The displacement of a medium in a sound wave is given by the equation `y_(1)=Acos(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.
(i) What are the wavelength and frequency of incident wave ? ________ (ii) Write the equation for the reflected wave. _________ (iii) In the resultant wave formed after reflection, find the maximum and minimum values of the particle speeds in the medium. ____________ (iv) Express the resultant wave as a superposition of a standing wave and a traveling wave. What are the positions of the antinodes of the standing wave ? What is the direction propagation of traveling wave?__________

Explanation

`(i)` Amplitude of the reflected wave `=sqrt(0.49A)=0.7A`
As the reflecting surface is a denser medium there will be a phase change of `pi`. Also, the direction of propagation will be reversed. Hence the equation of the reflected wave will be
`y=0.7A cos(-ax+bt+pi)`
or `y=-0.7A cos (bt-ax)`
`(ii)` Resultant wave
`y=A cos(ax+bt)-0.7A cos(bt-ax)`
`y=0.3A cos(ax+bt)-1.4A sinaxsinbt`
First term represents a progressive wave where as the second one is a standing wave.
`(iii)` For antinodes we have
`sinax=+-1`
`rArr ax=(2n-1)(pi)/(2)`, `x=(2n-1)pi//2a`
`(iv)` Maximum speed of a particle of medium will occur at antinode. Equation of motion of a particle at antinode.
`y=0.3Acos[(2n-1)(pi)/(2)+bt]-1.4Asin(2n-1)pi//2sinbt`
`(dy)/(dt)=-0.3bAsin[(2n-1)pi//2)+bt]-1.4Asin[(2n-1)pi//2]cosbt`
`(dy)/(dt)=1.7Abcosbt` (if `n` is even)
`(dy)/(dt)= -1.7Acosbt` (if `n` is odd)
`:. ` Maximum speed of the particle `=1.7Ab`