The displacement vector of a particle of mass m is given by r (t) = `hati A cos omega t + hatj B sin omega t`.
(a) Show that the trajectory is an ellipse.
(b) Show that F = `-m omega^(2)r`.

Explanation

(a) Displacement vector of the particle of mass m is given by
r(t) = `hati A cos omega t + hatjB sin omega t`
`therefore` Displacement along x-axis is ,
`x = A cos omega t `
or `" " (x)/(A) = cos omega t `
Displacement along y-axis is
and `" " y = B sin omega t `
or `" " (y)/(B) = sin omega t`
Squaring and then adding Eqs. (i) and (ii) , we get
`(x^(2))/(A^(2)) + (y^(2))/(B^(2)) = cos^(2) omega t + sin^(2) omega t = t`
This is an equation of ellipse .
Therefore , trajectory of the particle is an ellipse. (b) Velocity of the particle
`v = (dr)/(dt) = hati (d)/(dt) ( A cos omega t ) + hatj (d)/(dt) (B sin omega t)`
= `hati[A (- sin omega t ) . omega hatj + hatj[B(cos omegat) . omega]`
= `- hati A omega sin omega t + hatj B omega cos omega t `
Acceleration of the particle (a) = `(dv)/(dt)`
or `" " a = - hati A omega (d)/(d) (sin omega t) + hatj B omega (d)/(dt) (cos omegat)`
= `- hati A omega[cos omega t ] . omega + hatj B omega [- sin omega t0 . omega `
=`- hati A omega^(2) cos omega t - hatj B omega^(2) sin omega t`
=`- omega^(2) [hati A cos omega t + hatj B sin omega t]`
=`- omega^(2) r `
`therefore` Force acting on the particle ,
`F = ma = - m omega^(2)r. " " `Hence proved.