Consider two simple harmonic motion along x and y- axis having same frequencies but different amplitudes as `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.

Explanation

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-A sin 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.