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`?

Explanation

The given equation of stationary wave is
`y = (4cos) (pi x)/(3) sin 40 pi t`
`y = (2xx 2cos) (2pix)/(6) (sin)(2pi(120)t)/(6)`
We know that
` y = (2A cos)(2pix)/(lambda) (sin)(2pi v t)/(lambda)`
Comparing the equations - (6.158) and (6.159), we get
A = 2 cm, lambda = 6 cm and v = 120 cm/s
The comonent waves are
`y_(1) = (A sin) (2pi)/(lambda)(vt -x )`
`y_(2) = (Asin) (2pi)/(lambda) (vt + x)`
Distance between two adjacent nodes
`= (lambda)/(2) = (6)/(2)` = 3 cm
Particle velocity
`(dy)/(dt) = (4cos)(pix)/(3) (cos)(40pit) xx 40 pi = 160pi cos ((pix)/(3))cos 40pi t`