Irodov 1.95: An Ellipse Centered, Not Focused – Amit & Deepali Sinha Foundation

Problem Irodov 1.95

A particle of mass $m$ moves in the $xy$ plane with its coordinates given as functions of time by $x(t) = a\sin\omega t$ and $y(t) = b\cos\omega t$ , where $a$ , $b$ and $\omega$ are constants. Find the magnitude and the direction of the force acting on the particle.

Given

$x(t) = a\sin\omega t$
$y(t) = b\cos\omega t$

Find

the force

\vect F

acting on the particle: its magnitude and direction

Solution

Step 1Velocity components

Differentiate each coordinate once with respect to time.

\dot x = a\omega\cos\omega t, \qquad \dot y = -b\omega\sin\omega t.

Step 2Acceleration components

Differentiate once more, then recognize that the result is again proportional to the original coordinate.

\begin{align*}\ddot x &= -a\omega^2\sin\omega t = -\omega^2 x \\ \ddot y &= -b\omega^2\cos\omega t = -\omega^2 y\end{align*}

Step 3Newton’s second law in vector form

The force this step finds, drawn at P: it lies along - r, so it points back at the center of the ellipse and not at a focus. — The force this step finds, drawn at $P$ : it lies along $-\vect r$ , so it points back at the center of the ellipse and not at a focus.

Multiply each acceleration component by $m$ and assemble the two into one vector equation, using $\vect r = x\,\vect i+y\,\vect j$ for the particle’s position; this vector form already gives the force’s direction, opposite to $\vect r$ and so always pointing back toward the origin.

F_x = m\ddot x = -m\omega^2 x, \qquad F_y = m\ddot y = -m\omega^2 y.

\vect F = -m\omega^2\vect r

(1)

Step 4Magnitude of the force

Take the magnitude of the vector relation just found, using $|\vect r| = \sqrt{x^2+y^2}$ .

F = \left|-m\omega^2\vect r\right| = m\omega^2|\vect r|.

F = m\omega^2\sqrt{x^2+y^2}

(2)

Check

Dimensionally $[m\omega^2 r] = \text{kg}\cdot\text{s}^{-2}\cdot\text{m} = \text{N}$ , a force. If $a=b$ the path is a circle of radius $a$ and $F=m\omega^2 a$ at every point, the ordinary constant-magnitude centripetal force of uniform circular motion, which is what an ellipse with equal semi-axes must reduce to.

Remark

Because $\vect F=-m\omega^2\vect r$ is linear in the position vector measured from the origin, this is an isotropic two-dimensional harmonic oscillator: $x$ and $y$ each execute independent simple harmonic motion at the same frequency $\omega$ , and their combination closes into an ellipse centered on the force center itself. That centering is the signature of a linear restoring force, and it is what tells this problem apart from an inverse-square attraction, whose orbit is also an ellipse but with the attracting center sitting at a focus, off center, never at the middle.