dP/dx = 1 / (π √(a² − x²)) for −a < x < a. It is smallest at the centre and grows without limit at the ends. For your sliders a = cm, so at x = cm the density is per cm and the chance of finding the particle in a window dx = cm is dP = .
Problem statement
A particle performs harmonic oscillations along the x axis according to the law x = a cos ωt. Assuming that the probability P of the particle to fall within an interval from −a to +a is equal to unity, find how the probability density dP/dx depends on x. Here dP denotes the probability of the particle falling within an interval from x to x + dx. Plot dP/dx as a function of x.
We assume the motion is simple harmonic and undamped. This means a and ω stay fixed for all time.
We assume "probability" means the share of time spent in the interval, over a long time or over one full period. This means dP = Δt / T, where Δt is the time inside (x, x + dx) per period.
We assume dx is small. This means the speed is nearly the same across the window, so dt = dx / v.
Symbols: a is the amplitude, v is the speed at x, T = 2π/ω is the period, and ρ = dP/dx is the probability density in 1/cm. The page also gives the exact value of dP for a wide window.
The big idea in plain words
Think of a camera that takes a photo at a random moment. Where is the point most likely to be? Where it spends the most time, and that is where it moves slowest, near the two ends. In the middle it rushes past. So the chance is low at the centre and high near the ends. The size of the density is fixed by the rule that the total chance over the whole range is 1.
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.
Point rest frame
This frame does not apply. The point keeps accelerating, so a frame fixed to it is not inertial, and in it the point is always at the same place, which says nothing about the density.
The solution step by step
The motion repeats every period T. So the share of all the time that the point spends in the window (x, x + dx) is the time per period divided by T. In one period the point crosses the window twice, once going up and once coming down. Each crossing takes dt = dx / v, where v is the speed there.
The factor 2 is for the two crossings. Next we find the speed at x. From x = a cos ωt we get v = |dx/dt| = aω |sin ωt|. Using sin² = 1 − cos², and cos ωt = x/a, we get
Now put this speed in and use ωT = 2π, which removes both ω and T.
This is the answer, and it should add to 1 over the whole range. We check it by putting x = a sin θ, so dx = a cos θ dθ and √(a² − x²) = a cos θ.
For your sliders: v at the probe = cm/s, so each crossing lasts s, the time per period is s, and dP = . The density at the centre is the least, 1/(πa) = per cm.
Book answer: dP/dx = 1/(π√(a² − x²)), and the integral from −a to +a is 1. The page uses exactly this. The total of the exact bin shares is , against 1. A separate check samples the motion at 20000 equal time steps over one period and counts which bin the point is in. The largest gap from the exact share is . At the time shown, the area check gives against . A finite-difference slope of x gives cm/s against the formula cm/s.
Five methods
1. Force method
The force F = −mω²x gives the law x = a cos ωt, and the force is largest at the ends. That is why the point slows and turns round there. But the force does not give the density directly, so it only explains why the density is largest at the ends.
2. Energy method
Energy gives the speed at each place: (1/2) m v² + (1/2) m ω² x² = (1/2) m ω² a², so v = ω√(a² − x²). Then dP = 2 dx / (vT) gives the answer directly. This is the fastest route.
3. Graph method
On the phasor circle the point turns at a steady rate, so equal angles mean equal times. The window dx near the ends covers a wide angle, and near the centre a thin angle. The share of the circle in the violet arcs is dP. The page draws this.
4. Calculus method
The chance of being left of x is P(x) = 1/2 + arcsin(x/a)/π, and its derivative is the density. Differentiating arcsin(x/a) gives 1/√(a² − x²), so dP/dx = 1/(π√(a² − x²)).
5. Check method
Units: dP is a pure number and dx is a length, so dP/dx is in 1/cm, which matches 1/√(a² − x²). Limits: at x = 0 the density is 1/(πa), the smallest. At x = ±a it grows without limit, but the area stays finite, and it is 1 in total. The curve is symmetric about x = 0, as the motion is.
Fastest: the energy method. Most reliable: the calculus method, because the arcsine form also gives exact answers for wide windows.
What each slider changes
| Control | Changes the answer? | What it changes |
|---|---|---|
| Amplitude a | Yes | The range, ρ (proportional to 1/a at the centre), dP, the histogram and every graph |
| Period T | Not ρ or dP | ω, the speed at the probe, the time inside the window, the time axis and the motion speed |
| Probe position x | Yes | ρ at x, dP, the speed there, the shaded window; ρ is 0 outside the range |
| Window width dx | Yes (dP) | dP and the time in the window. ρ stays the same. |
| Histogram bins | No | Only the number of bars |
| Screen speed | No | Only how fast the animation runs compared with real time |
| Target density (given-answer mode) | Sets the answer | Solves a = √(x² + (1/(πρ))²) |
| Zoom, Auto-fit | No | Pixels per centimetre in the picture only |
Parameters
| Quantity | Symbol | Value used |
|---|---|---|
| Amplitude | a | cm |
| Period | T | s |
| Probe position | x | cm |
| Window width | dx | cm |
| Density at the probe | dP/dx | per cm |
| Chance in the window | dP | |
| Time in the window per period | Δt | s |
Exam traps
- Drawing the density as flat. The point does not spend equal time everywhere, because it is slow near the ends.
- Drawing the largest density at the centre. The density is smallest at the centre and large near ±a.
- Forgetting the factor 2 for the two crossings of the window. Without it the total would be 1/2.
- Leaving T or ω in the result. They cancel through ωT = 2π, so the density depends on a and x only.
- Saying the density is finite at x = ±a. It rises without limit there, but the total area is still 1.
- Using ρ·dx for a wide window. For a wide window use the arcsine form, dP = [arcsin(x2/a) − arcsin(x1/a)]/π.