A point moves in the plane xy according to the law x = a sin ωt, y = b cos ωt, where a, b, and ω are positive constants. Find: (a) the trajectory equation y (x) of the point and the direction of its motion along this trajectory; (b) the acceleration w of the point as a function of its radius vector r relative to the origin of coordinates.
(a) The path is the ellipse x²/a² + y²/b² = 1, traced clockwise. For a = cm and b = cm it is x²/ + y²/ = 1. (b) The acceleration is w = −ω² r. It points to the origin and has size ω²r, with ω² = s−2 here.
Problem statement
Problem 4.14. A point moves in the plane xy according to the law x = a sin ωt, y = b cos ωt, where a, b, and ω are positive constants. Find: (a) the trajectory equation y (x) of the point and the direction of its motion along this trajectory; (b) the acceleration w of the point as a function of its radius vector r relative to the origin of coordinates.
Interpretation. The x axis points to the right and the y axis points up, as in the usual drawing. Clockwise and anticlockwise are judged in that drawing. The radius vector r goes from the origin to the point. Part (a) asks for the path y(x) and the sense of motion along it. Part (b) asks for the acceleration written with the vector r.
Interpretation, the kind of motion. The motion is simple harmonic motion in two dimensions. The x part, x = a sin ωt, is a simple harmonic motion along the x axis. The y part, y = b cos ωt, is another one along the y axis. They have the same ω and are a quarter of a cycle apart. Adding two such motions at right angles gives an ellipse. This is elliptical SHM. It is the general case. When a = b the ellipse is a circle, and the motion is circular SHM, which is the same thing as uniform circular motion. Both obey w = −ω² r, so both have the same period 2π/ω. The difference is in the size of r. In the circle r never changes, so |w| = ω² r and the speed stay constant. In the ellipse r changes between b and a, so |w| changes along the same straight line |w| = ω² r. The test for simple harmonic motion is that the acceleration is proportional to the displacement and opposite to it. Along x this is wx = −ω² x, and along y it is wy = −ω² y. On the page the graph of wx against x and the graph of wy against y are each a straight line through the origin with slope −ω². So the motion is SHM along the x axis and SHM along the y axis at the same time, which together make SHM in the plane, w = −ω² r. The two lines are the same line, because the slope is the same. The x motion only visits the part from x = −a to a, and the y motion the part from y = −b to b. In the drawing the displacement and the acceleration along x are the yellow and pink arrows, and along y the lime and purple arrows. Each acceleration arrow points opposite to its displacement arrow. The graph of |w| against r shows this: a circle is one point on that line, and an ellipse is the stretch of the line between r = b and r = a.
We assume a = 6 cm, b = 4 cm and ω = 2 s−1, because the problem gives only symbols. All three are sliders. The page also shows the real period 2π/ω and slows the motion on the screen.
Symbols: a and b are the constants in the two coordinates, ω the angular frequency, r the radius vector, v the velocity, w the acceleration.
The big idea in plain words
Each coordinate is a simple oscillation, one with a sine and one with a cosine. Together they send the point round an ellipse. Each coordinate has an acceleration that points back to zero and grows with the coordinate. Together, the acceleration points back to the origin and grows with the distance from it.
The solution in each frame
Only the frame of the xy plane is used, so only one frame is needed.
The solution step by step
Step 1, part (a), the path. From x = a sin ωt we get sin²ωt = x²/a². From y = b cos ωt we get cos²ωt = y²/b². The sum of the two squares is 1, so x²/a² + y²/b² = 1. This is an ellipse with half-axes a along x and b along y. Written as y(x): y = ±(b/a)√(a² − x²), with the plus sign for the upper half and the minus sign for the lower half. For the numbers on screen: x²/ + y²/ = 1.
Step 2, part (a), the direction. At t = 0 the point is at x = 0, y = b, the top of the ellipse. A little later x = a sin ωt is positive and y is a little less than b. So at the top the point moves to the right. Moving right at the top is the clockwise way round. Another check: the velocity is vx = aω cos ωt, vy = −bω sin ωt, and the quantity x vy − y vx = −abω is negative for all time, which means clockwise.
Step 3, part (b), the acceleration. Differentiate again: wx = −aω² sin ωt = −ω²x and wy = −bω² cos ωt = −ω²y. As a vector, w = −ω² r. The acceleration points to the origin, and its size is ω²r.
Step 4, the numbers. With ω² = s−2, the size of w grows in a straight line with r. It is largest at the far ends of the ellipse and smallest where the path is nearest the origin.
Step 5, the cases. If a = b the path is a circle and the speed is constant, and w = −ω² r is the usual acceleration towards the centre. If a is not equal to b the speed changes, but the acceleration still points exactly to the origin. A larger ω makes the period 2π/ω shorter and the acceleration larger, with the same path.
A separate calculation uses only the coordinates x(t) and y(t). It finds the acceleration from small steps in time, and compares it with −ω² r. It also tests the ellipse equation and the sense of rotation at 120 points of the turn. The largest relative difference is .
Elliptical SHM and circular SHM compared
| Circle (a = b) | Ellipse (a ≠ b) | |
|---|---|---|
| Made of | Two SHMs at right angles, same ω, a quarter cycle apart | |
| Acceleration | w = −ω² r in both: it always points to the centre, and wx = −ω² x, wy = −ω² y | |
| Period | T = 2π/ω in both. It does not depend on a or b | |
| Distance r from the centre | Constant, r = a | Changes from b (smallest) to a (largest) |
| Size of the acceleration |w| = ω² r | Constant, ω² a | Changes from ω² b to ω² a |
| Speed | Constant, ω a | Largest (ω a) at the ends of the minor axis, smallest (ω b) at the ends of the major axis |
| Acceleration compared with velocity | Always at right angles | At right angles only at the ends of the axes. In between it has a part along the velocity, which changes the speed |
| Graph of |w| against r | One point on the line | The stretch of the line from r = b to r = a |
Five methods
1. Force method
Not suited: no mass or force is given. The problem is about motion only. As a remark, w = −ω² r means the force on a mass m would be −mω² r, a restoring force like a spring.
2. Energy method
Not suited, for the same reason.
3. Graph method
The first graph shows x and y against the phase ωt, a sine and a cosine. The second shows the acceleration parts, which are the same curves turned upside down and scaled by ω². The third shows |w| against r, a straight line through the origin with slope ω².
4. Calculus method
Differentiating twice, as in step 3.
5. Check method
Units: ω² r is s−2 times cm, which is cm/s². Limits: at t = 0 the point is at (0, b), and the acceleration is (0, −ω² b), pointing straight down to the origin. If a = b the path is a circle.
Fastest: differentiating twice. Most reliable: the numerical check.
What each slider changes
| Control | Changes the answer? | What it changes |
|---|---|---|
| Half-width a | Yes | The ellipse, the point, the speed, the acceleration and every curve. It does not change ω² |
| Half-height b | Yes | The ellipse, the point, the speed, the acceleration and every curve |
| Angular frequency ω | Yes | The period, the speed, the acceleration and the arrow lengths. It does not change the shape of the ellipse |
| Phase ωt | No | Where the point and the dots are |
| Given-answer mode | Sets ω | The page finds ω for a chosen period, and warns if ω falls outside 0.5 to 6 s−1 |
| Play and time bar | No | Play sweeps the phase through one turn in about 12 s and holds 3 s at ωt = 90 degrees. The bar sets the phase |
Parameters
| Quantity | Symbol | Value used |
|---|---|---|
| Half-width | a | cm |
| Half-height | b | cm |
| Angular frequency | ω | s−1 |
| Period | T | s |
Exam traps
- Writing the path as x² + y² = r². That is a circle, and it is true only when a = b.
- Saying anticlockwise because the point starts moving through the positive side. The start is at the top, and the point moves right.
- Saying the acceleration points along the velocity or along the path. It points to the origin, along −r.
- Forgetting the square: w = −ω² r, not −ω r.