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.

Answer

A

Explanation

(a) At `x=5cm`, the standing wave equation gives
`y=4sin((5pi)/15)cos(96pit)`
`=4sin(pi/3)cos (96pit)`
`=4xx sqrt3/2 cos(96pit)`
`:. Maximum displacement =`2sqrt3` cm
(b)The nodes are the points of permanent rest. Thus, they are those points for which
` sin((pix)/(15))=0`
i.e. `(pix)/(15) =npi`, where `n=0,1,2,3,4`...................
`x=15n`, i.e. at `x=0,15,30,45` and `60 cm`,
(c) The particle velocity is equal to
`((dely)/(delt)) = 4sin ((pix)/(15)) (96pi) (-sin 96pi t)`
=-`384 pi sin ((pix)/(15)) (-sin 96pit)`
At `x=7.5 cm` and `t=0.25 s`, we get
`(((dely)/(delt)))= -384 pi sin (pi/2)sin (24pi)=0`
(d)The equations of the component waves are
`y_(1) = 2sin (((pix)/15)+96pit)`
and `y_(2) = 2sin((pix)/(15) - 96pit)`
as we can see that `y=(y_1+y_2)` .