A standing wave of time period T is set up in a string clamped between two rigid supports. At t = 0

Question

A standing wave of time period T is set up in a string clamped between two rigid supports. At t = 0 antinode is at its maximum displacement 2A.

Accepted Answer

Sol.Equation of wave:y = 2A sinKx cosωt$\left( \omega = \frac{2\pi}{T} \right)$

KE = $\frac{1}{2}\mu dx\left( \frac{\partial y}{\partial t} \right)^{2}$

⇒ KE per unit length = $\frac{1}{2}\mu\left( \frac{\partial y}{\partial t} \right)^{2}$

= $\frac{1}{2}$μ(2Aω sinKx sin ωt)²

= 2μA²ω² sin² Kx sin²ωt

PE = $\frac{1}{2}Tdx\left( \frac{\partial y}{\partial x} \right)^{2}$

⇒ PE per unit length = $\frac{1}{2}T\left( \frac{\partial y}{\partial x} \right)^{2}$

= $\frac{1}{2}$T(2AK cos Kx cos ωt)²

= $\frac{1}{2}\left( \frac{\omega^{2}}{K^{2}}\mu \right)$ × 4A²K² cos² Kx cos²ωt $\left( v = \frac{\omega}{K} = \sqrt{\frac{T}{\mu}} \right)$

= 2 μω²A² cos² Kx cos²ωt

Hence,

Energy density = Total energy per unit length

= 2μA²ω² sin²Kx sin²ωt + 2μA²ω² cos² Kx cos²ωt

For Node, sin Kx = 0 & cos Kx = 1

E_Node = 0 + 2 μA²ω² cos ωt

At t = $\frac{T}{4}$ , E_Node = 2μA²ω² cos$\left( \frac{2\pi}{T} \times \frac{T}{4} \right)$ = 0

For Antinode, sin Kx = 1 & cos Kx = 0

E_Antinode = 2μA²ω² sin²ωt + 0

= $2\mu A^{2}\omega^{2}\sin^{2}\left( \frac{2\pi}{T} \times \frac{T}{4} \right)$

= 2μA²ω²

Hence E_Node≠ E_Antinodeat t = $\frac{T}{4}$

Similarly, check for t = $\frac{T}{2}$

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