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For a wave described by, y = A sin(omega t -kx),considerthe following points (i) x =0,(ii) x =pi/(4k) For aparticle at each ofthese.points at t = 0, describe whether the particle is moving or not and in what direction and describe whether the particle is speeding up, slowing down or instantaneously not accelerating. |
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Answer» Solution :`y =A sin(omegat -kx),` Particle velocity `v_p (X,t) =(delta y)/(delta t) = omegaA cos (ometat -kx)` and particle acceleration `a_p (x,t) = (delta^2y)/(deltat^2) =- OMEGA^2 A sin(omegat -kx)` i) `t =0 ,x=0 :v_p =+omegaA and a_p =0` i.e., particle is moving upwards but its acceleration is zero. II) At `t =0,x = PI/(4k), V_p =(A omega)/(sqrt2)`, upward and `a_p =(omega^2A)/sqrt2` upward |
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