The transverse displacement of a string clamped at its both ends is given by
`y(x, t) = 0.06 sin ((2pi)/3 x) cos(120pit)` 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. 324 N
B. 648 N
C. 832 N
D. 972 N

Answer

B

Explanation

The given equation is
`y(x,t)=0.06 sin((2pi)/3x)cos(120pit)`
Compare it with y(x, t) = 2a sin kx cos `omega`t
we get, k`=(2pi)/3or(2pi)/lamda=(2pi)/3orlamda=3 m`
and `omega = 120pi: or 2piupsilon = 120pi or upsilon =- 60 Hz = 60 s^(-1)`
Velocity of wave, `v = upsilonlamda = (60 s^(- 1)) (3 m) = 180 m s^(-1)`
Mass per unit length of the string,
`mu=(3xx10^(-2)kg)/(1.5m)=2xx10^(-2)kgm^(-1)`
Velocity of transverse wave in the string, v
`v=sqrt(T/mu)orv^(2)=T/muorT=v^(2)mu`
`T=(180ms^(-1))^(2)(2xx10^(-2)kgm^(-1))=648 N`