When a plane wave train transverses a medium, individual particles execute a periodic motion given
by the equation `y=5 sin" "pi/4 (4t +x/(16))`
where the lengths are expressed in centimeters and time in seconds. The phase difference for two positions of the same particles which are occupied at a time interval 0.8 s apart is-

Explanation

The equation of a wave motion is given by
` y = (A sin) (2pi)/(lambda) (vt + x)`
here `y = (4sin)(pi)/(2)(2t+(x)/(8))`
This equation can be written as
`y = (4sin)(2pi)/(32)(16t + x)`
Comparing equation-(6.16) with equation-(6.15), we get amplitude A = 4cm, wavelength `lambda` = 32 cm , wave velocity v = 16 cm/s.
V= 16 cm/s. Here frequency is given as `n = (v)/(lambda) = (16)/(32) = (1)/(2)` = 0.5 Hz.
(i) Phase of a particleat instant `r_(1)` is given by
`phi_(1) = (x)/(2)(2t_(1) +(x)/(8)`
The phase at instant `t_(2)` is given by
`phi_(2) = (pi)/(2)(2t_(2) + (x)/(8))`
The phase difference is given as
`phi_(1) - phi_(2) = (pi)/(2)[(2t_(1) + (x)/(8)) - (2t_(2) + (x)/(8))]`
= `pi(t_(1)-t_(2)) = pi(0.4) [As t_(1)- t_(2) = 0.4]` (ii) Phase different at an instant between two particles with path difference `Delta` A is
`phi = (2pi)/(lambda) xx Delta`
= `(2pi)/(32) xx 12`
= `(3pi)/(4)`