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A standing wave is maintained in a homogenous string of cross-sectional area S and density ρ. It is formed by the superposition of two waves density ρ, travelling in opposite directions given by the equation y_1=a \sin (\omega t-k x) and y_2=2a \sin (\omega t-k x). The total mechanical energy confined between the sections corresponding to the adjacent anti nodes is-

Option: 1

\frac{3 \pi \rho 5 \omega^2 a^2}{2 k}


Option: 2

\frac{\pi \rho 5 \omega^2 a^2}{2 k}


Option: 3

\frac{5 \pi s \rho \omega^2 a^2}{2 k}


Option: 4

\frac{2 \pi s \rho \omega^2 a^2}{k}


Answers (1)

best_answer

Distance between two successive anti-nodes is \frac{\lambda}{2} \text { and } \frac{\pi}{k}

? Volume between two antinodes will be \frac{\pi}{k}.s

Let u1 and u2 be the energy density due to two waves, then 

\begin{aligned} & E=\left(u_1+u_2\right) \times \text {volume } \\ &E=\left[\frac{1}{2} \rho \omega^2 a^2+\frac{1}{2} \rho \omega^2(2 a)^2\right] \frac{\pi}{k} \cdot s \\ & E=\frac{5}{2} \frac{\rho \omega^2 a^2 \pi s}{k} \end{aligned}

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Divya Prakash Singh

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