A 100 Hz sinusoidal wave is travelling in the positive x-direction along a string with a linear mass density of `3.5xx10^(3) kg//m` and a tension of 35 N. time t=0, the point x=0 has maximum displacement in the positive y-direction. Next, when this point has zero displacement the slope (iii) the expression which represents the displacement of string as a function of x (in metres) and t (in seconds).

Explanation

The speed of wave in string is
`v=sqrt(T)/(mu)=sqrt((35)/(3.5xx10^(-3))=100 m//s`
`lambda=(v)/(f)=(100)/(100)=1 m`
The shape of the string at t=0 s is as shown in figure

Let us assume that
`y=A sin (omegat-kx+theta)`
putting `t=0, x=0 and y=+A`
gives `theta=pi//2`
also `omega=2pif=2pixx100=200pi`
`k+(2pi)/(lambda)=(2pi)/(1)=2pi`
Hence `y=A sin(200pit-2pix+pi/2)`
it is given that
`(dely)/(delx)=(pi)/(20) at t=(T)/(4)=(1)/(400) [:'T=(1)/(f)=(1)/(100)]`
`(dely)/(delx)]_(t=(1)(400))=A2pi cros(200t-2pix+(pi)/(2))=-2Api=(pi)/(20)`
`A=(1)/(40)=0.025`
(i). Amplitude A=0.25 m`
(ii) wave speed =100 m//s`
(iii) wave equation is
`y=0.025sin(200pit-2pix+(pi)/(2))`