The transverse displacement of a string (clamped at both ends) is given by $y(x, t) = 0.06 \sin \left(\frac{2 \pi}{3} x\right) \cos (120 \pi t)$,where $x$ and $y$ are in $m$ and $t$ is in $s$. The length of the string is $1.5 \; m$ and its mass is $3.0 \times 10^{-2} \; kg$. Answer the following:
$(a)$ Does the function represent a travelling wave or a stationary wave?
$(b)$ Interpret the wave as a superposition of two waves travelling in opposite directions. What is the wavelength,frequency,and speed of each wave?
$(c)$ Determine the tension in the string.

(N/A) The general equation for a stationary wave is $y(x, t) = 2a \sin(kx) \cos(\omega t)$.
Comparing this with the given equation $y(x, t) = 0.06 \sin \left(\frac{2 \pi}{3} x\right) \cos (120 \pi t)$,we conclude that it represents a stationary wave.
$(b)$ $A$ stationary wave is the superposition of two waves: $y_1 = a \sin(\omega t - kx)$ and $y_2 = a \sin(\omega t + kx)$.
Comparing the given equation with $y = 2a \sin(kx) \cos(\omega t)$,we have $k = \frac{2 \pi}{3}$ and $\omega = 120 \pi$.
Since $k = \frac{2 \pi}{\lambda}$,we get $\lambda = \frac{2 \pi}{k} = \frac{2 \pi}{2 \pi / 3} = 3 \; m$.
Since $\omega = 2 \pi \nu$,we get $\nu = \frac{\omega}{2 \pi} = \frac{120 \pi}{2 \pi} = 60 \; Hz$.
The speed of each wave is $v = \nu \lambda = 60 \times 3 = 180 \; m/s$.
$(c)$ The speed of a transverse wave on a string is $v = \sqrt{\frac{T}{\mu}}$,where $\mu = \frac{m}{l} = \frac{3.0 \times 10^{-2}}{1.5} = 2 \times 10^{-2} \; kg/m$.
Thus,$T = v^2 \mu = (180)^2 \times (2 \times 10^{-2}) = 32400 \times 0.02 = 648 \; N$.