For ω = 4.00 s⁻¹, x0 = 25.0 cm and vx0 = 100 cm/s, the particle is at x = −29 cm with vx = −81 cm/s after 2.40 s. For your sliders the amplitude is cm, and at t* = s the particle is at x = cm with vx = cm/s.
Problem statement
A particle performs harmonic oscillations along the x axis about the equilibrium position x = 0. The oscillation frequency is ω = 4.00 s⁻¹. At a certain moment the particle has the coordinate x0 = 25.0 cm and the velocity vx0 = 100 cm/s. Find the coordinate x and the velocity vx of the particle 2.40 s after that moment.
We assume the oscillation is simple harmonic and undamped. This means a and ω stay fixed and the energy is constant.
We assume the "certain moment" is t = 0. This means the 2.40 s is counted from the moment when x = x0 and vx = vx0.
We assume ω is the angular frequency in rad/s. This means the period is T = 2π/ω.
Symbols: a is the amplitude, α is the starting phase, t* is the probe time (2.40 s in the book), x* and vx* are the position and velocity at t*.
The big idea in plain words
Every harmonic motion is a cosine with two unknown numbers: how big it is (a) and where in the cycle it starts (α). The start position and the start velocity are two facts, so they fix both numbers. Once a and α are known, the position and velocity at any later time come straight from the formulas.
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.
Particle rest frame
This frame does not apply. The particle is always accelerating, so a frame fixed to it is not inertial and gives no simpler way to find x and vx.
The solution step by step
We write the law as x = a cos(ωt + α). Differentiating gives the velocity.
At t = 0 the particle has x0 and vx0, so we put t = 0 in both equations.
To get a, we write cos α = x0/a and sin α = −vx0/(aω), square both and add. Because sin²α + cos²α = 1, the angle disappears.
To get α, we divide the second equation by the first.
The signs of cos α and sin α decide the quadrant. Here x0 is positive and vx0 is positive, so cos α is positive and sin α is negative. That puts α in the fourth quadrant, which is −45 degrees. Now we put in t* = 2.40 s.
Rounded to the book's two figures, x = −29 cm and vx = −81 cm/s. For your sliders: ωt* + α = rad, x = cm, vx = cm/s.
Book answer: x = −29 cm and vx = −81 cm/s. The live result for the book case is the same to the book's rounding. The PDF prints the amplitude formula without the square on x0, which is a typo, since its value 35.35 cm is only right with the square. At the time shown, the check vx² = and ω²(a² − x²) = (cm/s)² agree. The start energy ½vx0² + ½ω²x0² = equals ½ω²a² = (cm²/s²). A finite-difference slope of x gives cm/s against the formula cm/s.
Five methods
1. Force method
The motion follows wx = −ω² x, so the force is F = −mω² x. This is Hooke's law, and it is the reason the cosine law is right. The force reasoning tells us the form of the answer but not the numbers, so it needs the start values to finish.
2. Energy method
The energy per unit mass is E = ½vx² + ½ω²x², which stays fixed. At the start E = ½(100)² + ½(4.00)²(25.0)² = 10 000 cm²/s². At the turning points E = ½ω²a², so a² = 2E/ω² = 1250 cm² and a = 35.36 cm. It gives a quickly, but it cannot give the phase.
3. Graph method
On the (x, vx) ellipse the start is a point and the motion goes clockwise round the ellipse at the rate ω. After 2.40 s the phase has advanced by ωt* = 9.6 rad, which is 1.53 turns. Walk that far round the ellipse from the start point and read x and vx.
4. Calculus method
This is the step-by-step route: differentiate, use t = 0, find a and α, then evaluate at t*.
5. Check method
Units: x0 is in cm and vx0/ω is in cm, so they add. The amplitude must be at least |x0|, and here 35.36 is larger than 25.0, as it must be. The energy at t* must equal the start energy. Here ½vx² + ½ω²x² = ½(81.04)² + ½(16)(28.98)² = 3284 + 6716 = 10 000 cm²/s².
Fastest: the energy method for a, then the phase by calculus. Most reliable: the calculus method, with the energy check at the end.
What each slider changes
| Control | Changes the answer? | What it changes |
|---|---|---|
| Angular frequency ω | Yes | a, T, α, the phase at t*, x*, vx*, every graph |
| Start position x0 | Yes | a, α, x*, vx*, the energies and the ellipse |
| Start velocity vx0 | Yes | a, α, x*, vx*, the energies and the ellipse |
| Probe time t* | Yes, x* and vx* only | The probe marker on every graph, x*, vx*, wx*. The motion itself is the same. |
| Screen speed | No | Only how fast the animation runs compared with real time |
| Target amplitude (given-answer mode) | Sets the answer | Solves vx0 = ω √(a² − x0²) |
| Zoom, Auto-fit | No | Pixels per centimetre in the picture only |
Parameters
| Quantity | Symbol | Value used |
|---|---|---|
| Angular frequency | ω | rad/s (book: 4.00) |
| Period | T | s |
| Start position | x0 | cm (book: 25.0) |
| Start velocity | vx0 | cm/s (book: 100) |
| Probe time | t* | s (book: 2.40) |
| Amplitude | a | cm |
| Top speed | aω | cm/s |
Exam traps
- Taking a = x0. The particle is moving at t = 0, so it has not yet reached the turning point and a is larger than x0.
- Forgetting the square when finding a. The correct form is a = √(x0² + (vx0/ω)²).
- Using the calculator in degrees. The phase ωt* + α must be in radians, since ω is in rad/s.
- Taking α = +π/4 instead of −π/4. With vx0 positive the particle is moving away from the middle, so sin α is negative.
- Forgetting that ω is multiplied by t. Here ωt* = 9.6 rad is more than one full turn, so the phase wraps round.
- Mixing ω (rad/s) with the frequency f (Hz). T = 2π/ω = 1.57 s here.