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The vibrations of a string of length 60 cm fixed at both ends are represented by the equation y = 4 sin (( pi x)/(15)) cos ( 96 pi t) wherex and y are in cm and t in second . ii. Where are the nodes located along the strings ? iii. What is the velocity of the particle atx = 7.5 cm at t = 0.25 s? iv. Write down the equations of component waves whose superposition gives the above waves . |
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Answer» Solution :The given EQUATION for standing waves in the string is ` y = 4 sin ((pi X)/(15)) cos (96 pi t)`(i) i. The amplitude of the waves is given by ` A = 4 sin ( pi x)/(15)`(ii) Therefore , the maximum displacement or amplitude at ` x = 5 cm` is `A = 4 sin ( pi xx 5)/(15) = 4 sin (pi)/( 3)` ` = 4 sin 60^(@) = 4 xx ( sqrt( 3))/(2) = 2 sqrt(3) = 2xx 1.732 = 3.464 cm` ii. The position of zero displacement or nodes are given by `sin (pi x)/(15) = 0 or (pi x)/( 15) = r pi ( where r = 0 , 1 , 2, 3,...)` `rArr x = 15 r rArr x = 0, 0.15 cm , 0.30 cm , ....` iii. Differentiating Eq. (i) with respect to `t` , we get velocity of particle ` u = (DY)/(dt) = - 4 xx 96 pi sin ((pi x)/( 15)) sin (96 pi t)` Substituting ` x = 7.5 cm and t = 0.25 s`. `u = - 384 pi sin (( pi xx 7.5)/(15)) sin (96 pi xx 0.25)` ` = - 384 ( pi//2) sin (24 pi) = 0` Using the relation ` 2 sin A cos B = sin ( A + B) + sin ( A - B)` Equation (i) may be expressed as ` y = 2 [ sin {(pi x)/(15) + ( 96 pi t)} + 2 sin { (pi x)/(15) - (96 pi t)}]` ` = 2 sin {pi x)/(15) + 96 pi t + 2 sin {(pi x)/(15) - 96 pi t} = y_(1) + y_(2)` Therefore , the component waves are given by `{:(y_(1) = 2sin (96pit +(pix)/15)),(y_(2)=-2sin(96pit -(pix)/15)):}` |
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