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JEE Main 2026 · 5 Apr · Shift 1, Physics question 48

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

Answer and worked solution

Official answer: 349

Distance between successive crests is the wavelength:
λ=2πk=6.280.018≈348.9 cm.\lambda=\frac{2\pi}k=\frac{6.28}{0.018}\approx348.9\text{ cm}.

Answer: 349