Figure. Represents the displacement `y` versus distance `x` along the direction of propagation of a longitudinal wave. The pressure is maximum at position marked

Answer

`(a) 6m,0.25Hz,1.5m//s
(b) 1.5pi mm//s, 0.75pi^(2) mm//s© pi`

Explanation

`(a)` from `y-x` graph
wavelength `=lambda = 6m`
from `y-t` graph
Time period `=T=4 sec`
`rArr `frequency `=f=(1)/(4)=0.25Hz`
wave speed `=f lambda=0.25 xx 6 =1.50 m//s`
`(b)` maximum `=3 m m xx (pi)/(2) rad //sec = 1.5 pi m m //sec`
maximum acceleration `=w^(2) A =(pi^(2))/(4) xx 3m m=0.75 p^(2) mm //sec^(2).`
`(c ) k=(2pi)/(lambda)=(pi)/(3) m^(-1)`
`rArr w=(2pi)/(T)=(pi)/(2) rad // sec`
`y=3 sin ((pi)/(3)xx-(pi)/(2) t+theta_(0))`
`y(x=2,t=0)=0`
`rArr sin ((2pi)/(3)+theta_(0))=0`
`rArr theta _(0)=-(2pi)/(3) ` or `(pi)/(3)`
and `(dely)/(delt)(t=0,x=2) gt 0`
`rArr (-3pi)/(2) cos ((pi)/(3) x-(pit)/(2)+theta_(0)) gt 0`
`(` For `x=2,t=0)`
`rArr cos ((2pi)/(3)+theta_(0)) lt 0`
`rArr theta_(0)=(pi)/(3)`
`y=(x,t)=3 sin ((pix)/(3)-(pit)/(2)+(pi)/(3))`
`rArr `at `x=2 ` and `t=4 sec, (dely)/(delx)=pi`