1.

A transverse wave in a medium is described by the equationy = A sin 2(omegat - kx) . The magntiude of the maximum velocity of particles in the medium is equalthat of theof the wave velocity . If the value of A is

Answer»

` lambda/(2pi)`
`lambda/(4 pi) `
`lambda/pi`
`(2lambda)/pi`

Solution :The givenequation of the transverse wave is` y= A SIN2 ( OMEGAT -kx)`
Velocityof the particle= ` (dy)/(dt) = 2 A omegacos 2 (omegat - kx) `
Maximumvelocity` = 2 A OMEGA`
Velocityof the wave = ` ("COEFFICIENT of t ")/( "Coefficient of x ") = (2 Omega)/(2kg) = Omega/k`
As per QUESTION
` 2 A omega = omega/k or 2A = 1/k = lamda/(2pi) RightarrowA = lamda/(4pi)`


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