vx = −aω sin(ωt − π/4) and wx = −aω² cos(ωt − π/4). Against x: vx² = ω²(a² − x²), an ellipse, and wx = −ω² x, a straight line. For your sliders ω = rad/s, the top speed is cm/s and the top acceleration is cm/s².
Problem statement
A point oscillates along the x axis according to the law x = a cos(ωt − π/4). Draw the approximate plots (a) of the displacement x, the velocity projection vx and the acceleration projection wx against time t; (b) of vx and wx against the coordinate x.
We assume the law holds for all time. This means the amplitude a and the angular frequency ω never change, so there is no damping.
We assume the point moves only along the x axis. This means velocity and acceleration have only an x part, and we call them vx and wx.
We assume ω is positive, with ω = 2π / T, where T is the period (the time for one full oscillation).
Symbols: a is the amplitude (the largest distance from x = 0), ω is the angular frequency in radians per second, t is time, and δ = π/4 (45 degrees) is the phase shift. The slider lets you change δ to see its effect.
The big idea in plain words
Velocity tells how fast the position changes. It is the slope of the x graph. Acceleration tells how fast the velocity changes. It is the slope of the velocity graph. The slope of a cosine is a sine, and the slope of that is a cosine again. So all three graphs are the same wave, scaled and shifted along the time axis. For part (b) we remove the time between the equations, and a simple link between vx, wx and x appears.
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 is always accelerating, so a frame riding on it is not an inertial frame, and it would not help to plot x, vx and wx.
The solution step by step
First we find the velocity, because velocity is the rate of change of position. We use the rule that the derivative of cos(u) with respect to t is −sin(u) times the derivative of u. Here u = ωt − π/4, and the derivative of u is ω.
Next we find the acceleration in the same way, by differentiating the velocity. The derivative of sin(u) is cos(u) times the derivative of u.
This answers part (a). We now read the three graphs. All three are waves with the same period T. The velocity wave is the x wave moved a quarter period earlier, because −sin(u) = cos(u + π/2). The acceleration wave is the x wave turned upside down, because wx = −ω² x. The x wave has its first top at ωt = π/4, that is at t = T/8, so the whole pattern sits one eighth of a period later than a plain cos(ωt) graph.
For part (b) we remove t. We want vx in terms of x, because the problem asks for vx against x. We start from the two equations x = a cos(u) and vx = −aω sin(u), then use the fact that sin²(u) + cos²(u) = 1.
This is the equation of an ellipse with half-widths a (along x) and aω (along vx). The point goes round it clockwise, because when x is at its top (vx = 0) the speed then turns negative. The ellipse does not contain δ, because δ was inside u and u disappeared. So the phase shift π/4 changes only where the point starts, not the shape.
The acceleration against x needs no extra work, because we already have it.
This is a straight line through the origin, with a negative slope, running from wx = +aω² at x = −a to wx = −aω² at x = +a.
Book answer: vx = −aω sin(ωt − π/4), wx = −aω² cos(ωt − π/4), vx² = ω²(a² − x²) and wx = −ω² x. For your sliders the live check gives vx² on the left of (cm/s)² and ω²(a² − x²) on the right of (cm/s)², at the time shown. They are equal to rounding. A finite-difference slope of x gives cm/s against the formula cm/s.
Five methods
1. Force method
The result wx = −ω² x says the push on the point is proportional to its distance from x = 0 and points back towards it. That is a spring-like force. With Newton's second law F = m wx, the force is F = −(mω²) x, which is Hooke's law with spring constant k = mω². So the law x = a cos(ωt − π/4) is what a mass on a spring does, and the force reasoning gives the same line for wx against x.
2. Energy method
This gives part (b) directly. For a spring-like force the total energy is fixed: (1/2) m vx² + (1/2) k x² = (1/2) k a², because the speed is zero at x = a. With k = mω²:
3. Graph method
The first row of graphs shows x, vx and wx against t. The slope of the x curve at any time is vx, and the slope of the vx curve is wx, so you can read one from the other. The second row shows vx against x as an ellipse and wx against x as a line. The normalised graph at the right of the second row puts all three waves together: vx runs a quarter period ahead of x, and wx is half a period away from x.
4. Calculus method
This is the method used in the step-by-step section: differentiate twice for part (a), then remove t using sin² + cos² = 1 for part (b).
5. Check method
Units: aω has units of cm × s⁻¹ = cm/s, a speed. aω² is cm/s², an acceleration. Limits: at x = ±a the formula gives vx = 0, which is right for a turning point. At x = 0 it gives |vx| = aω, the top speed. The formula wx = −ω² x gives wx = ∓aω² at x = ±a, the top acceleration, as the graph shows. If ω tends to 0 the point stops moving, and every result shrinks to 0.
Fastest: the energy method for part (b), the calculus method for part (a). Most reliable: the calculus method, because every step is written out.
What each slider changes
| Control | Changes the answer? | What it changes |
|---|---|---|
| Amplitude a | Yes, proportional | x, vx, wx sizes, aω, aω², the track and ellipse sizes, every graph height |
| Period T | Yes, inverse | ω, f, aω (proportional to 1/T), aω² (proportional to 1/T²), the time axis (two periods), the motion speed |
| Phase shift δ | Partly | Start values x0 and vx0, the left or right shift of the three time curves, the time of the first maximum. It does not change the ellipse or the line, because t is removed. |
| Screen speed | No | Only how fast the animation runs compared with real time. The physics numbers stay the same. |
| Target max speed (given-answer mode) | Sets the answer | Solves a = vmax / ω |
| Zoom, Auto-fit | No | Pixels per centimetre in the picture only |
Parameters
| Quantity | Symbol | Value used |
|---|---|---|
| Amplitude | a | cm |
| Period | T | s |
| Angular frequency | ω | rad/s |
| Phase shift | δ | degrees (book: 45) |
| Top speed | aω | cm/s |
| Top acceleration | aω² | cm/s² |
Exam traps
- Writing vx = −aω cos(ωt − π/4). The derivative of a cosine is a sine, so the right form has sin.
- Forgetting the factor ω when differentiating. Each derivative multiplies by ω.
- Thinking the π/4 changes the shape of the ellipse or the line in part (b). It only moves the start point and shifts the time graphs.
- Drawing the ellipse with the wrong direction. It runs clockwise when vx is drawn upward and x to the right.
- Drawing the wx graph in step with x. It is exactly opposite to x, because wx = −ω² x.
- Mixing the angular frequency ω (rad/s) with the frequency f (Hz). ω = 2π f.