Two linear SHM of equal amplitudes `A` and frequencies `omega` and `2omega` are impressed on a particle along `x` and `y - axes` respectively. If the initial phase difference between them is `pi//2`. Find the resultant path followed by the particle.

Explanation

As per question, `x=asin omegat` …(i)
`y=asin(2omegat+pi//2)=acos2omegat`
`=a(1-2sin^(2)omegat)`
`=a[1-2x^(2)//a^(2)]=a-(2x^(2))/(a)` …(ii)
and `x^(2)=-(a)/(2)(y-a)` ….(iii)
when `x=0, `from (ii), `y=a`
when `y=0`, from (ii), `x=a//sqrt(2)`
The equation (iii) is of a parabola. Thus, finallly the particle will follow a parabolic path with vertex (0,a) and latus rectum `=asqrt(2)`