Here, the equation for transverse displacement is given by $y(x, t)=0.06 \sin\left(\frac{2\pi}{3} x\right) \cos(120\pi t)$...(i) (a) The displacement, which involves harmonic functions of x and t separately, represents a stationary wave and the displacement, which is harmonic function of the form $(vt \pm x)$, represents a travelling wave.Hence, the equation given above represents a stationary wave.(b) When a wave pulse$y'=A \sin \frac{2\pi}{\lambda}(vt-x)$travelling along X-axis is superimposed by the reflected pulse$y''=-A \sin \frac{2\pi}{\lambda}(vt-x)$from the other end, a stationary wave is formed and is given by$y=y'+y''$ $=-2A \sin \frac{2\pi}{\lambda} x \cos \frac{2\pi}{\lambda} v t$...(ii)Comparing the equations (i) and (ii), we have$\frac{2\pi}{\lambda}=\frac{2\pi}{3}$ or $\lambda=3\text{ m}$Also, $\frac{2\pi}{\lambda} v=120\pi$ or $v=60\lambda$ $=60\times 3=180\text{ m s}^{-1}$Now, frequency, $\upsilon=\frac{v}{\lambda}=\frac{180}{3}=60\text{ Hz}$(c) Velocity of transverse wave in a string is given by $v=\sqrt{\frac{T}{\mu}}$Here, $\mu=\frac{3.0\times10^{-2}}{1.5}=2\times10^{-2}\text{ kg m}^{-1}$Also, $v=180\text{ m s}^{-1}$$\therefore T=v^{2}\mu=(180)^{2}\times2\times10^{-2}=648\text{ N}$