Starting from the extreme position the mean speed is 3a/T = 0.5 m/s. Starting from the equilibrium it is 6a/T = 1 m/s. For your sliders (a = cm, T = s, f = ) the results are cm/s for case (a) and cm/s for case (b).
Problem statement
A point performs harmonic oscillations along a straight line with a period T = 0.60 s and an amplitude a = 10.0 cm. Find the mean velocity of the point averaged over the time interval during which it travels a distance a/2, starting from (a) the extreme position; (b) the equilibrium position.
We assume the motion is simple harmonic and undamped. This means a and T stay fixed.
We assume the time interval starts at t = 0 and ends when the point has first covered a/2. This means the point moves one way only, so the mean velocity is the distance divided by the time.
We assume the extreme start is x = a cos ωt and the equilibrium start is x = a sin ωt. This means the motion begins at the stated position at t = 0.
Symbols: f is the distance as a share of a (book: f = 1/2), t1 and t2 are the times to cover that distance in cases (a) and (b), and ω = 2π/T. The slider f lets you change the distance.
The big idea in plain words
The point is fastest at the middle and slowest at the ends. A stretch of a/2 starting at the middle is covered at high speed, so it takes a short time. The same stretch ending at the extreme is covered at low speed, so it takes a long time. Mean speed is distance divided by time, so the middle start gives the bigger mean. We find the times from the position law, then divide.
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 does not help find a time or a mean speed.
The solution step by step
Case (a). We write the motion as x = a cos ωt, which puts the point at x = a when t = 0. After it has covered a/2 it is at x = a − a/2 = a/2. Call that time t1.
We pick ωt1 = π/3 because t1 is before the quarter period, where ωt is below π/2. The speed is |vx| = aω sin ωt, and the mean speed is the total distance divided by the total time. The integral of the speed over the window is the distance a/2, so we can divide directly.
Case (b). Now the motion is x = a sin ωt, which puts the point at x = 0 when t = 0. It has covered a/2 when x = a/2. Call that time t2.
For your sliders, the general formulas are t1 = arccos(1 − f)/ω and t2 = arcsin(f)/ω, so ⟨v⟩ = f a / t. The live values are t1 = s, t2 = s, giving cm/s and cm/s.
Book answer: (a) 0.5 m/s and (b) 1 m/s. The live values for the book setting match these. The PDF writes the law of case (b) as x = aω sin ωt, which is a typo for x = a sin ωt, since its next line uses a sin. A numeric integral of the speed over the window gives cm for case (a) and cm for case (b), and both equal the distance f·a = cm. A finite-difference slope of x gives cm/s against the formula cm/s. The mean speeds are below the top speed aω = cm/s, as they must be.
Five methods
1. Force method
The force F = −mω²x gives the law x = a cos ωt or a sin ωt, which is what we used. The force alone does not give a time, so this method only supplies the position law.
2. Energy method
Energy gives the speed at each place, v = ω√(a² − x²). But the time needs the integral of dx/v, which is the same arccos or arcsin step as before. Energy alone does not shorten the route.
3. Graph method
On the speed against time graph the area under the curve in the window is the distance a/2. The mean speed is the height of a rectangle with the same area and the same width. The page draws that dashed rectangle for both cases.
4. Calculus method
This is the step-by-step route: solve the position law for the time, then use mean = distance / time. You can also write ⟨v⟩ = (1/t) ∫ |vx| dt and the integral gives a/2.
5. Check method
Units: a/T is cm/s, a speed. Limits: when f is very small, case (b) gives a mean speed that tends to aω, the top speed, because the point barely leaves the middle. Case (a) tends to 0. When f = 1 both cases cover the whole amplitude in T/4 and give 4a/T.
Fastest: the graph method gives the ratio 2 at a glance. Most reliable: the calculus method, followed by the numeric area check.
What each slider changes
| Control | Changes the answer? | What it changes |
|---|---|---|
| Amplitude a | Yes, proportional | Distance f·a, both mean speeds, the track, every graph height |
| Period T | Yes, inverse | Both window times, both mean speeds, the time axis, the motion speed |
| Distance share f | Yes | Distance f·a, both window times, the shaded areas, both mean speeds |
| Screen speed | No | Only how fast the animation runs compared with real time |
| Target mean speed, case (a) (given-answer mode) | Sets the answer | Solves T = 2π f a / (arccos(1 − f) × mean speed) |
| Zoom, Auto-fit | No | Pixels per centimetre in the picture only |
Parameters
| Quantity | Symbol | Value used |
|---|---|---|
| Amplitude | a | cm (book: 10.0) |
| Period | T | s (book: 0.60) |
| Distance share | f | (book: 0.5) |
| Distance covered | f·a | cm |
| Angular frequency | ω | rad/s |
| Window time, case (a) | t1 | s |
| Window time, case (b) | t2 | s |
| Top speed | aω | cm/s |
Exam traps
- Using the mean of the two end speeds. The speed is not a straight line in time, so that average is wrong.
- Using a/2 divided by T/4. The distance a/2 is not covered in a quarter of the period.
- Giving both cases the same time. The same distance takes T/6 from the extreme and T/12 from the middle.
- Choosing the wrong angle, such as 2π/3, when cos ωt = 1/2. The first time uses the smallest angle, π/3.
- Forgetting that the answer is in m/s when a is in cm. 50 cm/s is 0.5 m/s.
- Mixing ω (rad/s) with the frequency f (Hz). Here ω = 2π/T = 10.47 rad/s.