Physics · Transverse waves

JEE Main 2026 — 5 April, Morning Shift — Question 21

A transverse wave on a string is described by $y = 3\sin(36t + 0.018x + \pi/4)$ , where x, y are in cm and t in seconds. The least distance between the two successive crests in the wave is ______ cm. (Nearest integer) ( $\pi = 3.14$ )

Answer: 349

Numerical answer — enter this value.

Step-by-step solution

Wave number $k = 0.018$ cm⁻¹, $\lambda = 2\pi/k = 2\times3.14 / 0.018 = 348.89$ cm ≈ 349 cm.

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Exam: JEE Main 2026
Paper: 5 April, Morning Shift
Subject: Physics
Chapter: Transverse waves
Topic: Introduction, wave parameters and wave Equation