A simple harmonic plane progressive wave is travelling along +ve X-axis having amplitude A and $ω = π r a d / s$ . At t=1s, particle at x=2m and x=3m are moving in downward and upward direction respectively at position $y = + A / 2$ . The smallest time $(i n s)$ , when maximum power transfer will take place at x=11/4m is.

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Detailed Solution

Calculation of K:
$Δ Q_{P Q} = K Δ x_{P Q} \Rightarrow \frac{2 π}{3} = k \times 1 \Rightarrow K = \frac{2 π}{3}$
Max. power transfer will take place when element is at meant position (i.e. y=0)

$Δ t = \frac{Δ x}{v} = \frac{k Δ x}{w} = \frac{2 π .1}{3. π} = \frac{2}{3} S ∴ t = t_{i} + Δ t = 1 + \frac{2}{3} = \frac{5}{3} S$

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