Question

The equation of a wave on a string of linear mass density `0.04 kgm^(-1)` is given by
`y = 0.02(m) sin[2pi((t)/(0.04(s)) -(x)/(0.50(m)))]`.
Then tension in the string is
A. `12.5N`
B. `0.5N`
C. `6.25N`
D. `4.0N`

Answer 1

Correct Answer - C
`y=0.02(m) sin[2pi((t)/(0.04(s))-(x)/(0.50(m)))]`
Comparing it with the equation of wave
`y=rsin ((2pi)/(lambda))(upsilont-x)=rsin 2pi((upsilont)/(lambda)-(x)/(lambda))`
We have, `(upsilon)/(lambda)=(1)/(0.04)` and `lambda=0.50`
`:. upsilon=(lambda)/(0.04)=(0.50)/(0.04)=(25)/(2)m//s`
If `T` is tension in the string , then
`upsilon=sqrt((T)/(m))` or`T= upsilon^(2)m=((25)/(2))^(2)xx0.04=6.25N`