Two travelling waves of equal amplitudes and equal frequencies move in opposite direction 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 and amplitude of the component travelling waves . (b) Find the node closest to the origin in the region ` x gt 0`. ( c) Find the antinode closest to the origin in the region `x gt 0` . (d) Find the amplitude of the particle at ` x = 2.33 cm`.

Explanation

The standing wave is formed by the superposition of the waves
`y_(1) = (A)/(2) sin ( omega t - kx) and y_(2) = (A)/(2) sin ( omega t + kx)`
The wave velocity (magnitude) of either of the wave is
`v = ( omega)/(k) = ( 78.5 s^(-1))/( 1.57 cm^(-1)) = 50 cm//s, Amplitude = 0.5 mm`
(b) For a node , ` cos kx = 0`.
The smallest positive `x` satisfying is given by
`kx = pi //2`
or , ` x = (pi)/( 2 k) = ( 3.14)/( 2 xx 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`
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) = 1.57 [ 2 + (1)/(3)]`
` = pi + (pi)/(6) = (7 pi)/(6)`
Thus , the amplitude will be
`( 1.0 mm) | cos ( pi + pi//6)| = (sqrt(3))/(2) mm = 0.86 mm`