The transverse displacement of a string clamped at its both ends is given by
`y(x, t) = 0.06 sin ((2pi)/3 x) cos(l20pit)` where x and y are in m and t in s. The length of the string is 1.5 m and its mass is `3 xx 10^(-2)` kg. The tension in the string is

Options

A. 648 N
B. 724 N
C. 832 N
D. 980 N

Answer

A

Explanation

`m=(m')/(l) = (3xx10^(-2))/(1.5) = 2xx10^(-2) kg//m`
`k=(2pi)/(3) therefore (2pi)/(lamda)= (2pi)/(3)`
`therefore lamda = 3m` and
`omega = 2pin = 120 pi therefore n=60`
`v=nlamda`
`=60xx3 = 180m//s`
`v=sqrt(T/m) = sqrt((T)/(2xx10^(-2))`
`therefore T=v^2 xx2xx10^(-2) = 180 xx180 xx 2 xx 10^(-2)`
`=18xx18xx2=36xx18 =648 N`