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A standing wave xi= a sin kx. Cos omegat...

A standing wave `xi= a sin kx. Cos omegat ` is maintained in a homogeneous rod with cross `-` sectional area `S` and density `rho`. Find the total mechanical energy confined between the sections corresponding to the adjacent displacement nodes.

Text Solution

Verified by Experts

The correct Answer is:
A, B, D
Amplitude at a distance x is `A = a sin kx`
First node can be obtained at x=0,
and the second at `x = pi//k`
At position x, mass of the element PQ is
`dm = (rhoS) dx`
Its amplitude is `A = a sin kx`
Hence mechanical energy stored in this element is

(energy of particle in SHM) ` dE = 1/2 (dm) A^2 omega^2 `
or ` dE = 1/2 (rho SA^2omega^2)dx`
`= 1.2 (rhoSa^2omega^2sin^2kx)dx`
Therefore, total energy stored between two
adjacent nodes will be
`E = int_(x=0)^(x = pi//k) dE`
Solving this, we get
`E = (piSrhoomega^2a^2)/(4k)` .
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