JEE Main 2026 — Apr 5, Shift 1 — Physics solution

JEE Main2026Apr 5, Shift 1Physics

A transverse wave on a string is described by $y = 3 sin (36 t + 0.018 x + π /4)$ . 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) ( $π = 3.14$ )

Your answer

Solution

The least distance between successive crests is one full wavelength $λ = \frac{2 π}{k}$ .

From the wave equation, $k = 0.018$ cm $^{- 1}$ , so $λ = \frac{2 π}{0.018} = \frac{2 \times 3.14}{0.018} \approx 348.89$ cm.

Rounded to the nearest integer: $349$ cm.

← Back to full paper

Practice more Physics

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →