Welcome to Lecture 123 (Master Index Entry: L117) of our comprehensive JEE Physics Master Series! Advancing deeper into oscillatory wave patterns in Unit 10: Oscillations and Waves, we present the foundational resonance masterclass: Reflection of Waves and Standing Waves on Stretched Strings — Boundary Phase Shifts, Normal Modes, Harmonics, and Melde’s Experiment. When a progressive wave encounters a structural boundary, reflection creates counter-propagating waves that superpose to form standing (stationary) wave patterns. In this masterclass, we analyze the phase mechanics of rigid versus free boundary reflections, derive the standing wave equation with decoupled spatial and temporal coordinates, identify nodes and antinodes, establish the complete harmonic spectrum of clamped strings, examine Melde’s longitudinal and transverse resonance laws, and master high-yield JEE Advanced problem patterns.
Detailed Concept Card: Wave Reflection & Standing Waves in Strings
1. Boundary Conditions & Phase Shifts in Wave Reflection
When a travelling wave pulse reaches the boundary of an elastic medium, the boundary conditions dictate the amplitude and phase of the reflected wave:
| Boundary Type | Physical Boundary Condition | Phase Shift ($\Delta\phi$) | Waveform Behavior | Mathematical Form |
|---|---|---|---|---|
| Rigid / Fixed End (Denser Medium) | Displacement must remain zero: $y(0, t) = 0$. By Newton’s third law, the support exerts an inverted reaction force. | $$\mathbf{\Delta\phi = \pi\text{ rad} \ (180^\circ)}$$ (Path change: $\lambda/2$) |
An incident crest reflects as a trough (inverted pulse). | $$y_i = A\sin(kx – \omega t)$$ $$y_r = -A\sin(kx + \omega t)$$ |
| Free / Open End (Rarer Medium) | Transverse force must be zero: $\frac{\partial y}{\partial x} = 0$. The boundary particle is free to move. | $$\mathbf{\Delta\phi = 0\text{ rad} \ (0^\circ)}$$ (No path change) |
An incident crest reflects as an upright crest. | $$y_i = A\sin(kx – \omega t)$$ $$y_r = A\sin(kx + \omega t)$$ |
Reflection & Transmission Coefficients at a String Junction ($\mu_1 \to \mu_2$):
$$\mathbf{A_r = \left(\frac{v_2 – v_1}{v_2 + v_1}\right) A_i = \left(\frac{\sqrt{\mu_1} – \sqrt{\mu_2}}{\sqrt{\mu_1} + \sqrt{\mu_2}}\right) A_i}, \quad \mathbf{A_t = \left(\frac{2v_2}{v_2 + v_1}\right) A_i = \left(\frac{2\sqrt{\mu_1}}{\sqrt{\mu_1} + \sqrt{\mu_2}}\right) A_i}$$
2. Mathematical Formulation of Standing (Stationary) Waves
When an incident wave $y_1(x, t) = A \sin(kx – \omega t)$ superposes with a reflected wave from a rigid boundary at $x = 0$, $y_2(x, t) = -A \sin(-kx – \omega t) = A \sin(kx + \omega t)$:
$$y_{\text{net}} = y_1 + y_2 = A [\sin(kx – \omega t) + \sin(kx + \omega t)] = \mathbf{[2A \sin(kx)] \cos(\omega t)}$$
Taking the spatial amplitude function $\mathbf{A(x) = 2A \sin(kx)}$:
$$\mathbf{y_{\text{net}}(x, t) = A(x) \cos(\omega t)}$$
- Complete Decoupling of Variables: Unlike progressive waves where variables appear coupled as $(kx \pm \omega t)$, in standing waves, the spatial distribution $A(x)$ and temporal oscillation $\cos(\omega t)$ are completely separated.
- Zero Energy Propagation: Energy is trapped between nodal boundaries. In each half-cycle, energy oscillates continuously between kinetic energy of moving elements and potential strain energy of deformed elements.
3. Nodes and Antinodes
| Feature / Parameter | Nodes ($N$) | Antinodes ($A$) |
|---|---|---|
| Physical Definition | Points of permanent zero displacement and zero velocity. | Points of maximum transverse oscillation amplitude ($2A$). |
| Mathematical Condition | $$A(x) = 0 \implies \sin(kx) = 0 \implies kx = n\pi$$ | $$|A(x)| = 2A \implies |\sin(kx)| = 1 \implies kx = (2n+1)\frac{\pi}{2}$$ |
| Spatial Positions | $$\mathbf{x = n \frac{\lambda}{2} = 0, \frac{\lambda}{2}, \lambda, \frac{3\lambda}{2}, \dots}$$ | $$\mathbf{x = (2n+1)\frac{\lambda}{4} = \frac{\lambda}{4}, \frac{3\lambda}{4}, \frac{5\lambda}{4}, \dots}$$ |
| Elastic Strain & Tension | Maximum strain ($\frac{\partial y}{\partial x} = \text{max}$). | Zero strain ($\frac{\partial y}{\partial x} = 0$). |
| Consecutive Separation | $$\text{Distance between two consecutive nodes} = \mathbf{\frac{\lambda}{2}}$$ | $$\text{Distance between node and adjacent antinode} = \mathbf{\frac{\lambda}{4}}$$ |
Phase Relationships Across Nodes: All particles located between two consecutive nodes vibrate in the exact same phase. Particles on opposite sides of a node vibrate in exact anti-phase ($180^\circ$ phase difference).
4. Normal Modes of a String Fixed at Both Ends
For a string of length $L$ fixed rigidly at both $x = 0$ and $x = L$, nodes must exist at both boundaries: $\sin(kL) = 0 \implies kL = n\pi$:
$$\mathbf{L = n \frac{\lambda_n}{2} \implies \lambda_n = \frac{2L}{n}} \quad (n = 1, 2, 3, \dots)$$
The allowed resonant frequencies are:
$$\mathbf{f_n = \frac{v}{\lambda_n} = n \left(\frac{v}{2L}\right) = \frac{n}{2L} \sqrt{\frac{T}{\mu}}}$$
- Fundamental Mode ($n = 1$, 1st Harmonic): $$\lambda_1 = 2L, \quad \mathbf{f_1 = \frac{v}{2L} = \frac{1}{2L}\sqrt{\frac{T}{\mu}}}$$ (String vibrates in 1 loop with 2 nodes at ends and 1 antinode in the center).
- Second Harmonic ($n = 2$, 1st Overtone): $$\lambda_2 = L, \quad \mathbf{f_2 = 2 f_1 = \frac{v}{L}}$$ (String vibrates in 2 loops with 3 nodes and 2 antinodes).
- Third Harmonic ($n = 3$, 2nd Overtone): $$\lambda_3 = \frac{2L}{3}, \quad \mathbf{f_3 = 3 f_1 = \frac{3v}{2L}}$$ (String vibrates in 3 loops with 4 nodes and 3 antinodes).
- General Harmonic Series: All integer multiples of the fundamental frequency are present: $$\mathbf{f_1 : f_2 : f_3 : \dots = 1 : 2 : 3 : \dots}$$
5. Melde’s Experiment & Sonometer Laws
- Melde’s Transverse Arrangement: The tuning fork prong vibrates perpendicular to the string length. $$f_{\text{string}} = f_{\text{fork}} \implies \mathbf{p \sqrt{T} = \text{constant}} \implies \mathbf{p^2 T = \text{constant}}$$ where $p$ is the number of loops and $T$ is the string tension.
- Melde’s Longitudinal Arrangement: The tuning fork prong vibrates parallel to the string length. The tension fluctuates at $f_{\text{fork}}$, so the string executes parametric resonance at half the fork frequency: $$f_{\text{string}} = \frac{1}{2} f_{\text{fork}} \implies \mathbf{4 p^2 T = \text{constant}}$$
- Laws of Stretched Strings (Sonometer):
- Law of Length: $f \propto \frac{1}{L}$ (for constant $T, \mu$).
- Law of Tension: $f \propto \sqrt{T}$ (for constant $L, \mu$).
- Law of Mass: $f \propto \frac{1}{\sqrt{\mu}} \propto \frac{1}{r\sqrt{\rho}}$ (for constant $L, T$).
Four Solved Examples (Graded Basic to Advanced)
Example 1: Basic Application — Resonance Harmonics & Nodal Mapping
Problem Statement: A uniform piano wire of length $L = 2.0\text{ m}$ and total mass $m = 20.0\text{ g} = 0.020\text{ kg}$ is clamped rigidly at both ends under a mechanical tension $T = 100.0\text{ N}$.
(a) Determine the linear mass density $\mu$ and the speed $v$ of transverse waves on the wire.
(b) Calculate the fundamental resonance frequency $f_1$.
(c) Find the frequencies of the second and third harmonics.
(d) If the wire vibrates in its third harmonic mode with an antinode amplitude $2A = 6.0\text{ mm}$, state the number and exact positions of all nodes and antinodes along the wire.
Solution:
(a) Linear Density & Speed:
$$\mu = \frac{m}{L} = \frac{0.020\text{ kg}}{2.0\text{ m}} = \mathbf{0.010\text{ kg/m}}$$
$$v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{100.0\text{ N}}{0.010\text{ kg/m}}} = \sqrt{10000} = \mathbf{100.0\text{ m/s}}$$
(b) Fundamental Frequency:
$$\mathbf{f_1 = \frac{v}{2L} = \frac{100.0\text{ m/s}}{2(2.0\text{ m})} = \frac{100.0}{4.0} = 25.0\text{ Hz}}$$
(c) Second & Third Harmonics:
$$\mathbf{f_2 = 2 f_1 = 2(25.0\text{ Hz}) = 50.0\text{ Hz}}$$
$$\mathbf{f_3 = 3 f_1 = 3(25.0\text{ Hz}) = 75.0\text{ Hz}}$$
(d) Nodes & Antinodes in Third Harmonic ($n = 3$):
The wavelength for $n = 3$ is:
$$\lambda_3 = \frac{2L}{3} = \frac{2(2.0\text{ m})}{3} = \frac{4.0}{3}\text{ m} \approx 1.333\text{ m}$$
Loop length is $\frac{\lambda_3}{2} = \frac{2.0}{3}\text{ m} \approx 0.667\text{ m}$.
– Nodes (4 nodes): Located at $x = 0, \frac{\lambda_3}{2}, \lambda_3, \frac{3\lambda_3}{2} = L$:
$$\mathbf{x_N = 0\text{ m}, \quad \frac{2}{3}\text{ m} \ (\approx 0.667\text{ m}), \quad \frac{4}{3}\text{ m} \ (\approx 1.333\text{ m}), \quad 2.0\text{ m}}$$
– Antinodes (3 antinodes): Located at loop midpoints:
$$\mathbf{x_A = \frac{1}{3}\text{ m} \ (\approx 0.333\text{ m}), \quad 1.0\text{ m}, \quad \frac{5}{3}\text{ m} \ (\approx 1.667\text{ m})}$$
Example 2: Intermediate Application — Component Travelling Waves & Particle Velocity
Problem Statement: A standing wave on a stretched string is represented mathematically 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 equations of the two identical constituent travelling waves whose superposition forms this standing wave.
(b) Find the wavelength $\lambda$, frequency $f$, and propagation speed $v$ of the component waves.
(c) Calculate the distance between two consecutive nodes and between two consecutive antinodes.
(d) Calculate the transverse velocity of a string particle located at $x = 0.50\text{ m}$ at time $t = 0.0125\text{ s}$.
Solution:
(a) Constituent Progressive Waves:
Using the trigonometric identity $2 \sin A \cos B = \sin(A – B) + \sin(A + B)$:
$$y(x, t) = 0.03 [\sin(0.5\pi x – 40\pi t) + \sin(0.5\pi x + 40\pi t)]$$
The two component progressive waves are:
$$\mathbf{y_1(x, t) = 0.03 \sin(0.5\pi x – 40\pi t)\text{ m} \quad (\text{Travelling along } +x)}$$
$$\mathbf{y_2(x, t) = 0.03 \sin(0.5\pi x + 40\pi t)\text{ m} \quad (\text{Travelling along } -x)}$$
(b) Wave Parameters:
$$k = 0.5\pi\text{ rad/m} \implies \lambda = \frac{2\pi}{k} = \frac{2\pi}{0.5\pi} = \mathbf{4.0\text{ m}}$$
$$\omega = 40\pi\text{ rad/s} \implies f = \frac{\omega}{2\pi} = \frac{40\pi}{2\pi} = \mathbf{20.0\text{ Hz}}$$
$$\mathbf{v = f \lambda = (20.0\text{ Hz})(4.0\text{ m}) = 80.0\text{ m/s}}$$
(c) Nodal Spacing:
$$\mathbf{d_{NN} = \frac{\lambda}{2} = \frac{4.0\text{ m}}{2} = 2.0\text{ m}}$$
$$\mathbf{d_{AA} = \frac{\lambda}{2} = 2.0\text{ m}}$$
(d) Transverse Particle Velocity:
$$v_p = \frac{\partial y}{\partial t} = 0.06 \sin(0.5\pi x) [-40\pi \sin(40\pi t)] = -2.4\pi \sin(0.5\pi x) \sin(40\pi t)$$
At $x = 0.50\text{ m}$ and $t = 0.0125\text{ s} = \frac{1}{80}\text{ s}$:
$$\sin(0.5\pi \times 0.50) = \sin(0.25\pi) = \sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$$
$$\sin\left(40\pi \times \frac{1}{80}\right) = \sin\left(\frac{\pi}{2}\right) = 1.0$$
$$\mathbf{v_p = -2.4\pi \left(\frac{1}{\sqrt{2}}\right)(1.0) = -\frac{2.4\pi}{\sqrt{2}} \approx -1.697\pi\text{ m/s} \approx -5.33\text{ m/s}}$$
Example 3: Advanced Application — Sonometer Wire Tuning with Buoyant Load Immersion
Problem Statement: A sonometer wire of vibrating length $L_1 = 80.0\text{ cm}$ is kept under tension by a suspended solid brass cylinder of mass $M = 4.0\text{ kg}$ and density $\rho_{\text{brass}} = 8000\text{ kg/m}^3$. In this state, it resonates in its fundamental mode with a tuning fork of frequency $f_0$. When the brass cylinder is completely immersed in a beaker of an unknown liquid, the vibrating length must be shortened to $L_2 = 72.0\text{ cm}$ to restore resonance with the exact same tuning fork ($g = 10\text{ m/s}^2$).
(a) Determine the ratio of the effective tensions $\frac{T_2}{T_1}$.
(b) Formulate the buoyant force equation and calculate the density $\sigma$ of the liquid.
(c) If the wire length had remained fixed at $80.0\text{ cm}$ upon immersing the cylinder, find the new fundamental frequency $f’$ in terms of $f_0$.
Solution:
(a) Tension Ratio:
The fundamental frequency of a sonometer wire is $f = \frac{1}{2L}\sqrt{\frac{T}{\mu}}$. Since the frequency $f = f_0$ and wire density $\mu$ are constant:
$$\frac{\sqrt{T_1}}{L_1} = \frac{\sqrt{T_2}}{L_2} \implies \frac{T_2}{T_1} = \left(\frac{L_2}{L_1}\right)^2 = \left(\frac{72.0\text{ cm}}{80.0\text{ cm}}\right)^2 = \left(\frac{9}{10}\right)^2 = \mathbf{0.81}$$
(b) Liquid Density:
In air, the tension is $T_1 = Mg$.
When immersed in the liquid of density $\sigma$, buoyant force acts upwards:
$$T_2 = Mg – F_B = Mg – \left(\frac{M}{\rho_{\text{brass}}}\right) \sigma g = Mg \left(1 – \frac{\sigma}{\rho_{\text{brass}}}\right)$$
$$\frac{T_2}{T_1} = 1 – \frac{\sigma}{\rho_{\text{brass}}} = 0.81 \implies \frac{\sigma}{\rho_{\text{brass}}} = 1 – 0.81 = 0.19$$
$$\mathbf{\sigma = 0.19 \rho_{\text{brass}} = 0.19(8000\text{ kg/m}^3) = 1520\text{ kg/m}^3}$$
(c) Frequency with Constant Length:
With $L$ unchanged at $80.0\text{ cm}$:
$$\mathbf{f’ = f_0 \sqrt{\frac{T_2}{T_1}} = f_0 \sqrt{0.81} = 0.90 f_0}$$
Example 4: JEE Advanced Scenario — Composite Welded String with Nodal Junction
Problem Statement: A composite horizontal string is fabricated by welding two uniform wires at a junction $x = 0$. Segment 1 ($x \le 0$) has linear mass density $\mu_1 = 0.010\text{ kg/m}$, while segment 2 ($x \ge 0$) has linear mass density $\mu_2 = 0.090\text{ kg/m}$. Both segments are maintained under a uniform continuous tension $T = 36.0\text{ N}$. The composite wire is clamped at $x = -L_1$ and $x = +L_2$. A standing wave of frequency $f = 100\text{ Hz}$ is excited such that the junction $x = 0$ is a node.
(a) Determine the wave speed $v_1$ in segment 1 and $v_2$ in segment 2.
(b) Calculate the wavelength $\lambda_1$ and $\lambda_2$ in the two segments.
(c) Determine the minimum permissible lengths $L_{1,\text{min}}$ and $L_{2,\text{min}}$ of the segments.
(d) Formulate the general relation between the number of loops $n_1$ and $n_2$ for any shared resonant frequency.
Solution:
(a) Wave Speeds:
$$\mathbf{v_1 = \sqrt{\frac{T}{\mu_1}} = \sqrt{\frac{36.0\text{ N}}{0.010\text{ kg/m}}} = \sqrt{3600} = 60.0\text{ m/s}}$$
$$\mathbf{v_2 = \sqrt{\frac{T}{\mu_2}} = \sqrt{\frac{36.0\text{ N}}{0.090\text{ kg/m}}} = \sqrt{400} = 20.0\text{ m/s}}$$
(b) Wavelengths:
Since frequency $f = 100\text{ Hz}$ is invariant across the welded junction:
$$\mathbf{\lambda_1 = \frac{v_1}{f} = \frac{60.0\text{ m/s}}{100\text{ Hz}} = 0.60\text{ m} = 60.0\text{ cm}}$$
$$\mathbf{\lambda_2 = \frac{v_2}{f} = \frac{20.0\text{ m/s}}{100\text{ Hz}} = 0.20\text{ m} = 20.0\text{ cm}}$$
$$\frac{\lambda_1}{\lambda_2} = \frac{v_1}{v_2} = \frac{60.0}{20.0} = 3.0$$
(c) Minimum Segment Lengths:
Because the clamped ends and the central junction $x = 0$ are nodes:
$$L_1 = n_1 \frac{\lambda_1}{2}, \quad L_2 = n_2 \frac{\lambda_2}{2}$$
For minimum length, each segment vibrates in a single loop ($n_1 = 1, n_2 = 1$):
$$\mathbf{L_{1,\text{min}} = \frac{\lambda_1}{2} = \frac{0.60\text{ m}}{2} = 0.30\text{ m} = 30.0\text{ cm}}$$
$$\mathbf{L_{2,\text{min}} = \frac{\lambda_2}{2} = \frac{0.20\text{ m}}{2} = 0.10\text{ m} = 10.0\text{ cm}}$$
(d) General Loop Ratio:
Equating the frequencies $f = \frac{n_1 v_1}{2 L_1} = \frac{n_2 v_2}{2 L_2}$:
$$\frac{n_1}{n_2} = \left(\frac{L_1}{L_2}\right) \left(\frac{v_2}{v_1}\right) = \mathbf{\frac{1}{3} \left(\frac{L_1}{L_2}\right)}$$
Practice Worksheet: 10 Target JEE Problems
Master standing wave equations, nodal placements, and sonometer laws. Complete all 10 problems before consulting the detailed solutions at the end.
Section I: Single Correct Choice Questions (Q1–Q5)
Q1. When a transverse wave pulse travelling along a stretched string reflects from a rigid, clamped boundary, the reflected pulse experiences a phase shift of:
(A) $0$
(B) $\pi/2$
(C) $\pi$
(D) $2\pi$
Q2. The spatial distance between two successive nodes in a standing wave of wavelength $\lambda$ is:
(A) $\lambda$
(B) $\lambda/2$
(C) $\lambda/4$
(D) $2\lambda$
Q3. In a standing wave established on a string, which of the following physical quantities attains its maximum value at a node?
(A) Transverse particle velocity
(B) Transverse particle displacement
(C) Elastic strain and tension variation
(D) Kinetic energy density
Q4. A string of length $L$ fixed rigidly at both ends vibrates in its third harmonic mode. The total number of nodes and antinodes formed along the string (including both fixed ends) are, respectively:
(A) 4 nodes, 3 antinodes
(B) 3 nodes, 2 antinodes
(C) 3 nodes, 4 antinodes
(D) 4 nodes, 4 antinodes
Q5. In Melde’s experiment, when the tuning fork is operated in the longitudinal arrangement (fork prongs vibrating parallel to the string), the frequency of vibration of the string is:
(A) Equal to the fork frequency $f_{\text{fork}}$
(B) Twice the fork frequency $2 f_{\text{fork}}$
(C) Half of the fork frequency $\frac{1}{2} f_{\text{fork}}$
(D) Four times the fork frequency $4 f_{\text{fork}}$
Section II: One or More Than One Correct Questions (Q6–Q8)
Q6. For the standing wave $y(x, t) = 2A \sin(kx) \cos(\omega t)$ formed on a string clamped at both ends:
(A) No net energy is transported across any nodal cross-section.
(B) All particles situated between two consecutive nodes vibrate in identical phase.
(C) Particles in adjacent loops separated by a node vibrate in anti-phase ($180^\circ$).
(D) Every particle executes simple harmonic motion with the same frequency $\omega$ but position-dependent amplitude $2A|\sin(kx)|$.
Q7. A sonometer wire under tension $T$ vibrates in its fundamental mode with frequency $f$. If the vibrating length of the wire is halved ($L’ = L/2$) and the mechanical tension is quadrupled ($T’ = 4T$):
(A) The fundamental frequency quadruples ($f’ = 4f$).
(B) The transverse wave speed doubles ($v’ = 2v$).
(C) The wavelength of the fundamental mode is halved ($\lambda’ = \lambda/2$).
(D) The linear mass density $\mu$ of the wire remains unchanged.
Q8. When a wave pulse travelling on a light string under tension $T$ encounters a junction connecting it to a heavier string under the same tension:
(A) The speed of the wave in the heavier string is smaller than in the light string.
(B) The reflected wave pulse undergoes a phase reversal of $\pi$ radians ($180^\circ$).
(C) The transmitted wave pulse undergoes zero phase shift ($0^\circ$).
(D) The amplitude of the transmitted wave pulse is smaller than the incident amplitude.
Section III: Numerical / Integer Type Questions (Q9–Q10)
Q9. A string of length $L = 1.20\text{ m}$ clamped at both ends vibrates with a standing wave frequency of $f = 250\text{ Hz}$. If this frequency corresponds to the 5th harmonic mode ($n = 5$), calculate the propagation speed $v$ (in $\text{m/s}$) of transverse waves on the string.
Q10. In a sonometer experiment, a wire of length $L = 50.0\text{ cm}$ resonates in its fundamental mode with a tuning fork of frequency $f_0$. If the tension in the wire is increased by $44\%$, by how many centimeters must the vibrating length of the wire be increased to restore resonance with the exact same tuning fork?
Comprehensive Solutions & Answer Key
Answer Key
| Question No. | Correct Answer | Key Concept / Formula Tested |
|---|---|---|
| Q1 | (C) | Reflection from fixed rigid end introduces an inversion of $\pi$ radians |
| Q2 | (B) | Distance between consecutive nodes is $\lambda/2$ |
| Q3 | (C) | At a node: displacement is zero, but spatial slope (strain) is maximum |
| Q4 | (A) | Third harmonic has 3 loops $\implies$ 4 nodes (including ends) and 3 antinodes |
| Q5 | (C) | Melde’s longitudinal mode: tension varies at $f_{\text{fork}} \implies f_{\text{string}} = f_{\text{fork}}/2$ |
| Q6 | (A, B, C, D) | Standing wave kinematics, nodal phase barriers, and zero energy transport |
| Q7 | (A, B, C, D) | $f = \frac{1}{2L}\sqrt{T/\mu} \implies L/2$ and $4T$ gives $2 \times 2 = 4\times$ frequency |
| Q8 | (A, B, C, D) | Reflection from denser medium: inversion, transmission upright, $A_t < A_i$ |
| Q9 | 120 | $f_5 = 5v/(2L) \implies v = 250(2.40)/5 = 120\text{ m/s}$ |
| Q10 | 10 | $L \propto \sqrt{T} \implies L’ = 50\sqrt{1.44} = 60\text{ cm} \implies \Delta L = 10\text{ cm}$ |
Detailed Step-by-Step Solutions
Solution to Q1:
When a transverse wave hits a rigid support, the support cannot move ($y = 0$). To satisfy this boundary condition, the reflected pulse must cancel the incident pulse at the wall at all times. By Newton’s third law, the support exerts an equal and opposite reaction force, sending back a pulse that is inverted relative to the incident pulse. This corresponds to a phase shift of $\mathbf{\pi\text{ radians} \ (180^\circ)}$.
Hence, the correct choice is (C).
Solution to Q2:
In a standing wave, nodes occur at positions $x_n = n \frac{\lambda}{2}$. The separation between two consecutive nodes $x_{n+1}$ and $x_n$ is:
$$x_{n+1} – x_n = (n + 1)\frac{\lambda}{2} – n\frac{\lambda}{2} = \mathbf{\frac{\lambda}{2}}$$
Hence, the correct choice is (B).
Solution to Q3:
In a standing wave $y(x, t) = 2A \sin(kx) \cos(\omega t)$:
– At a node, $\sin(kx) = 0$, so displacement $y = 0$ and velocity $v_p = \frac{\partial y}{\partial t} = 0$ at all times.
– The longitudinal strain is given by the spatial derivative:
$$\frac{\partial y}{\partial x} = 2A k \cos(kx) \cos(\omega t)$$
At a node, $\cos(kx) = \pm 1$, meaning the spatial gradient (strain) reaches its maximum possible magnitude ($2Ak$). Since elastic force and tension variations are directly proportional to strain, they are maximum at the nodes.
Hence, the correct choice is (C).
Solution to Q4:
For a string clamped at both ends, the $n$-th harmonic mode consists of $n$ vibrating segments (loops). For the third harmonic ($n = 3$):
– Number of loops = 3.
– Number of antinodes = number of loops = 3.
– Number of nodes = number of loops $+ 1$ = $3 + 1 = 4$ (located at $x = 0, L/3, 2L/3, L$).
Hence, the correct choice is (A).
Solution to Q5:
In Melde’s longitudinal arrangement, the fork prongs oscillate parallel to the string. During one full oscillation of the fork, the string is slackened and tightened twice, causing the string tension to oscillate at frequency $f_{\text{fork}}$. The string executes parametric resonance at half the excitation frequency:
$$f_{\text{string}} = \frac{1}{2} f_{\text{fork}}$$
Hence, the correct choice is (C).
Solution to Q6:
Let us evaluate each statement:
(A) Since medium elements at nodes never move ($v_p = 0$), the rate of energy transfer across a node is $P = -T \left(\frac{\partial y}{\partial x}\right) v_p = 0$. (A) is correct.
(B) Between two nodes, $\sin(kx)$ retains a single sign, so all particles pass through their extremes and equilibrium simultaneously. (B) is correct.
(C) Crossing a node changes the sign of $\sin(kx)$, introducing a phase reversal of $\pi$ ($180^\circ$). (C) is correct.
(D) Each particle undergoes SHM with displacement amplitude $|A(x)| = 2A|\sin(kx)|$ at frequency $\omega$. (D) is correct.
Hence, all choices (A, B, C, D) are correct.
Solution to Q7:
The fundamental frequency is $f = \frac{1}{2L}\sqrt{\frac{T}{\mu}}$.
With $L’ = L/2$ and $T’ = 4T$:
$$f’ = \frac{1}{2(L/2)} \sqrt{\frac{4T}{\mu}} = 2 \left[\frac{2}{2L}\sqrt{\frac{T}{\mu}}\right] = 4 f$$
Statement (A) is correct.
Wave speed: $v’ = \sqrt{T’/\mu} = \sqrt{4T/\mu} = 2v$. Statement (B) is correct.
Wavelength: $\lambda’ = 2L’ = 2(L/2) = L = \lambda/2$. Statement (C) is correct.
Linear mass density $\mu$ is an intrinsic material property and is unchanged. Statement (D) is correct.
Hence, all choices (A, B, C, D) are correct.
Solution to Q8:
For a wave incident from a lighter string ($\mu_1$) to a heavier string ($\mu_2 > \mu_1$):
(A) $v_2 = \sqrt{T/\mu_2} < v_1 = \sqrt{T/\mu_1}$. (A) is correct.
(B) $A_r = \left(\frac{v_2 – v_1}{v_2 + v_1}\right) A_i < 0$, which signifies a phase shift of $\pi$. (B) is correct.
(C) $A_t = \left(\frac{2v_2}{v_2 + v_1}\right) A_i > 0$, indicating no phase shift ($0^\circ$). (C) is correct.
(D) Since $v_2 < v_1$, $\frac{2v_2}{v_2 + v_1} < 1$, so $A_t < A_i$. (D) is correct.
Hence, all choices (A, B, C, D) are correct.
Solution to Q9:
The frequency of the $n$-th harmonic for a clamped string is:
$$f_n = n \left(\frac{v}{2L}\right)$$
For the 5th harmonic ($n = 5$), with $L = 1.20\text{ m}$ and $f_5 = 250\text{ Hz}$:
$$250 = 5 \left(\frac{v}{2(1.20)}\right) = \frac{5v}{2.40}$$
Solving for $v$:
$$\mathbf{v = \frac{250 \times 2.40}{5} = 50 \times 2.40 = 120\text{ m/s}}$$
Answer: 120
Solution to Q10:
To maintain resonance with the same tuning fork ($f = \text{constant}$):
$$\frac{\sqrt{T}}{L} = \frac{\sqrt{T’}}{L’}$$
Given $T’ = T + 0.44 T = 1.44 T \implies \sqrt{T’} = \sqrt{1.44}\sqrt{T} = 1.20\sqrt{T}$.
$$L’ = L \left(\frac{\sqrt{T’}}{\sqrt{T}}\right) = (50.0\text{ cm})(1.20) = 60.0\text{ cm}$$
The required increase in vibrating length is:
$$\mathbf{\Delta L = L’ – L = 60.0\text{ cm} – 50.0\text{ cm} = 10\text{ cm}}$$
Answer: 10