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

From the knowledge of theory, if equation of statinary wave is
`y=2Acoskxsinomegat,` then
the component waves are
`y_(1)=Asin(omegat-kx)` and `y_(2)=Asin(omegat+kx)`
`:.` Amplitude of component waves
`=A=(2A)/(2)=(4)/(2)=2cm`
velocity of component waves
`v=(omega)/(k)=(40pi)/(pi//3)=120cm//s=1.2m//s`
(b) Wavelength, `lambda=(2pi)/(k)=(2pi)/(pi//3)=6cm.`
Distance between adjacent nodes
`=(lambda)/(2)=(6)/(2)=3cm`
(c) Particle velocity
`upsilon=(dy)/(dt)=(d)/(dt)[4cos((pix)/(3))sin40pit]`
`160picos((pix)/(3))cos40pit`
For `x=3cm ` and `t=(1)/(8)s`
`upsilon=160picos((pixx3)/(3))cos40pixx(1)/(8)`
`=160pi(-1)(-1)=160picm//s`
`=1.6pim//s`