The vibrations of a string of length 60 cm fixed at both ends are represented by the equation `y=4sin((pix)/15) cos (96 pi t)`, where x and y are in cm and t in seconds.
(a)What is the maximum displacement of a point at `x = 5cm`?
(b)Where are the nodes located along the string?
(c)What is the velocity of the particle at x=7.5cm and t=0.25s?
(d)Write down the equations of the component waves whose superposition gives the above wave.

Explanation

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)):}`