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In (Q. 10.) the relative deformation amplitude of medium is
A. `0.02pi`
B. `0.08pi`
C. `0.06pi`
D. None of these

1 Answer

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Best answer
Correct Answer - B
The relative deformation is
`(dely)/(delx) =(del)/(delx)[4 sin (omega t - kx)]`
`=-4k cos (omegat-kx)`
Th relative deformation amplitude
`((dely)/(delx))_("max")=ak=4 xx (2pi)/(100) = 0.08 pi`

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