes oscillate in the same phase.
(B) Particles located in adjacent loops oscillate in antiphase (phase difference of $\pi\text{ rad}$).
(C) At the nodes, the strain in the string is maximum when the string is flat.
(D) Mechanical energy flows continuously across the antinodes from left to right.
Q8 (One or More than One Correct): A wave traveling along a lighter string 1 strikes the junction with a heavier string 2 ($\mu_2 > \mu_1$). Then:
(A) The speed of the wave in string 2 is less than that in string 1.
(B) The frequency of the transmitted wave equals the frequency of the incident wave.
(C) The transmitted wave is in phase with the incident wave.
(D) The reflected wave is in phase with the incident wave.
Q9 (Numerical Value): A wire of length $1.2\text{ m}$ clamped at both ends is under a tension that produces a transverse wave speed of $240\text{ m/s}$. The frequency of its third harmonic (second overtone) in Hertz is ________.
Q10 (Numerical Value): A wave pulse of amplitude $6\text{ mm}$ travels on string 1 with wave speed $20\text{ m/s}$ towards a junction with string 2. If the wave speed in string 2 is $60\text{ m/s}$, the amplitude of the transmitted wave $A_t$ in millimeters is ________.
7. Answer Key & Detailed Step-by-Step Solutions
Answer Key:
Q1: (C) | Q2: (B) | Q3: (C) | Q4: (B) | Q5: (B) | Q6: (B) | Q7: (A, B, C) | Q8: (A, B, C) | Q9: 300 | Q10: 9
Detailed Solutions
Solution 1:
A rigid boundary represents an infinitely dense medium ($v_2 = 0$). By the reflection coefficient formula:
$$A_r = \left(\frac{v_2 – v_1}{v_1 + v_2}\right) A_i = \left(\frac{0 – v_1}{v_1 + 0}\right) A_i = -A_i$$
The negative sign corresponds to an inverted pulse, which is a phase change of $\pi\text{ rad}$. Hence, (C) is correct.
Solution 2:
A string vibrating in $n$ loops has length $L = n \left(\frac{\lambda}{2}\right)$.
For $n = 4$:
$$L = 4\left(\frac{\lambda}{2}\right) = 2\lambda \implies \lambda = \frac{L}{2}$$
Hence, (B) is correct.
Solution 3:
In any standing wave, nodes occur where $k x = n\pi \implies x = n\frac{\lambda}{2}$, and antinodes occur where $k x = (2n + 1)\frac{\pi}{2} \implies x = (2n + 1)\frac{\lambda}{4}$.
The separation between an adjacent node and antinode is $\frac{\lambda}{4} – 0 = \frac{\lambda}{4}$. Hence, (C) is correct.
Solution 4:
From $y = 10 \cos(\pi x) \sin(20\pi t)$, the spatial envelope is $A(x) = 10 \cos(\pi x)$.
Nodes occur where $\cos(\pi x) = 0 \implies \pi x = (2n + 1)\frac{\pi}{2} \implies x = n + 0.5\text{ cm}$.
Consecutive nodes occur at $x_1 = 0.5\text{ cm}, x_2 = 1.5\text{ cm}, x_3 = 2.5\text{ cm}$.
Distance between adjacent nodes: $\Delta x = 1.5 – 0.5 = \mathbf{1.0\text{ cm}}$. Hence, (B) is correct.
Solution 5:
For a string clamped at both ends: $f_{\text{both}} = \frac{v}{2L}$.
For a string clamped at one end and free at the other: $f_{\text{one}} = \frac{v}{4L}$.
Ratio: $\frac{f_{\text{both}}}{f_{\text{one}}} = \frac{v / (2L)}{v / (4L)} = \frac{4}{2} = 2 : 1$. Hence, (B) is correct.
Solution 6:
For a string fixed at both ends, all harmonics are integer multiples of the fundamental: $f_n = n f_1$.
The difference between any two consecutive harmonics is the fundamental frequency:
$$\Delta f = f_{n+1} – f_n = (n+1)f_1 – n f_1 = f_1$$
$$f_1 = 200\text{ Hz} – 150\text{ Hz} = \mathbf{50\text{ Hz}}$$
Hence, (B) is correct.
Solution 7:
(A) True: Between two consecutive nodes, $\sin(kx)$ does not change sign, so all points oscillate synchronously with identical phase.
(B) True: Crossing a node flips the sign of $\sin(kx)$, introducing an exact $\pi$ phase shift.
(C) True: Strain is $\frac{\partial y}{\partial x} \propto \cos(kx)$, which reaches its maximum magnitude at the nodes ($\sin kx = 0 \implies \cos kx = \pm 1$) when the string passes through the equilibrium line.
(D) False: Standing waves transport zero net energy through nodes or antinodes. Hence, (A, B, C) are correct.
Solution 8:
(A) True: $v = \sqrt{T/\mu}$. Since $\mu_2 > \mu_1$ under identical tension, $v_2 < v_1$.
(B) True: Frequency is determined by the source driver and is strictly invariant across boundaries.
(C) True: $A_t = \frac{2 v_2}{v_1 + v_2} A_i > 0$, so transmission is always in phase.
(D) False: $A_r = \frac{v_2 – v_1}{v_1 + v_2} A_i < 0$ because $v_2 < v_1$, so the reflected wave undergoes a $\pi$ phase reversal. Hence, (A, B, C) are correct.
Solution 9:
Fundamental frequency: $f_1 = \frac{v}{2L} = \frac{240}{2(1.2)} = \frac{240}{2.4} = 100\text{ Hz}$.
Third harmonic ($n = 3$, second overtone):
$$f_3 = 3 f_1 = 3 \times 100 = \mathbf{300\text{ Hz}}$$
Answer is 300.
Solution 10:
Using the transmission amplitude formula:
$$A_t = \left(\frac{2 v_2}{v_1 + v_2}\right) A_i$$
Given $v_1 = 20\text{ m/s}$, $v_2 = 60\text{ m/s}$, and $A_i = 6\text{ mm}$:
$$A_t = \left(\frac{2 \times 60}{20 + 60}\right)(6\text{ mm}) = \left(\frac{120}{80}\right)(6) = \left(\frac{3}{2}\right)(6) = \mathbf{9\text{ mm}}$$
Answer is 9.