A transverse displacement of vibrating string is $y=0.06 \sin \left(\frac{2 \pi}{3}\right) x \cos (120 \pi t)$. If the mass per unit length of a string is $4 \times 10^{-2} \mathrm{~kg} / \mathrm{m}$, then the tension in the string will be

Given the transverse displacement $y = 0.06 \sin\left(\frac{2\pi}{3}x\right) \cos(120\pi t)$, comparison with the standard standing wave form $y = A \sin(kx) \cos(\omega t)$ identifies the wave number $k = \frac{2\pi}{3}$ rad/m and angular frequency $\omega = 120\pi$ rad/s.

The wave speed follows from $v = \frac{\omega}{k} = \frac{120\pi}{\frac{2\pi}{3}} = 180$ m/s.

For a string with mass per unit length $\mu = 4 \times 10^{-2}$ kg/m, the speed of transverse waves obeys $v = \sqrt{\frac{T}{\mu}}$. Solving for tension yields $T = v^2 \mu = (180)^2 (4 \times 10^{-2}) = 1296$ N.

The tension in the string is 1296 N.