The amplitude is a/2 and the period is T = π/ω. The speed obeys vx² = 4ω² x (a − x). For your sliders a = cm, so the amplitude is cm, ω = rad/s, T = s and the top speed is cm/s.
Problem statement
A point moves along the x axis according to the law x = a sin²(ωt − π/4). Find (a) the amplitude and period of the oscillations, and draw the plot x(t); (b) the velocity projection vx as a function of the coordinate x.
We assume the law holds for all time. This means a and ω never change, so there is no damping.
We assume the point moves only along the x axis. This means velocity has only an x part, called vx.
We assume a and ω are positive. This means x never goes below 0, because a square is never negative.
Symbols: a is the constant in the law, ω is a rate in radians per second, and δ = π/4 (45 degrees) is the phase shift. The slider lets you change δ.
The big idea in plain words
A sine squared is never negative, so the point never goes to the left of x = 0. It moves between 0 and a, which means it swings about the middle point x = a/2. A square also doubles the rate: sin² repeats twice as fast as sin. So the period is half of the plain sine period, and the swing is a/2 on each side of the middle.
The solution in each frame
Lab frame
The axis is fixed to the lab and the point moves along it. The full solution is in the next section.
Point rest frame
This frame does not apply. The point keeps accelerating, so a frame riding on it is not inertial, and it would not help to find the amplitude or the speed.
The solution step by step
Part (a). We use the identity sin²θ = (1 − cos 2θ) / 2, with θ = ωt − π/4. It comes from cos 2θ = 1 − 2 sin²θ.
This matches the form x − x_mid = A sin(Ωt + φ) of a simple oscillation. So the middle is a/2, the amplitude is A = a/2 and the rate is Ω = 2ω. The period is the time for the angle 2ωt to grow by 2π.
The plot x(t) is a wave between x = 0 and x = a. At t = 0 the point is at x = a sin²(π/4) = a/2 and it moves toward x = 0, since vx(0) = −aω.
Part (b). We differentiate x = (a/2)(1 − sin 2ωt). The constant a/2 has zero slope, and the derivative of sin 2ωt is 2ω cos 2ωt.
Now we write sin 2ωt with x. From x = (a/2)(1 − sin 2ωt) we get sin 2ωt = 1 − 2x/a. Put this in.
This is an ellipse centred at x = a/2, with half-width a/2 along x and half-height aω along vx. The speed is zero at x = 0 and x = a, and largest at x = a/2, where |vx| = aω = cm/s.
Book answer: amplitude a/2, period π/ω, and vx² = 4ω² x (a − x). The PDF prints the last result without the square, a typo, since its own line before it has vx². For your sliders the live check gives vx² on the left of (cm/s)² and 4ω² x (a − x) on the right of (cm/s)². They are equal to rounding. A finite-difference slope of x gives cm/s against the formula cm/s.
Five methods
1. Force method
Differentiating twice gives wx = −4ω²(x − a/2). The push is proportional to the distance from the middle point x = a/2 and points back to it. That is Hooke's law with k = 4mω², and the spring rest point is a/2. The period is then 2π√(m/k) = π/ω, the same as before.
2. Energy method
For a spring-like force, (1/2) m vx² + (1/2) k (x − a/2)² = (1/2) k (a/2)². With k = 4mω² this gives vx² = 4ω² [(a/2)² − (x − a/2)²] = 4ω² x (a − x). It is the fastest route to part (b).
3. Graph method
The x(t) graph is a wave between 0 and a, so the amplitude is half the gap, a/2. Measure the time between two tops to read the period. The ellipse in the second row is centred at x = a/2, so the speed is zero at both ends.
4. Calculus method
This is the step-by-step route: half-angle identity, one derivative, then remove t.
5. Check method
Units: aω is cm/s, a speed. Limits: at x = 0 and x = a the formula gives vx = 0, right for turning points. At x = a/2 it gives |vx| = aω, the top speed. If ω tends to 0 the period grows without limit and the point stops, as the formula says.
Fastest: the energy method for part (b). Most reliable: the calculus method, where every step is written out.
What each slider changes
| Control | Changes the answer? | What it changes |
|---|---|---|
| Constant a | Yes, proportional | Amplitude a/2, the span of x, vx, wx, aω, the track, the ellipse and the line, every graph height |
| Period T = π/ω | Yes, inverse | ω, f, aω, aω², the time axis (two periods), the motion speed |
| Phase shift δ | Partly | Start values x0 and vx0, the shift of the time curves, the time of the first x = 0. It does not change the ellipse or the line, because t is removed. Amplitude and period do not depend on it. |
| Screen speed | No | Only how fast the animation runs compared with real time |
| Target top speed (given-answer mode) | Sets the answer | Solves a = (aω) / ω |
| Zoom, Auto-fit | No | Pixels per centimetre in the picture only |
Parameters
| Quantity | Symbol | Value used |
|---|---|---|
| Constant in the law | a | cm |
| Amplitude | a/2 | cm |
| Period | T = π/ω | s |
| Angular frequency | ω | rad/s |
| Phase shift | δ | degrees (book: 45) |
| Top speed | aω | cm/s |
| Top acceleration | 2aω² | cm/s² |
Exam traps
- Giving the amplitude as a. The point only moves between 0 and a, so the amplitude is a/2.
- Giving the period as 2π/ω. Squaring the sine doubles the rate, so T = π/ω.
- Taking the middle of the motion to be x = 0. It is x = a/2.
- Writing vx² = ω²(a² − x²), the result for a plain cosine. Here the right form is 4ω² x (a − x).
- Thinking the phase π/4 changes the amplitude or period. It only shifts the curve.
- Mixing ω (rad/s) with the frequency f (Hz). Here ω = π/T and f = 1/T.