InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
Two plane wavespropagate in a homogeneous elasticmedium, one along the x axis and the other along the y axis : xi _(1) = a cos ( omega t - kx), xi _(2) = a cos ( omega t - ky ) .Find the motion patternfo particles in the plane xy if both waves. (a) are tansverse and their oscillation directions coincide, (b) are longitudinal. |
|
Answer» Solution :`(a)` Equation of the resultant wave, `xi=xi_(1)+xi_(2)=2 a cos k ((y-x)/(2))XOA {omegat-(k(x+y))/(2)}, ` `=a'cos { omegat-(k(x+y))/2}`,where `a^(') = 2 a cos k^(') ((y-x)/2)` Now, the equation of wave pattern is, `x+ y=k`, (aConst.) For sought plots see the answer `-` sheet of the problem book. For antinodes, i.e. MAXIMUM intensity `cos((k(y-x))/(2))=+-1= cos n pi` or, `+-(x-y)=(2npi)/(k) = n lambda` or `y=x+- nlambda, n=0,1,2,.....` Hence, the particles of the medium at the POINTS, lying ono the solide straight lines `(y=x+-n lambda)` , oscillate with maximum amplitude.ltbr. For nodes, i.e. minimum intensity, `cos ((k(y-x))/(2))=0` or `+- (k(y-x))/( 2)=(2n+1) ( pi)/(2)` or , `y=x+- ( 2n +1) lambda//2`, and hence the particles at the points, lying on dotted lines do not oscillate. `(b)` When the waves are longitudinal, For sought plots see the answer `-` sheet of the problem book. `k(y-x)=cos^(-1)((xi_(1))/(a))- cos^(-1)((xi_(2))/( a))` or, `(xi_(1))/( a)= cos { k(y-x)+cos ^(-1) ((xi_(2))/(a))}` `=(xi_(2))/( a) cos k(y-x) - SIN k y( y-x) sin(cos ^(-1)((xi_(2))/(a)))` `=( xi_(2))/(a) cos k ( y-x)- sin k ( y-x) SQRT(1-(xi_(2)^(2))/( a^(2)))...(1)` from `(1)`, if `sin k ( y-x) = 0 =cos ( 2n +1) ( pi)/(2)` `(xi_(1)^(2))/(a)=1-xi_(2)^(2)//a^(2)`, acircle. Thus the particles, at the points, where `y= x+-(n+- 1//4) lambda` , will oscillate along circles, In general, all other particles will move along ellipse. |
|