Exams › JEE Advanced › Physics
A 1.5 m long steel rod is clamped at both ends and undergoes longitudinal standing wave vibrations. The speed of longitudinal waves in steel is 5000 m/s. The rod vibrates in its third harmonic mode. Which of the following statements are correct? (A) The fundamental frequency of longitudinal standing waves is 5000/3 Hz. (B) The displacement equation along the rod can be written as S = A sin(2*pi*x) cos(10000*pi*t). (C) The positions of displacement nodes are x = 0, 0.5 m, 1 m, and 1.5 m. (D) The positions of maximum stress along the rod are x = 0, 0.5 m, 1 m, and 1.5 m.
- The fundamental frequency of longitudinal standing waves is 5000/3 Hz.
- The displacement equation along the rod can be written as S = A sin(2*pi*x) cos(10000*pi*t).
- The positions of displacement nodes are x = 0, 0.5 m, 1 m, and 1.5 m.
- The positions of maximum stress along the rod are x = 0, 0.5 m, 1 m, and 1.5 m.
Correct answer: The positions of displacement nodes are x = 0, 0.5 m, 1 m, and 1.5 m.
Solution
Both ends clamped => displacement nodes at ends. Fundamental: L = lambda₁/2 => f1 = v/2L = 5000/3 Hz (A correct). Third harmonic: f3 = 3f1 = 5000 Hz, lambda₃ = v/f3 = 1 m. k = 2*pi/lambda₃ = 2*pi rad/m. omega₃ = 2*pi*f3 = 10000*pi. Displacement S = A sin(2*pi*x) cos(10000*pi*t) (B correct). Nodes at sin(2*pi*x) = 0 => x = 0, 0.5, 1.0, 1.5 m (C correct). Maximum stress (strain) occurs at displacement antinodes, not nodes, so D is WRONG. Stress max at midpoints between nodes: x = 0.25, 0.75, 1.25 m.
Related JEE Advanced Physics questions
⚔️ Practice JEE Advanced Physics free + battle 1v1 →