Oscillations · Lesson 3 of 8
Velocity, Acceleration and the Restoring Force
“Derive velocity and acceleration, interpret their graphs and signs, and test whether a force produces SHM.”
• Differentiate the SHM position function with the chain rule and interpret the derivatives physically. • Obtain velocity and acceleration from reference-circle projections. • Use v² = ω²(A² − x²), with the direction supplied separately. • Relate the signs of position, velocity, acceleration and force at specified points. • Apply the constant-coefficient criterion a = −ω²x and distinguish damping and nonlinear forces.
Velocity is a rate of change, not a second position
Velocity measures how quickly position changes: over a short interval Δt, the average velocity is Δx/Δt. The derivative dx/dt is the limit as that interval shrinks to zero, giving the slope of the tangent to the position–time graph. Positive slope means motion along +x; negative slope means motion along −x. Acceleration is the corresponding rate of change of velocity, dv/dt. It can be nonzero at an instant when velocity is zero, as when an oscillator reverses at an extreme.
Write θ = ωt + φ. The derivative of cos θ with respect to θ is −sin θ, while dθ/dt = ω because ω and φ are constants. The chain rule multiplies those rates: d[cos θ]/dt = −sin θ × ω. Differentiating once more uses d[sin θ]/dt = ω cos θ. The extra factor ω at each step is essential; omitting it gives both wrong values and wrong dimensions.
The reference particle’s speed is ωA and its velocity is tangent to the circle. Projecting that vector onto x gives −ωA sin θ. Its centripetal acceleration has constant magnitude (ωA)²/A = ω²A and points toward the circle’s centre; its x-component is −ω²A cos θ. Thus geometry agrees with differentiation. The actual linear particle has a varying speed and acceleration magnitude even though these two reference-vector magnitudes are constant.
Speed depends on distance from equilibrium
The identities x/A = cos θ and v/(ωA) = −sin θ allow time to be eliminated. Square both and add; sin²θ + cos²θ = 1 gives x²/A² + v²/(ω²A²) = 1. Rearranging yields v² = ω²(A² − x²). Taking a square root gives speed, not the direction. At the same x the particle can be moving either way, so a signed velocity needs the current phase or direction of travel.
At x > 0 acceleration is negative; at x < 0 it is positive. Acceleration always points toward equilibrium, but velocity may point toward or away from equilibrium. On the outward part of a trip they oppose each other and speed decreases. On the inward part they point the same way and speed increases. At equilibrium acceleration and restoring force vanish, yet velocity has its greatest magnitude. At an extreme the velocity vanishes for an instant, yet acceleration and force have their greatest magnitudes.
For φ = 0 the displacement is cos(ωt), the normalized velocity is −sin(ωt) = cos(ωt + π/2), and the normalized acceleration is −cos(ωt) = cos(ωt + π). Velocity is a quarter-cycle out of phase with displacement, while acceleration is half a cycle out of phase. A negative phase or time shift can describe the same relationship, so inspect the written functions rather than memorising a sign for “phase difference”. The slope of the x–t graph gives v; the slope of the v–t graph gives a. Their different vertical units mean their raw heights cannot be compared as if they measured the same physical quantity.
Problem
In SI units x = 5 cos(2πt + π/4). At t = 1.5 s, find displacement, velocity, speed and acceleration.
- 1.Identify A = 5 m and ω = 2π rad s⁻¹. The period is 1 s. The phase at 1.5 s is 3π + π/4 = 13π/4, equivalent to 5π/4 modulo 2π.
- 2.cos(5π/4) = −1/√2, so x = −5/√2 = −3.53553 m.
- 3.sin(5π/4) = −1/√2, so v = −(2π)(5)(−1/√2) = +22.2144 m s⁻¹. Speed is its magnitude, 22.2144 m s⁻¹, approximately 22 m s⁻¹ to two significant figures.
- 4.a = −(2π)²x = +139.577 m s⁻², approximately 140 m s⁻². Its positive sign is correct because x is negative: the body is accelerating toward the centre.
- 5.Retain unrounded values until the last step. Speed cannot be negative, even though a signed velocity can be.
The force criterion is linear and restoring
Newton’s second law connects the motion to its cause: Fnet = ma = −mω²x. Define k = mω², a positive constant with unit N m⁻¹. The result is Fnet = −kx. The magnitude is proportional to equilibrium displacement and the direction is opposite to it. Both parts are essential. A force that merely points inward is not enough; if it varies as x³, for example, the motion is generally not sinusoidal and its period can depend on amplitude.
Starting with a sinusoid and differentiating twice produces the force law. Conversely, the differential equation x″ + (k/m)x = 0 has sine and cosine solutions with ω = √(k/m). Their arbitrary coefficients encode the two initial conditions. This converse is the standard solution of a second-order differential equation, not an instruction to integrate −kx as if x were a constant. Substituting A cos(ωt + φ) back into the equation checks it directly without requiring a full differential-equations course.
| Acceleration law | SHM about x = 0? | Reason |
|---|---|---|
| a = 0.7x | No | Positive coefficient drives the particle away from equilibrium |
| a = −200x² | No | Not linear; negative even for negative x, so not restoring on both sides |
| a = −10x | Yes | Constant negative coefficient; ω = √10 rad s⁻¹ |
| a = 100x³ | No | Nonlinear and points away from x = 0 |
In a real system the force can include higher powers of displacement. Near a stable equilibrium the linear term may dominate, allowing an approximate SHM description over a limited amplitude range. A bowl and a magnet near stable alignment are examples of this small-displacement idea. “Linear oscillator” describes a force linear in displacement, not a claim that the displacement–time graph is a straight line. A constant offset force can shift the equilibrium and must be balanced before applying the criterion.
Check signs using the equilibrium midpoint
Take endpoints A and B separated by 10 cm, choose the midpoint as x = 0 and the direction A to B as positive. Then A is x = −5 cm and B is x = +5 cm. The acceleration and force signs follow x alone; the velocity sign follows the stated direction. At the endpoints v is zero. This avoids the common error of using the distance from A as though it were displacement from equilibrium.
| State | x / cm | Sign of v | Sign of a | Sign of F |
|---|---|---|---|---|
| At A | −5 | 0 | + | + |
| At B | +5 | 0 | − | − |
| Midpoint moving toward A | 0 | − | 0 | 0 |
| 2 cm from B moving toward A | +3 | − | − | − |
| 3 cm from A moving toward B | −2 | + | + | + |
| 4 cm from B moving toward A | +1 | − | − | − |
For exact SHM the amplitude is constant. A viscous resistance proportional to velocity adds a term −bv, with b in kg s⁻¹, so the net force is no longer just −kx. For an underdamped oscillator, the amplitude envelope has the standard form A₀ exp[−bt/(2m)]; this stated result identifies the damping timescale 2m/b, rather than supplying a full damping derivation. Over an interval Δt much smaller than 2m/b the envelope changes little, so the motion may be approximated locally by a constant-amplitude sinusoid. Do not label the complete decaying motion as exact SHM. A weakly damped system also has a frequency close to its undamped natural frequency.
Quick Quiz
Quiz
At the rightmost extreme of ideal SHM, what are v and a?
The particle is instantaneously at rest, but a = −ω²A is directed toward equilibrium.
Which acceleration law is SHM about x = 0?
It has the required constant negative coefficient multiplying x.
If x = A cos(ωt), which is the velocity?
Differentiating cosine gives minus sine and the chain rule supplies ω.
At x = A/2, how large is the speed?
|v| = ω√(A² − A²/4) = √3 ωA/2.
At x < 0 while moving toward equilibrium, which signs apply?
Moving right means v > 0, and a = −ω²x is also positive.
Practice Problems
- Starting from x = A cos(ωt + φ), derive velocity and acceleration, explaining every chain-rule factor. Check the SI units at each step.
- For x = 0.08 cos(3t + π/6) m, find x, v and a at t = 0. State their directions and identify the maximum speed and acceleration magnitude.
- A body moves between A and B, 10 cm apart, with A to B positive. Find the signs of velocity, acceleration and restoring force at A; at B; at the centre moving toward A; 2 cm from B moving toward A; 3 cm from A moving toward B; and 4 cm from B moving toward A.
- Classify a = 0.7x, a = −200x², a = −10x and a = 100x³ for motion about x = 0. Assume consistent SI coefficients and explain both proportionality and direction.
- An oscillator has A = 0.05 m and ω = 12 rad s⁻¹. Find its speed and acceleration at x = −0.03 m. Give the two possible signed velocities and say what additional information would choose one.
- Sketch normalized x, v and a over one cycle for φ = 0. Mark T/4 intervals and use graph slopes to check the velocity and acceleration signs.
- A damped oscillator has m = 0.5 kg and b = 0.02 kg s⁻¹. Use the stated envelope to find the amplitude timescale. Explain whether a 0.1 s observation can reasonably treat its envelope as constant, and why the complete motion is not exact SHM.
Key Takeaways
• Differentiation gives v = −ωA sin(ωt + φ) and a = −ω²x; each chain-rule factor matters. • Acceleration and restoring force oppose displacement, while velocity depends on direction of travel. • Speed is greatest at equilibrium and zero at the extremes; acceleration magnitude behaves oppositely. • The SHM test requires a constant linear restoring coefficient, not merely a force toward the centre. • Damping and nonlinear terms must be accounted for before claiming exact SHM.