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.
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.