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Consider two simple harmonic motion along x and y- axis having same frequencies but different amplitudesas x=A sin (omega t+varphi) (along x axis) and y= B sin omega t( along y axis). then show that (x^(2))/(A^(2))+(y^(2))/(B^(2))-(2xy)/(AB) cos varphi = sin^(2) varphi and also discuss the special cases when varphi=(pi)/(2) and A=B Note : when a particle is subjected to two simple harmonic motion at right angle to each other the particle may move along different paths. |
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Answer» Solution :Given : `x=A sin (OMEGA t- phi) ""...(1)` `y= B sin omega t""...(2)` In equation (1) use, `sin (A-B)= sin A COS B+ cos A sin B` (1) `rArr x= A sin omega t. + A phi A cos omega t. sin phi x-Asin omega t. cos phi= A cos omega t sin phi` Squaring on both sides we get, `(x-A sin omega t. cos phi)^(2)=A^(2) cos^(2) omega t sin^(2) phi ....(3)` In equation (2) ` sin omega c` can be re- written as, `(Y)/(b)` [ from equation (2)]. ALSO, use, `cos^(2) omega t=I- sin^(2) omega t` in equation (3) `:. (3)` becomes `(x-A(y)/(B). cos phi)^(2)=A^(2)(1-(y^(2))/(B^(2))) sin ^(2) phi` On expansion. `x^(2)+(A^(2)y^(2))/(B^(2)) cos^(2) phi-(2xAy)/(B) cos phi` `=A^(2) sin^(2) phi-(A^(2)y^(2))/(B^(2)) sin phi....(4)` `x^(2)+(A^(2)y^(2))/(B^(2))( sin^(2) phi+ cos^(2) phi)-(2xyA)/(B) cos phi` `=A^(2) sin^(2) phi "" ( -: byA^(2))` we get, `(x^(2))/(A^(2))+(y^(2))/(B^(2))*1 -(2xy)/(AB) cos phi= sin^(2) phi ""...(5)` HENCE proved. Special cases : `varphi =(pi)/(2)` and A=B in equation (5) `(x^(2))/(A^(2))+(y^(2))/(A^(2))=1 rArr x^(2)+y^(2)=A^(2)` The above equation of a circle whose centre is origin. |
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