Stationary waves produced in a string of length 60 cm is described by `y = 4 sin ((pix )/(15))cos (96pi t)` (where x and y are in cm and is in s). Find (1) position of nodes (il) positions of antinodes (iii) maximum displacement of a particle at * 5 cm (iv) equations of component waves of given stationary wave.

Explanation

Comparing `y = 4 sin ((pix )/(15)) cos (96 pi t )`
with `y = 2 A sin (kx) cos (omega t ) ` we get,
`2A = 4 implies A = 2 cm`
`k = (pi)/(15) (rad)/(cm) = (2pi)/(lamda ) implies lamda = 30 cm`
`omega = 96 pi rad//s`
(i) Nodes are located at `0, (lamda)/(2) , lamda, (3 lamda)/(2), 2 lamda,...`
`=15 cm, 30 cm, 45 cm, 60 cm,...`
(ii) Antinodes are located at `(lamda)/(4) , (3lamda)/(4), (5lamda)/(4), (7lamda)/(4),...`
`=7.5 cm , 22.5 cm, 37.5 cm 52.5 cm,...`
(iii) Amplitude of stationary wave at `x = 5 cm = 2 A sin (kx)`
`= 4 sin ((pi)/(15) xx 5)`
`= 4 sin ((pi)/(3)) = 4 xx 0.8660 = 3.464 cm`
(iv) We have `y = 4 sin ((pix )/( 15)) cos (96 pi t )`
`therefore y = 2 xx 2 sin alpha cos beta`
(Where `alpha = (pi x )/(15) and beta = 96 pi t )`
`=2 {sin (alpha + beta) + sin (alpha -beta)}`
(By formula)
`=2 { sin ((pi x )/( 15) + 96 pi t ) + 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)`
Where `y _(1) = 2 sin ((pi x )/(15) + 96 pi x ) cm`
`y _(2) = 2 sin ((pi x )/( 15) - 96 pi t ) cm`
are the two component waves of given stationary wave.