If the equation of transverse wave is $Y = 2 sin (k x - 2 t)$ , the the maximum particle velocity is

4 u n i t

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2 u n i t

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Zero

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6 u n i t

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Solution

The correct option is B $4 u n i t$
Equation of wave
$y = 2 sin (k x - 2 t)$
Comparing with standard equation
$y = A sin (k x - ω t)$
$A = 2, ω = 2$
$∴$ Maximum particle velocity
$v_{m a x} = A ω = 2 \times 2$
$= 4 u n i t$

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