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

(i) Displacement vector of the particle of mass m is given by
`vec(r ) (t) = hat(i) A cos omega t hat(j) B sin omega t` `:.` Displacement along x-axis is,
`x = A cos omega t` or `(X)/(A) = cos omega t` and displacement along y-axis is,
`y = B sin omega t` or `(y)/(B) = sin omega t`
Squaring and then adding Eqs. (i) and (iii), we get
`(x^(2))/(A^(2)) + (y^(2))/(B^(2)) = cos^(2) omega t + sin^(2) omega t = 1`
This is an equation of ellipse. Therefore, trajectory of the particle is an ellipse.
(ii) Velocity of the particle
`vec(v ) = (d vec(r))/(dt) = hat(i) (d)/(dt) (A cos omega t) + hat(j) (d)/(dt) (B sin omega t)`
`= hat(i) [A (-sin omega t).omega]. hat(j) [B sin omega t).omega] = - hat(i) A omega sin omega t + hat(j) B omega cos omega t`
Acceleration of the particle `(vec(a)) = (d vec(v))/(dt)`
or ` a = - hat(i) A omega (d)/(d) (sin omega t) + hat(j) B omega (d)/(dt) (cos omega t)`
`- hat(i) A omega [cos omega t]. omega + hat(j) B omega^(2) sin omega t` `= - omega^(2) [hat(i) A cos omega t + hat(j) B sin omega t]` `= - omega^(2) vec(r )`
`:.` Force acting on the particle, `vec(F) = m vec(a) =- m omega^(2) vec(r )`