A tranverse wave propagating on a stretched string of linear density `3 xx 10^(-4)` kg per m is represented by the equation
`" " y = 0.2sin(1.5x + 60t)`
Where , x is in metre and t is in second. The tension in the string (in newton) is

Options

A. 0.24
B. 0.48
C. `1.20`
D. `1.80`

Answer

b

Explanation

Equation of wave,
`y=0.2sin[1.5x+60t]`
Comparing with standard equation, we get
`y=Asin[kx+omegat]`
`k=1.5=(2pi)/(lambda)`
and `omega=60=(2pi)/(T)`
`therefore` Velocity of wave , `v=(omega)/(k)`
`=(60)/(1.5)=40m//s`
Velocity of wave in stretched string,v `=sqrt((T)/(m))`,
where, m is the linear density,T is tension in the string.
So , `T=v^(2)m=(40)^(2)xx3xx10^(-4)`
`=0.48N`