The following equation represents standing wave set up in a medium ,
`y = 4 cos (pi x)/(3) sin 40 pi t`
where `x and y` are in cm and t in second. Find out the amplitude and the velocity of the two component waves and calculate the distance adjacent nodes . What is the velocity of a medium particle at ` x = 3 cm` at time `1//8 s`?

Answer

`2 cm , 120 cm//s , 3 cm , 160 pi cm//s`

Explanation

The given equation of stationary wave is
`y = 4 cos (pi x)/( 3) sin 40 pi t`
`y = 2 xx 2 cos ( 2 pi x)/( 6) sin ( 2 pi (120) t)/(6)`
We know that
`y = 2 A cos ( 2pi x)/( lambda) sin (2pivt)/(lambda)` (ii)
Comparing Eqs.(i) and (ii) , we get
`A = 2 cm , lambda = 6 cm and v = 120 cm//s`
The component waves are
`y_(1) = A sin ( 2pi)/( lambda) ( v t - x)`
and `y_(2) = A sin ( 2 pi)/( lambda)( v t + x)`
Distance between two adjacent nodes
`= (lambda)/( 2) = (6)/(2) = 3 cm`
Particle velocity
`(dy)/( dt) = 4 cos ( pi x)/(3) cos ( 40 pi t) xx 40 pi`
` = 160 pi cos((pi x)/(3)) cos 40 pi t`
At `x = 3 , t = 1//8` the particle velocity is given by
` = 160 pi cos pi cos(40 pi xx (1)/(8)) = 160 pi cm//s`