Answer
$\tau = 0.036~N$
Work Step by Step
$y = (y_m) ~sin(kx-\omega t)$
$y = (0.021~m) ~sin[(2.0 ~m^{-1})x-(30 ~s^{-1}) t]$
We can find the wave speed:
$v = \frac{\omega}{k}$
$v = \frac{30~s^{-1}}{2.0~m^{-1}}$
$v = 15~m/s$
We can find the tension in the string:
$v = \sqrt{\frac{\tau}{\mu}}$
$v^2 = \frac{\tau}{\mu}$
$\tau = v^2 \mu$
$\tau = (15~m/s)^2 (1.6\times 10^{-4} kg/m)$
$\tau = 0.036~N$
