Over the first 3/8 of a period: (a) the mean velocity projection is 2√2 aω/(3π) ≈ 0.300 aω; (b) the modulus of the mean velocity vector is the same, 2√2 aω/(3π); (c) the mean speed is 2(4 − √2) aω/(3π) ≈ 0.549 aω. For your sliders the three results are , and cm/s.
Problem statement
At the moment t = 0 a point starts oscillating along the x axis according to the law x = a sin ωt. Find (a) the mean value of its velocity vector projection vx; (b) the modulus of the mean velocity vector |⟨v⟩|; (c) the mean value of the velocity modulus ⟨v⟩ averaged over 3/8 of the period after the start.
We assume the motion is simple harmonic and undamped. This means a and ω stay fixed.
We assume the averaging window is 0 to 3T/8. This means t = 0 is the start and the time interval is τ = 3T/8.
We assume the motion is along the x axis only. This means the velocity vector has only an x part, so |⟨v⟩| = |⟨vx⟩|.
Symbols: g is the window as a share of the period (book: g = 3/8), τ = gT is the window, x_f is the position at the end of the window, and ω = 2π/T. The slider lets you change g.
The big idea in plain words
Three different averages hide behind the word "mean". The mean of the velocity with its sign is just the displacement over the time. The size of that mean is a number. The mean of the speed is the total distance over the time, and distance counts the trip back as well as the trip out. In 3/8 of a period the point goes out to x = a and comes part of the way back, so distance is more than displacement and the mean speed is larger than the size of the mean velocity.
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 it would not help to average the velocity.
The solution step by step
From x = a sin ωt, the velocity is vx = aω cos ωt, because the derivative of sin ωt is ω cos ωt. The window is 0 to 3T/8, and ω = 2π/T, so at the end of the window ωt = 2π × 3/8 = 3π/4.
Part (a). The mean of vx over time is ∫vx dt divided by the time. The integral of vx is the change in x, so we can use the end positions.
Part (b). The point moves on a straight line, so the velocity vector is v = vx i and its mean is ⟨vx⟩ i. The modulus is the size of that number.
Part (c). Speed is |vx|, which equals aω cos ωt for t up to T/4 and −aω cos ωt from T/4 to 3T/8, because the point turns round at x = a. We add the distance on each leg. The out leg is a. The back leg is a − a sin(3π/4) = a(1 − 1/√2).
For your sliders: window = s, displacement = cm, distance = cm, so (a) = cm/s, (b) = cm/s and (c) = cm/s.
Book answers: (a) and (b) 2√2 aω/(3π), (c) 2(4 − √2) aω/(3π). The live ratios to aω at the book setting are 0.3002 and 0.5487, which match. The PDF's working states "using ω = 2 ω/T" in part (c), which is a typo for 2π/T. A numeric integral of |vx| over the window gives cm against the formula distance cm. 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 sin ωt, which we used. It does not directly give a mean, so it only supplies the motion.
2. Energy method
Energy gives the speed at each place, v = ω√(a² − x²), but we want a time average, not a place average. Averaging over place gives a different number, so this method does not suit the problem.
3. Graph method
Plot vx against t. The signed area under it is the displacement, and its height over the window gives the mean vx. Plot speed against t. The area under it is the distance, and the mean speed is the height of the equal-area rectangle. The page draws both.
4. Calculus method
This is the step-by-step route: use ∫vx dt = Δx for the signed mean, and split |vx| at the turning point for the mean speed.
5. Check method
Units: aω is cm/s, a speed. Limits: for g up to 1/4 the point never turns, so distance equals displacement and the mean speed equals |mean vx|. At g = 1/2 or 1 the mean vx is zero, while the mean speed is not. The mean speed can never be above the top speed aω, and 0.549 is below 1.
Fastest: the graph method for (a) and (b), because the displacement is read straight off. Most reliable: the calculus method, with the numeric area check.
What each slider changes
| Control | Changes the answer? | What it changes |
|---|---|---|
| Amplitude a | Yes, proportional | Displacement, distance, all three means, the track, every graph height |
| Period T | Yes, inverse | Window length gT, all three means, the time axis, the motion speed |
| Window share g | Yes | Window length, end position, displacement, distance, all three means, the shaded areas |
| Screen speed | No | Only how fast the animation runs compared with real time |
| Target mean speed (given-answer mode) | Sets the answer | Solves T = a × (∫|cos|) / (g × mean speed) |
| Zoom, Auto-fit | No | Pixels per centimetre in the picture only |
Parameters
| Quantity | Symbol | Value used |
|---|---|---|
| Amplitude | a | cm |
| Period | T | s |
| Window share | g | (book: 0.375) |
| Angular frequency | ω | rad/s |
| Window length | τ = gT | s |
| Top speed | aω | cm/s |
Exam traps
- Giving the same answer to (a) and (c). The signed mean is displacement over time; the mean speed uses the whole distance.
- Forgetting the turn at x = a. Only the first T/4 is outward, so the last T/8 is the trip back.
- Using the displacement a/√2 as the distance. The distance is a + (a − a/√2).
- Writing a negative mean speed. Speed is |vx| and cannot be negative.
- Using the wrong angle at the end of the window. The phase is 2π × 3/8 = 3π/4, not 3π/8.
- Mixing ω (rad/s) with the frequency f (Hz). ω = 2π/T.