The equation of a wave on a string of linear mass density 0.04 ~kg ~m^-1 is given by y=0.02(~m) sin [2 π (t/(0.04(~s))-x/(0.50(~m)) ) ] . The tension in the string is

Correct answer: (D) 6.25 ~N. T=μ v^2=μ (ω^2)/(k^2)=0.04 ((2 π / 0.004)^2)/((2 π / 0.50)^2)=6.25 ~N

Q90 / 90 · JEE Main 2010 · Physics

The equation of a wave on a string of linear mass density $0.04 \mathrm{~kg} \mathrm{~m}^{-1}$ is given by $y=0.02(\mathrm{~m}) \sin \left[2 \pi\left(\frac{\mathrm{t}}{0.04(\mathrm{~s})}-\frac{\mathrm{x}}{0.50(\mathrm{~m})}\right)\right]$. The tension in the string is

$\mathrm{T}=\mu \mathrm{v}^2=\mu \frac{\omega^2}{\mathrm{k}^2}=0.04 \frac{(2 \pi / 0.004)^2}{(2 \pi / 0.50)^2}=6.25 \mathrm{~N}$

Hence, the correct answer is option: 'D'.