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For the wave equation, $y = (4 cm) \sin [πt + 2 πx]$
Here, t is in second and x in metres.
Column-I Column-II
A) at x=0, particle velocity is maximum at t = P) 0.25 s
B) at x=0, particle acceleration is maximum at t = Q) 1.0 s
C) at x=0.5 m, particle velocity is maximum at t = R) 0.75 s
D) at x=0.5 m, particle acceleration is maximum at t = S) 1.5 s
T) 1/3 s

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$A \to S; B \to Q; C \to S; D \to Q$

$A \to R; B \to R; C \to T; D \to T$

$A \to R; B \to T; C \to R; D \to Q$

$A \to Q; B \to S; C \to Q; D \to S$

answer is A.

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Detailed Solution

$\begin{array}{l} v_{p} = \frac{\partial y}{\partial t} = (4 π) cm / scos [πt + 2 πx] \\ a_{p} = \frac{\partial^{2} y}{\partial t^{2}} = (- 4 π^{2}) cm / s^{2} \sin [πt + 2 πx] \end{array}$
Now, substitute the values of t and x

at x=0, particle velocity is maximum at t =1.0 s

at x=0, particle acceleration is maximum at t = 1.5 s

at x=0.5 m, particle velocity is maximum at t = 1.0 s

at x=0.5 m, particle acceleration is maximum at t = 1.5 s

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