A particle is subjected to two mutually perpendicular simple harmonic motions such that its `X` and `y` coordinates are given by `X=2 sin omegat` , `y=2 sin (omega+(pi)/(4))`
The path of the particle will be:

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)` in equation (5)
`(x^(2))/(A^(2))+(y^(2))/(B^(2))=1`
The above equation of an ellipse whose centre is origin.