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 stationary waves is (are)

Options

A. `y(t)=Asin(2pix)/6cos(50pit)/3`
B. `y(t)=Asin(pix)/3cos(100pit)/3`
C. `y(t)=Asin(5pix)/6cos(255pit)/3`
D. `y(t)=Asin(5pix)/2cos250pit`

Answer

D

Explanation

The fixed end is a node while the free end is an antinode.
Therefore, at `x=0` is a node and at `x=3 ` m is an antinodel.
Possible modes of vibration are
`L=(2n+1)(lamda)/(4)` or `lamda=(4L)/(2n+1)=(12)/(2n+1)" "(because" "L=3"m"("Given"))`
`" "k=(2pi)/(lamda)=(2pi)/(12//(2n+1))=((2n+1)pi)/(6) `
`" "omega=vk=100(2n+1)(pi)/(6)=((2n+1)50pi)/(3)`
For `n=0, k=(pi)/(6), omega=(5pi)/(3)`
`n =1 , k =(pi)/(2), omega=50pi`
`n=2, k=(5pi)/(6), omega=(250pi)/(3)`
`{:(,.,,.,,.),(,.,,.,,.),(,.,,.,,.),(,.,,.,,.),(,.,,.,,.):}`
`n=7, k=(5pi)/(2), omega=250pi`
`{:(,.,,.,,.),(,.,,.,,.),(,.,,.,,.),(,.,,.,,.),(,.,,.,,.):}` so on
For `n =2 `
`" "y(t)=Asin""(pix)/(6 )cos""(50pit)/(3)`
For `n=2`
` y(t)=Asin""(5pix)/(6)cos""(250pit)/(3)`
For `n=7`
` y(t)=Asin""(5pix)/(2)cos250pit`