Q.15.24 One end of a long string of linear mass density 8.0\times 10^{-3}kg\: m^{-1} is connected to an electrically driven tuning fork of frequency 256\: Hz. The other end passes over a pulley and is tied to a pan containing a mass of 90\; 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\: cm. Write down the transverse displacement y as function of x and t that describes the wave on the string.

Answers (1)
S Sayak

y(x,t)=Asin(\omega t\pm kx+\phi )

A=0.05 m

Tension in the string is T=mg

\\T=90\times 9.8\\ T=882N

The speed of the wave in the string is v

\\v=\sqrt{\frac{T}{\mu }}\\ v=\sqrt{\frac{882}{8\times 10^{-3}}}\\ v=332ms^{-1}

Angular frequency of the wave is

\\\omega =2\pi \nu \\ \omega =2\pi \times 256\\ \omega =1608.5rad/s

\\k=\frac{2\pi }{\lambda }\\ k=\frac{2\pi\nu }{v}\\ k=4.84m^{-1}

Since at t=0, the left end (fork end) of the string x=0 has zero transverse displacement (y=0) and is moving along the positive y-direction, the initial phase is zero. \left (\phi =0\ rad \right )

Taking the left to the right direction as positive we have

y(x,t)=0.05sin(1608.5t-4.84x)

Here t is in seconds and x and y are in metres.

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