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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 isA. `12.5N`B. `0.5N`C. `6.25N`D. `4.0N` |
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Answer» 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` |
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