`(a)` The given equation is ,
`xi=a cos ( omegat - kx)`
So at ` t=0,`
`xi= a cos kx `
Now, `(d xi)/(dt)=-a omega sin ( omegat- kx ) `
and `(d xi)/(dx)=a omega sin kx, `at` t=0`
Also, `(d xi)/(dx)=+a k sin ( omega t - kx )`
and at ` t = 0`,
`(d xi)/( dx)=-a k sin kx. `
Hence all the graphs are similar having different amplitude, as shown in the answere`-` sheet of the problem book.
`(b)` At the points, where ` xi= 0`, the velocity direction is positive, i.e.,along `+ve x-` axis in the case of longitudinal and `+ ve y-`axis in the case of transverse waves, Where `(d xi)/(dt)` is positive and vice versa.
For sought plots see the answer`-` shet of the problem book.