Find the trajectory equation y (x) of a point if it moves according to the following laws: (a) x = a sin ωt, y = a sin 2ωt; (b) x = a sin ωt, y = a cos 2ωt.
(a) y² = 4x²(1 − x²/a²), the same as y = ±2x√(1 − x²/a²). The path is a figure eight.(b) y = a(1 − 2x²/a²) = a − 2x²/a. The path is an arc of a parabola. With a = cm this is the curve on the screen. Both answers are given below, one under each tab.
Problem statement
Problem 4.15. Find the trajectory equation y (x) of a point if it moves according to the following laws: (a) x = a sin ωt, y = a sin 2ωt; (b) x = a sin ωt, y = a cos 2ωt.
Interpretation. The point moves in a plane with coordinates x and y. In both laws x = a sin ωt. In law (a) y = a sin 2ωt, so y oscillates twice as fast as x. In law (b) y = a cos 2ωt. The trajectory equation y(x) is the relation between y and x with the time removed. The two tabs above the picture show the two laws; the first tab is (a).
We assume a = 5 cm and ω = 2 s−1, because the problem gives only symbols. Both are sliders. They change the size and the speed, never the form of the equation.
Symbols: a is the amplitude, ω the angular frequency, t the time.
The big idea in plain words
Each law gives x and y as functions of time. To find the path, remove the time. The first law links x to sin ωt. The second law is written with 2ωt, so use a double angle rule to bring it back to ωt. Then put x/a in place of sin ωt.
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, law (a), use x. From x = a sin ωt, sin ωt = x/a. Then cos ωt = ±√(1 − x²/a²), because sin² + cos² = 1. The two signs belong to the two times in a period when the point is at the same x.
Step 2, law (a), the double angle. The rule sin 2A = 2 sin A cos A comes from sin(A + A) = sin A cos A + cos A sin A. So y = a sin 2ωt = 2a sin ωt cos ωt.
Step 3, law (a), remove t. y = 2a (x/a) (±√(1 − x²/a²)) = ±2x√(1 − x²/a²). Squaring: y² = 4x²(1 − x²/a²).
Step 4, law (a), the shape. For each x between −a and a there are two values of y, equal in size and opposite in sign. At x = 0 both are zero, so the curve crosses itself at the origin. It is a figure eight, with its widest points at x = ±a (where y = 0) and its tallest points where y has size a at x = ±a/√2.
Step 1, law (b), use x. From x = a sin ωt, sin²ωt = x²/a².
Step 2, law (b), the double angle. The rule cos 2A = 1 − 2 sin²A comes from cos 2A = cos²A − sin²A and cos²A = 1 − sin²A. So y = a cos 2ωt = a(1 − 2 sin²ωt).
Step 3, law (b), remove t. y = a(1 − 2x²/a²) = a − 2x²/a. This is a parabola with its top at (0, a). There is no square root, because sin²ωt = x²/a² goes in directly.
Step 4, law (b), the shape. Only x between −a and a is reached. At x = ±a, y = −a. The point starts at the top, slides to one end, comes back through the top to the other end, and returns. It runs to and fro along the arc and never makes a loop.
Step 5, the cases. Law (a) has two values of y for each x, so it needs the plus and minus sign. Law (b) has one value of y for each x. A larger a makes the curve larger in both directions, and a larger ω makes the point go round faster without changing the curve.
A separate calculation takes 721 phases of one turn, finds x and y from the law, and tests the trajectory equation of the chosen tab on each pair. The largest relative difference is .
Five methods
1. Force method
Not suited: no force or mass is given.
2. Energy method
Not suited, for the same reason.
3. Graph method
The first graph shows x and y against the phase. The second shows y against x, the trajectory itself, with the point on it. The third shows the speed against the phase.
4. Calculus method
Not needed for the path. The velocity (used for the arrow) is vx = aω cos ωt and vy = 2aω cos 2ωt for law (a), or −2aω sin 2ωt for law (b).
5. Check method
Test the point at t = 0. In (a) it is at (0, 0), and the equation gives y² = 0. In (b) it is at (0, a), and the equation gives y = a. At ωt = 90 degrees, x = a, and (a) gives y = 0 while (b) gives y = −a. Both agree with the equations.
Fastest: the double angle rule. Most reliable: the numerical check.
What each slider changes
| Control | Changes the answer? | What it changes |
|---|---|---|
| Tab (a) or (b) | Yes | The law, the curve, the equation, the working and every graph |
| Amplitude a | Yes | The size of the curve, the speed and every curve. It does not change the form of the equation |
| Angular frequency ω | No | The period and the speed. It does not change the path |
| 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 |
|---|---|---|
| Amplitude | a | cm |
| Angular frequency | ω | s−1 |
| Period | T | s |
| Trajectory | y(x) |
Exam traps
- Writing y = sin 2x for law (a). The 2 multiplies ωt, not x.
- Forgetting the ± sign in (a). The path has two values of y for each x.
- Putting a square root in (b). cos 2ωt = 1 − 2 sin²ωt has no root once sin²ωt = x²/a² is used.
- Thinking (b) is a closed loop. The point goes to and fro along an arc.