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

(i) `2 sqrt(3)cm`
(ii) `1.49 xx10^(-3)m`
(iii) 0
(iv) `y_(1)=2sin((pix)/(15)-96pit)and y_(2)=2sin((pix)/(15)+96pit)`

Explanation

(i) `(y_(max))_(x=5cm)=4 sin((pix)/(15))=4sin((5pi)/(15))=2sqrt(3)`
(ii) `(2pi)/(lamda)=(pi)/(15)rArr lamda=30cm`
`:.` Position of nodes
`=15cm,30cm`….
(iii) `v=-4(96pi)sin((pix)/(15))sin(96pit)`
at`x=7.5 cm,t=0.25sec`
`v=-4 xx 96 pi sin((7.5pi)/(15)) sin((96pi)/(4))=0`
(iv) `Y=4 sin((pix)/(15))cos(96pit)`
`=2sin((pix)/(15)-96t)+2sin((pix)/(15)+96pit)`.