A wave on a string is described by `y(x,t)=A cos (kx-omegat)`. (a) Graph `y,v_(y)`, and `a_(y)` as function of x for time t=0. (b) consider the following points on the string: (i) x=0,(ii) x=pi//4k, (iii) x=pi//2k, (iv)x=3pi//4k, (v)x=pi//k, (vi)x=5pi//4k, (vii) x=3pi//2k, (viii) x=7pi//4k`. For a particle at each of these points at t=0, describes in words whether the particle is moving and in what direction, and whether the particle is speeding up, slowing down, or instantaneously not acceleraring.

Explanation

`y=A cos(kx-omegat)`
`v_(y)=(dy)/(dt)=Aomega sin(kx-omegat)`
`a_(y)=(dy_(y)/(dt)=-omega^(2) A cos(kx-omegat)=-omega^(2)y`
for `t=0`
`y=A cos(kx)=A cos((2pi)/(lambda)x)` ltbr) `v_(y)=Aomega sin(kx0=Aomega sin((2pi)/(lambda)x)`
`a_(y)=-omega^(2)A cos(kx)=-omega^(2)A cos((2pi)/(lambda))x)=-omega^(2)y`
(i) At `x=0:y=A, v_(y)=0`, `a_(y)=-omega^(2)A,` the particle is at positive extreme position. It is at rest and accelerating downward at maximum acceleration.
(ii) At `x=(pi)/(4k)=(pilambda)/(4(2pi))=(lambda)/(8)`
`y=A cos((2pi)/(lambda)(lambda)/(8))=(A)/sqrt(2)`, `v_(y)=(Aomega)/sqrt(2)`, `ay=(-omega^(2)A)/sqrt(2)`
the particle is above mean position, moving upward and slowing down. acceleration is in downward direction.
(iii) At`x=(pi)/(2k)=(pilambda)/(2(2pi))=(lambda)/(4)`
here` y=0`, `v_(y)omegaA`, `a_(y)=0`
the particle is at mean position and moving upwards wire maximum velocity. acceleration of particle is zero.
(iv). At `x=(3pi)/(4k)=(3pi)/(8)`
here `y=-(A)/sqrt(2)`, `v_(y)=(Aomega)/sqrt(2)`, `a_(y)=(omega^(2)A)/sqrt(2)`
the particle is below mean position, having velocity acceleration both in upward direction. thus particle speeding up.
(v).At `x=(pi)/(k)=(lambda)/(2)`
here `y=-A,v_(y)=0,a_(y)=omega^(2)A`
`the particle is at negative extreme position it instantaneously at rest and having acceleration in upward direction.
(vi) At `x=(5pi)/(4k)=(5pi)/(8)` ,
here `y=(-A)/sqrt(2)`, `v_(y)=-(Aw)/sqrt(2)`, `a_(y)=(w^(2)A)/sqrt(2)`
the particle is below mean position, moving downward and acceleration upward, thus slowing down.
(vii) At `x=(3pi)/(2k)=(3lambda)/(4)`
here `y=0, v_(y)=-Aomega`, `a_(y)=0`
The particle is at mean position and moving downward with maximum velocity. acceleration of particle is zero.
(viii). At `x=(7pi)/(4k)=(7lambda)/(8)`
here `y=(A)/sqrt(2)`, `v_(y)=(Aomega)/sqrt(2), a_(y)=-(omega^(2)A)/sqrt(2)`
the particle is above mean position, having velocity and acceleration both in downward direction. thus particle is speeding up.