A particle of mass `m` free to move in the `x - y` plane is subjected to a force whose components are `F_(x) = - kx` and `F_(y) = - ky`, where `k` is a constant. The particle is released when `t = 0` at the point `(2, 3)`. Prove that the subsequent motion is simple harmonic along the straight line `2y - 3x = 0`.

Answer

A

Explanation

`F = - kxhat(i) - kyhat(j)`

`F = 0` at `(0, 0)`
When it is displaced to a point `P` whose position vector is
`r = xhat(i) + yhat(j)`
Force on it is `F = - k(xhat(i) + yhat(j)) = - kr`
Since, `F prop - r`, motion is simple harmonic. At `t = 0` particle is at `(2,3)`

`(y)/(x) = (3)/(2)` or `2y - 3x = 0`
i.e. the particle will oscillate simple harmonically along this line.