1.

A transverse wave is represented by the equation y = y_(0) sin" " (2pi)/(lambda)(vt - x). For what value of lambda is the maximumparticle velocity equal to two times the wave velocity ?

Answer»

`lambda = y_(0)` /2
`lambda = 2y_(0)`
`lambda = (2 pi y_(0))/(2) `
`lambda = y_(0)`/4 .

Solution :y = `y_(0) sin" " (2pi)/(lambda) (vt- x)`
`(DY)/(dt)= y_(0) COS" " (2pi)/(lambda) (vt - x) [ (2pi)/(lambda) . V ]`
For `(dy)/(dt ) ` to be maximum , cos `(2pi)/(lambda)` (vt - x) = 1.
`therefore ((dy)/(dt))_("max") = y_(0) . (2pi)/(lambda) .v `
since `v_(max) = 2v rArr y_(0) . (2pi)/(lambda). v = 2 v `
`rArr "" lambda = pi y_(0) `


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