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
`v = (omega)/(k) = (78.5s^(-1))/(1.57 cm^(-1)) = 50 cm//s,` Amplitude `= 0.5 mm`.
(b) For a node, `cos kx = 0`.
The smallest positive `x` satisying this relation is given by
`kx = pi//2` or, `x = (pi)/(2k) = (3.14)/(2 xx 1.57 cm^(-1)) = 1 cm`
For an antinode, `|cos kx| = 1`.
The smallest positive `x` satisfying this relation is given by
`kx = pi` or, `x = (pi)/(k) = 2 cm`
(d) 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) | = (sqrt(3))/(3)mm = 0.86 mm`.