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A longitudnal wave is represented by x=x_(0) sin 2pi(nt - x/lambda). The maximum particle velocity will be four times the wave velocity if |
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Answer» `lambda = (pi x_(0))/4` `x =x_(0) sin2pi(nt -x/lambda)` `x =x_(0)sin(2pint -(2PIX)/lambda)` Compare it with the standard wave equation, `x =A sin(omega t-kx)` we get, `omega =2pin, K =(2pi)/lambda` Wave velocity, `v= omega/k =(2pi n)/(2pi//lambda) = nlambda`.......(i) Particle velocity, `v_(p) =(dx)/(dt) = 2pin x_(0) cos(2pint -(2pix)/lambda)` Maximum particle velocity, `(v_(p))_("max") = 2pinx_(0)`......(ii) According to given problem, `(v_(p))_("max") = 4v` Using (i) and (ii), we get `2pinx_(0) = 4nlambda` `rArr lambda = (2pi nx_(0))/(4n) = (pi x_(0))/2` |
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