One end of a taut string of length `3m` along the x-axis is fixed at `x = 0`. The speed of the waves in the string is `100ms^(-1)`. The other end of the string is vibrating in the y-direction so that stationary waves are set up in the string. The possible wavelength`(s)` of these sationary waves is (are)

Options

A. `y(t)=A"sin"(pix)/(6)"cos"(50pit)/3`
B. `y(t)=A"sin"(pix)/(3)"cos"(100pit)/(3)`
C. `y(t)=A"sin"(5pix)/(6)"cos"(250pit)/(3)`
D. `y(t)=A"sin"(5pix)/(2)"cos"250pit`

Answer

A, C, D

Explanation

The fixed end is a node while the first end is an antinode. Therefore at x=0 is a node and at x=3 m is an antinode possible mode of variation are :
`L=(2n+1)(lambda)/(4)` where n=0,1,2,3,….. Or ` lambda=(4L)/(2n+1)=(12)/(2n+1)`
(L=3 m(given))
`k=(2pi_/(lambda)=(2pi)/(12//(2n+1))(pi)/6=((2n+1)pi)/3`
We have statonary wave equation`y=A sin kx cos omegat`
For n=0:`y(t)=A sin (pix)/(6) cos(50pit)/(3) ` For n=2: `y(t)=A sin (5pix)/(6) cos (20pit)/(3)`
For n=7: `y(t)=A sin (5pix)/(2)cos 250pit`