A `100 Hz` sinusoidal wave is travelling in the positive x-direction along a string with a linear mass density of `3.5 xx10^(-3) kg//m` and a tension of `35 N`. At time `t= 0`, the point `x=0` , has maximum displacement in the positive y-direction. Next when this point has zero displacement, the slope of the string is `pi//20`. Which of the following expression represent (s) the displacement of string as a function of `x`(in metre) and `t` (in second)

Options

A. `y=0.025sin(200pit-2pix+(pi)/(4))`
B. `y=0.025sin(2pix-200pit+(pi)/(2))`
C. `y=0.025sin(2pix-200pit+(pi)/(4))`
D. `y=0.025sin(200pit-2pix+(pi)/(2))`

Answer

D

Explanation

The speed of wave in string is
`V=sqrt((T)/(mu))=sqrt((35)/(3.5xx10^-3))=100 m//s`
`lamda=(v)/(f)=(100)/(100)=1m`
Let us assume that
`t=0`,`x=0` and `y=+A` `because` `theta=(pi)/(2)`
Hence `y=Asin(200pit-2pix+(pi)/(2))`
It is given that
`(dely)/(delx)=(pi)/(20)` at `t=(T)/(4)=(1)/(400)sec`
`(dely)/(delx)]_(t=(1)/(400)S)=A2picos(200pit-2pix+(pi)/(2))`
or `2Api=(pi)/(20)impliesA=(1)/(40)=0.025m`
Wave equation is
`y=0.025sin(200pit-2pix+(pi)/(2))`