Two travalling wavews of equal amplitudes and equal frequencies move in opposite directions along a string. They interfere to produce a standing wave having the equation `y = A cos kx sin omega t` in which `A = 1.0 mm, k = 1.57 cm^(-1) and omega = 78.5 s^(-1)` (a) Find the velocity of the component travelling waves. (b) Find the node closet to the origin in the x gt 0. (c ) Find the antinode closet to the origin in the region x gt 0 (d) Find the amplitude of the particle at x = 2.33cm.

Explanation

(a) The standing wave is formed by the superposition of the waves ` y_1 = (A)/(2) sin (omegat - kx)` and `
`y_2 = (A)/(2) sin (omegat + kx).`
The wave velocity (magnitude) of either of the waves is
`upsilon = (omega)/(k) = (78.5s^(-1))/(1.57 cm^(-1)) = 50 cm s^(-1).`
(b) For a node, cos kx = 0. The smallest positive x satisfying this relation is given by
`kx =(pi)/(2)`
or `x =(pi)/(2k) = (3.14)/(2xx 1.57 cm^(-1)) =1 cm.`
(c ) For an antinode, |cos kx| =1. The smallest positive x satisfying this relation is given by
`kx = (pi)/(2)`
or `x =(pi)/(k) =2cm.
The amplitude of vibration of the particle at x is given by |A cos kx|. For the given point,
`kx = (1.57 cm ^(-1) (2.33 cm) = (7)/(6)pi = pi +(pi)/(6)`
Thus, the amplitude will be (1.0 mm) `| cos (pi + pi//6) | = (sqrt3)/(2) mm = 0.86mm.`