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MHT CET 2024 (4 May, Shift 2) · Stationary Waves & Vibrating Strings

Stationary Waves & Vibrating Strings question from MHT CET 2024 (4 May, Shift 2)

Q36MHT CET 2024 · 4 MAY · SHIFT 2Easy

Stationary Waves & Vibrating Strings

A stationery wave is represented by $y = 12\cos\left(\frac{\pi}{6}x\right)\sin(8\pi t)$ , where $x \,\&\, y$ are in cm and $t$ in second. The distance between two successive antinodes is

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Answer: (c)

$k = \frac{\pi}{6} = \frac{2\pi}{\lambda} \Rightarrow \lambda = 12\text{ cm}$ .
Successive antinodes are $\frac{\lambda}{2} = 6\text{ cm}$ apart.
So the answer is (c) 6 cm.

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