For a wave described by `y = A sin (omegat - kx)`, consider the following or not and in what direction and describe whether the particle is speeding up, slowing sown or instantanteously not accelerating?

Explanation

`y =A sin (omegat - kx)`
Particle velocity `v_p (x, t)=(dely)/(delt) =omegaA cos (omegat - kx)` and particle acceleration
`a_p (x, t)=(del^(2)y)/(delt) = -omega^(2)A sin(omegat - kx)`
(a) `t =0 , x= 0: v_p =+ omegaA` and `a_(p) =0`
i.e. particle is moving upward but its acceleration is zero.
(b) `t =0`, `x = (pi)/(4k)` ` x=(pi)/(4k)`
`:. kx=(pi)/(4)`
`v_(p) =omegaA cos (-(pi)/(4)) = + (omegaA)/(sqrt(2)`
and ` a_(p)= -omega^(2)A sin (-(pi)/(4)) = + (omega^(2)A)/sqrt(2)`
Velocity of particle is positive, i.e. the particle is moving upward (along positive y- direction). Further `(v_(P))` and `(a_(p)) are in the same direction (both are positive). Hence, the particle is speeding up.
(C) `t =0 , x =(pi)/(2k) , x=(pi)/(2k) :. kx = (pi)/(2)`
`v_(p) =omegaA cos(-pi//2) = 0`
`a_(p) = -omega^(2) A sin (-pi//2) =omega^(2) A`
i.e. particle is stationary or at its exterme position `(y = - A)`. So, it is speeding up at this instant.
(d) `t =0 , x= (3pi)/(4k) , x = (3pi)/(4k) :. kx = (3pi)/(4)`
`:. v_(p) = omegaA cos (-(3pi)/(4)) = -(omegaA)/(sqrt(2))`
`a_(p) = -omega^(2)A sin(-(3pi)/(4)) = + -(0megaA)/sqrt(2)`
`a_(p) = - omega^(2) A sin (-(3pi)/(4))= + (omega^(2)A )/sqrt(2)`
Velocity of particle is negative, i.e. the particle is moving downwards. Further `v_(p)` and `(a_(P))` are in opposite direction , i.e. the particle is slowing down.