Question

One end of a long string of linear mass dnesity `8.0xx10^(-3)kgm^(-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 describest the wave on the string.

Answer 1

Here,`m=8.0xx10^(-3)kgm^(-1), v=256Hz, T-90kg=90xx9.8-882N`
Amplitude of wave, `r=5.0cm -0.05m.`
As, the wave propagating along the string is a transverse travelling wave, the velocity of the wave is given by
`upsilon=sqrt((T)/(m))=sqrt((882)/(8.0xx10^(3)))=3.32xx10^(2)ms^(-1)`
`omega =2piv=2xx(22)/(7)xx256=1.61xx10^(3)rad//s`
`lambda=(upsilon)/(v)=(3.32xx10^(2))/(256)m`.
Propagation constant, `k=(2pi)/(lambda)=(2xx3.142xx256)/(3.32xx10^(2))=4.84m^(-1)` As, the wave is propagating along positive x direction, the equation of the wave is
`y(x,t)=rsin(omegat-kx)=0.05sin(1.61xx10^(3)t-4.84x)`
Here, x, y are in metre and t is in second.