For a stationary wave, y = 10cos (π x/15)sin (48π t) cm, the distance between a node and the successive antinode is (a) 30 cm (b) 22.5 cm (c) 15 cm (d) 7.5 cm

Answer: (d). k = π/15 = (2π)/(λ) ⇒ λ = 30 cm. Node to next antinode = (λ)/(4) = 7.5 cm. So the answer is (d) 7.5 cm.

For a stationary wave, y = 10cos (π x/15)sin (48π t) cm, the distance… | MHT CET 2026 (19 May, Shift 1)

Q1MHT CET 2026 · 19 MAY · SHIFT 1Easy

Stationary Waves & Vibrating Strings

For a stationary wave, $y = 10\cos(\pi x/15)\sin(48\pi t)$ cm, the distance between a node and the successive antinode is

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Answer: (d)

$k = \frac{\pi}{15} = \frac{2\pi}{\lambda} \Rightarrow \lambda = 30\text{ cm}$ .
Node to next antinode $= \frac{\lambda}{4} = 7.5\text{ cm}$ .
So the answer is (d) $7.5$ cm.

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