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.

Stationary waves produced in a string of length 60 cm is described by

1.	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.
Answer» Solution :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.