ω = √[(v1² − v2²)/(x2² − x1²)] and a = √[(v1²x2² − v2²x1²)/(v1² − v2²)]. For your readings x1 = cm, v1 = cm/s, x2 = cm and v2 = cm/s, the result is ω = rad/s and a = cm.
Problem statement
Find the angular frequency and the amplitude of the harmonic oscillations of a particle, if at the distances x1 and x2 from the equilibrium position its velocity equals v1 and v2 respectively.
We assume the oscillation is simple harmonic and undamped. This means ω and a stay fixed, so the same two numbers fit both readings.
We assume x1 and x2 are distances from the equilibrium position, and v1 and v2 are the speeds there. This means only the squares matter, so the direction of motion is not needed.
We assume the readings do come from one oscillation. This means the farther point must be the slower one, and the page warns when they are not.
Symbols: ω is the angular frequency, a is the amplitude, T = 2π/ω is the period. The example values x1 = 3 cm, v1 = 8 cm/s, x2 = 4 cm, v2 = 6 cm/s are our choice. The book gives no numbers, because the answer is a formula.
The big idea in plain words
A harmonic oscillator has one law that links its speed and its place: the farther from the middle, the slower. That law has two unknown numbers, ω and a. One reading gives one equation, which is not enough. Two readings give two equations, and two equations fix two unknowns.
The solution in each frame
Lab frame
The axis is fixed to the lab, with x = 0 at the equilibrium position. The full solution is in the next section.
Particle rest frame
This frame does not apply. The particle is always accelerating, so a frame fixed to it is not inertial, and it gives no way to link speed and place.
The solution step by step
First we need the speed law. From x = a cos(ωt + φ) we get vx = −aω sin(ωt + φ). Then x² = a² cos² and vx² = a²ω² sin², and since sin² = 1 − cos² we find vx² = ω²(a² − x²). Now write it for both readings.
Subtract the second from the first. The term ω²a² is the same in both and cancels.
Now find a. Multiply the first equation by x2² and the second by x1², then subtract. The terms ω²x1²x2² cancel.
For the example the numbers are neat: ω² = (64 − 36)/(16 − 9) = 4, so ω = 2 rad/s, and a² = (64 × 16 − 36 × 9)/28 = 25, so a = 5 cm. The period is T = 2π/ω = s.
The book answer is the pair of formulas above, and the page uses exactly those. A live test checks that ω²(a² − x1²) gives v1² and ω²(a² − x2²) gives v2², and that ½v² + ½ω²x² is the same at both readings: and against ½ω²a² = (cm²/s²). At the time shown, vx² = and ω²(a² − x²) = (cm/s)². A finite-difference slope of x gives cm/s against the formula cm/s.
Five methods
1. Force method
A force F = −mω²x gives wx = −ω²x. The force law fixes the form of the motion, but it does not use the readings, so it cannot give the numbers alone. It tells us the speed law is the right tool.
2. Energy method
The energy per unit mass E = ½v² + ½ω²x² is the same at both readings. Setting the two equal gives ½v1² + ½ω²x1² = ½v2² + ½ω²x2², which is the ω formula again. The total E = ½ω²a² then gives a. This is the fastest route.
3. Graph method
Plot v² against x². The law v² = ω²a² − ω²x² is a straight line with slope −ω² and intercept ω²a². Two readings are two points, and a line through them gives both numbers. The second row of graphs shows this line.
4. Calculus method
This is the step-by-step route: differentiate once to get the speed, remove the time, then solve two equations.
5. Check method
Units: v²/x² has units s⁻², so its square root is a rate in s⁻¹. Limits: a² − x1² = v1²/ω² is never negative, so the amplitude is at least as big as both readings. If v1 = v2 with x1 ≠ x2 the formula fails, which matches the physics, because a point farther out cannot be as fast.
Fastest: the energy method. Most reliable: the graph method, because the line shows at once if the two readings do not fit.
What each slider changes
| Control | Changes the answer? | What it changes |
|---|---|---|
| x1, x2 | Yes | ω, a, T, the marked lines, the ellipse, the v² against x² line and every graph |
| v1, v2 | Yes | ω, a, T and every graph |
| Screen speed | No | Only how fast the animation runs compared with real time |
| Target ω and target a (given-answer mode) | Sets the answer | Solves v1 and v2 from v = ω√(a² − x²), then recovers ω and a again as a check |
| Zoom, Auto-fit | No | Pixels per centimetre in the picture only |
Parameters
| Quantity | Symbol | Value used |
|---|---|---|
| First distance | x1 | cm |
| Speed at x1 | v1 | cm/s |
| Second distance | x2 | cm |
| Speed at x2 | v2 | cm/s |
| Angular frequency | ω | rad/s |
| Amplitude | a | cm |
| Period | T | s |
Exam traps
- Subtracting the speeds instead of their squares. The formula needs v1² − v2², not v1 − v2.
- Swapping the order in the denominator. It is x2² − x1², and the numerator is v1² − v2², so the signs match.
- Reading the answer as the period. The formula gives ω, and the period is T = 2π/ω.
- Taking a = the larger of x1 and x2. The amplitude is bigger than both, unless one reading is at a turning point.
- Expecting a result from any two readings. They must satisfy: the farther point is slower.
- Using readings that are very close together. A small error in a reading then causes a big error in ω and a.