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).

Options

A. `y=0.025cos(200pit-2pix)`
B. `y=0.5cos(200pit-2pix)`
C. `y=0.25cos(100pit-10pix)`
D. `y=0.5cos(100pit-10pix)`

Answer

A

Explanation

Let `y=Asin(omegat-kx+phi)`
`C=(omega)/(k)=sqrt((T)/(mu))implies (omega)/(k)=sqrt((35)/(3.5xx10^(-3)))=100m//s` . .. (i)
At t=0, x=0, y=+A` implies phi=pi//2 implies y=Acos(omegat-kx)`
slope`=(deltay)/(deltax)=+Ak sin(omegat-kx)`
And particle comes from `t=0,y+A` to `y=0` at `t=T//4=(pi)/(2v)`
`implies AKsin(omega,(pi)/(2omega)-0)=pi//20` . .. given
`implies A((2pi))/(lamda)=(pi)/(20)implies lamda=40A` . . (ii)
Also frequency=100Hz (given)
`implies omega=200pi` . . (ii)
`implies`From (i) and (iii) k=`2piimplies lamda=1m. implies A=(1)/(40)m=2.5cm`
`implies `Equation of wave is 2.5cos(`200pit-2pix`)