A standing wave is maintained in a homogeneous string of cross-sectional area s and density $ρ$ . It is formed by the superposition of two waves travelling in opposite directions given by the equation $y_{1} = a \sin (ω t - k x)$ and $y_{2} = 2 a \sin (ω t + k x)$ . The total mechanical energy confined between the sections corresponding to the adjacent antinodes is

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$\frac{3 π s ρ ω^{2} a^{2}}{2 k}$

$\frac{2 π s ρ ω^{2} a^{2}}{2 k}$

$\frac{π s ρ ω^{2} a^{2}}{2 k}$

$\frac{5 π s ρ ω^{2} a^{2}}{2 k}$

answer is C.

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Detailed Solution

Distance between two adjacent antinodes is $\frac{λ}{2},$ we know $k = \frac{2 π}{λ} \Rightarrow \frac{λ}{2} = \frac{π}{k}$ volume of string between two adjacent antinodes $V = \frac{π}{k} s ∴ E = (V) (u_{1} + u_{2})$ Here u is the energy density (energy per unit volume) which is equal to $\frac{1}{2} ρ A^{2} ω^{2}$ $∴ E = (\frac{π}{k}) (s) [\frac{1}{2} ρ a^{2} ω^{2} + \frac{1}{2} ρ {(2 a)}^{2} ω^{2}]$

$= \frac{5 π s ρ a^{2} ω^{2}}{2 k} .$ Hence option ‘C’ is correct