InterviewSolution
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InterviewSolution
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One end of a long string of linear mass density 8.0 xx 10^(-3) kgm^(-1) is connected to an electrically driven tuning fork of frequency 256Hz. The other end passes over a pulley and is tiedto a pan containing a mass of 90 kg. The pulley end absorbs all the incoming energy so that reflected waves at thisend have negligible amplitude. At t=0, the leftend ( 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. |
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Answer» Solution :Here , `mu= 8.0 xx 10^(-3) kg//m,v = 256Hz, T = 50 kg= 90 xx 9.8 = 882 N`. Amplitudeof wave, r= 5.01m = 0.05m . As the wave propagation along the string isa transverse travelling wave, the velocity of the wave is given by `v= sqrt((T)/(mu)) = sqrt((882)/( 8.0 xx 10^(-3))) = 3.32 xx 10^(2)m//s ` `omega = 2piv =2 xx(22)/(7) xx 256= 1.61 xx10^(3 ) rad//s` `lambda =(v)/(v) = ( 3.32 xx10^(2))/(256) m`. Propagation constant `k = ( 2pi)/( lambda)` `= ( 2 xx 3.142 xx 256)/( 332 xx 10^(2)) = 4.84m^(-1)` As the wave is propagating along x direction, the equation of the wave is`y ( x,t) = r sin ( omega t - kx) ` `=0.05 sin ( 1.61 xx 10^(3) t- 4.84x)` Here x,yare in mt & t in SEC. |
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