The transverse displacement of a string (clamped at its 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 $t$ 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.

The general equation representing a stationary wave is given by the displacement function

$y(x, t)=2 a \sin k x \cos \omega t$

This equation is similar to the given equation:

$y(x, t)=0.06 \sin \left(\frac{2}{3} x\right) \cos (120 \pi t)$

Hence, the given function represents a stationary wave.

A wave travelling along the positive $x$ -direction is given as:

$y_{1}=a \sin (\omega t-k x)$

The wave travelling along the negative $x$ -direction is given as:

$y_{2}=a \sin (\omega t+k x)$

The superposition of these two waves yields:

$y=y_{1}+y_{2}=a \sin (\omega t-k x)-a \sin (\omega t+k x)$

$=a \sin (\omega t) \cos (k x)-a \sin (k x) \cos (\omega t)-a \sin (\omega t) \cos (k x)-a \sin (k x) \cos (\omega t)$

$=-2 a \sin (k x) \cos (\omega t)$

$=-2 a \sin \left(\frac{2 \pi}{\lambda} x\right) \cos (2 \pi v t)\dots(i)$

The transverse displacement of the string is given as

$y(x, t)=0.06 \sin \left(\frac{2 \pi}{3} x\right) \cos (120 \pi t)\dots(ii)$

Comparing equations ( $i$ ) and $(ii)$, we have:

$\frac{2 \pi}{\lambda}=\frac{2 \pi}{3}$

Wavelength, $\lambda=3 \,m$

It is given that:

$120 \pi=2 \pi v$

Frequency, $v=60 \,Hz$

Wave speed, $v=v \lambda$

$=60 \times 3=180 \,m / s$

The velocity of a transverse wave travelling in a string is given by the relation:

$v=\sqrt{\frac{T}{\mu}}$

Where,

Velocity of the transverse wave, $v=180 \,m / s$

Mass of the string, $m=3.0 \times 10^{-2} \,kg$

Length of the string, $l=1.5 \,m$

Mass per unit length of the string, $\mu=\frac{m}{l}$

$=\frac{3.0}{1.5} \times 10^{-2}$

$=2 \times 10^{-2}\, kg\, m ^{-1}$

Tension in the string $=T$

$T=v^{2} \mu$

$=(180)^{2} \times 2 \times 10^{-2}$

$=648\, N$