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

Answer

A

Explanation

`because` `omega = 2pif = (2pi)(100)`
`=(200pi) rad//s`
`v=(omega)/(k)`
`:. k = (omega)/(v) = omegasqrt((mu)/(T)`
`= (200pi)sqrt((3.5 xx 10^(-3))/(35)`
`=2pi m^(-1)`
At zero displacement,
`v_(P) =omegaA = -v(dely)/(delx)`
`v((dely)/(delx))`
`:. |A| = v(((dely)/(delx)))/(omega) = ((sqrt(T)/(mu))("slope"))/(omega)`
`=((sqrt((35)/(3.5xx10^(-3))))(pi//20))/(200pi)`
`0.025 m`