The displacement of the medium in a sound wave is given by the equation `y_(1) = A cos(ax + bt)` where `A`, `a and b` are positive constants. The wave is reflected by an obstacle situated at `x = 0`. The intensity of the reflected wave is `0.64` times that of the incident wave.
(a) What are the wavelength and frequency of incident wave?
(b) Write the equation for the reflected wave.
( c ) In the resultant wave formed after reflection, find the maximum and minimum values of the particle speeds in the medium.
(d) Express the resultant wave as a superposition of a standing wave and a travelling wave. What are the positions of the antinodes of the standing wave ? What is the direction of propagation of travelling wave?

Answer

(a) `(2pi)/(a), (a)/(b)` (b) `- 0.8 A cos (bt - ax), (c) 1.8 Ab, 0.2 Abm (d) `[2A sin b t sin ax + 0.2 A cos (bt - ax)]`

Explanation

Given wave is
`y = A cos(ax + bt)`
(a) We use (omega) = b and `(2pi)/(lambda) = a`
`lambda = (2pi)/(a)` and `n = (b)/(2pi)`
Wave speed
`v = nlambda = (b)/(a)`
(b) Equation of reflected wave is
`y_(r) = 0.8 A cos(bt - ax - pi) = 0.8 A cos(bt - ax)` Here phase of `pi` is added due to reflection an obstacle.
(c) After superposition maximum and minimum amplitudes are
`A_(max) = 1.8 b and A_(min) = 0.2b`
Thus maximum and minimum particle speeds are
`v_(max) = A_(max) omega = 1.8 Ab`
`v_(min) = A_(min) omega = 0.2Ab`
(d) Resulting wave is given as
`y_(R) = y + y_(r)`
= `A cos (bt + ax) - 0.8 A cos(bt - ax) + [0.2 A cos(bt -ax)`
= `0.2 A cos (bt - ax)]`
= `2A sin bt sin ax + 0.2 A cos (bt - ax)`