One end of a long string of linear mass density $8.0 \times 10^{-3} \mathrm{~kg} \mathrm{~m}^{-1}$ is connected to an electrically driven tuning fork of frequency $256 \mathrm{~Hz}$. The other end passes over a pulley and is tied to a pan containing a mass of $90 \mathrm{~kg}$. The pulley end absorbs all the incoming energy so that reflected waves at this end have negligible amplitude. At $t=0$, the left end (fork end) of the string $x=0$ has zero transverse displacement $(y=0)$ and is moving along positive $y$-direction. The amplitude of the wave is $5.0 \mathrm{~cm}$. Write down the transverse displacement y as a function of $x$ and $t$ that describes the wave on the string.

Linear mass density of the string is given as $\mu=8.0 \times 10^{-3} \mathrm{~kg} \mathrm{~m}^{-1}$

Frequency of the tuning fork is given as $=256 \mathrm{~Hz}$

Mass on the pan is given as $90 \mathrm{~kg}$

Tension on the string will be, $T=90 \times 9.8=882 \mathrm{~N}$

Amplitude is given as $A=0.05 \mathrm{~m}$

For a transverse wave, the velocity can be calculated as,

$v=\sqrt{\frac{T}{\mu}}=\sqrt{\frac{882}{8 \times 10^{-3}}}$

$=332 \mathrm{~m} / \mathrm{s}$

Angular frequency, $\omega=2 \pi \mathrm{f}$

$=2 \times 3.14 \times 256=1608.5 \mathrm{rad} / \mathrm{sec}$

Wavelength will be $\lambda=\mathrm{v} / \mathrm{f}=332 / 256=1.296 \mathrm{~m}$

Propagation constant will be $\mathrm{k}=2 \pi / \lambda=(2 \times 3.14) / 1.296$

$=4.845 \mathrm{~m}^{-1}$

The general equation of the wave is

$y(x, t)=A \sin (\omega t-k x)$

On Substituting all the values we get

$y(x, t)=A \sin (1608.5 t-4.845 x)$

$x$ and $y$ are in metre and $t$ is in seconds.