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How do the sinusoidal expressions for position, velocity and acceleration describe an oscillator, and how do we extract amplitude, phase and the maxima from them?

Topic 7.3 Representing and Analyzing SHM: write the sinusoidal position, velocity and acceleration of an oscillator, relate their amplitudes and phases, and read the motion from graphs and initial conditions.

A focused answer to AP Physics C: Mechanics Topic 7.3, covering the sinusoidal expressions for position, velocity and acceleration of an oscillator, the relationships among their amplitudes (vmax and amax), the phase relationships, reading amplitude and phase from initial conditions, and interpreting SHM graphs, with calculus-based worked examples.

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  1. What this topic is asking
  2. The three sinusoidal functions
  3. Maximum speed and acceleration
  4. Phase relationships
  5. Reading amplitude and phase from initial conditions
  6. Try this

What this topic is asking

The College Board (Topic 7.3) wants you to write the sinusoidal position, velocity and acceleration of an oscillator, to relate their amplitudes and phases, and to read the motion from graphs and initial conditions. This is the analysis side of SHM: given x(t)=Acos(ωt+ϕ)x(t) = A\cos(\omega t + \phi), differentiate to get vv and aa, and interpret where the speed and acceleration are largest.

The three sinusoidal functions

All three quantities oscillate sinusoidally at the same angular frequency ω\omega, but with different amplitudes and phases. The position has amplitude AA; the velocity has amplitude AωA\omega; the acceleration has amplitude Aω2A\omega^2. Each factor of ω\omega comes from a differentiation. This is the calculus heart of analyzing an oscillator: you are handed one of the three functions and asked to produce the others by differentiating (or integrating).

Maximum speed and acceleration

From the amplitudes of the three functions, the maximum speed and maximum acceleration are

vmax=Aω,amax=Aω2.v_{max} = A\omega, \qquad a_{max} = A\omega^2.

These occur at different points in the cycle. The speed is greatest at the equilibrium position (x=0x = 0), where all the energy is kinetic; the acceleration is greatest at the extremes (x=±Ax = \pm A), where the restoring force (and hence a=ω2xa = -\omega^2 x) is largest. So when the oscillator races through the center it has zero acceleration, and when it pauses at a turning point it has maximum acceleration. Reading off these maxima from the amplitude and frequency is a standard exam task.

Phase relationships

The three functions are shifted in phase. Because the velocity is the sine (the derivative of the cosine), it leads the position by a quarter cycle (π/2\pi/2): the velocity peaks when the position is zero. The acceleration is ω2x-\omega^2 x, exactly out of phase (π\pi) with the position: it is most negative when the displacement is most positive. Visualizing these phase relationships, position and acceleration mirror-imaged, velocity a quarter-turn ahead, lets you sketch all three graphs from any one of them.

Reading amplitude and phase from initial conditions

The constants AA and ϕ\phi are not part of the physics of the oscillator; they encode how it was started. If the oscillator is released from rest at maximum displacement x0x_0, then A=x0A = x_0 and ϕ=0\phi = 0 (pure cosine). If it is launched from equilibrium with speed v0v_0, then ϕ=±π/2\phi = \pm\pi/2 (a sine) and A=v0/ωA = v_0/\omega. In general you evaluate x(0)x(0) and v(0)v(0) from the expressions and solve the two equations for AA and ϕ\phi. The amplitude can also be found from A=x02+(v0/ω)2A = \sqrt{x_0^2 + (v_0/\omega)^2}, which combines the initial position and velocity.

Try this

Q1. An oscillator has amplitude 0.150.15 m and angular frequency 8.08.0 rad/s. Calculate its maximum speed and maximum acceleration. [2 points]

  • Cue. vmax=Aω=1.2v_{max} = A\omega = 1.2 m/s; amax=Aω2=9.6a_{max} = A\omega^2 = 9.6 m/s squared.

Q2. State where in its cycle an SHM oscillator has zero velocity and maximum acceleration. [2 points]

  • Cue. At the extremes of the motion (x=±Ax = \pm A), where the displacement (and restoring force) is largest.

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2024 (style)6 marksSection II (FRQ, calculus). An oscillator has position x(t)=0.10cos(4.0t)x(t) = 0.10\cos(4.0t) (SI units). (a) Derive expressions for the velocity and acceleration as functions of time. (b) Determine the maximum speed and maximum acceleration. (c) State the position when the speed is maximum and when the acceleration is maximum. (d) Determine the first time after t=0t = 0 at which the kinetic energy equals the potential energy.
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A 6-point calculus FRQ analyzing an oscillator.

(a) Velocity and acceleration (2 points): v=dxdt=0.40sin(4.0t)v = \dfrac{dx}{dt} = -0.40\sin(4.0t); a=dvdt=1.6cos(4.0t)a = \dfrac{dv}{dt} = -1.6\cos(4.0t).
(b) Maxima (2 points): vmax=Aω=(0.10)(4.0)=0.40v_{max} = A\omega = (0.10)(4.0) = 0.40 m/s; amax=Aω2=(0.10)(4.0)2=1.6a_{max} = A\omega^2 = (0.10)(4.0)^2 = 1.6 m/s squared.
(c) Positions (1 point): speed is maximum at x=0x = 0 (equilibrium); acceleration is maximum at x=±Ax = \pm A (the extremes).
(d) Equal energies (1 point): K=UK = U when x=±A/2x = \pm A/\sqrt{2}, i.e. cos(4.0t)=1/2\cos(4.0t) = 1/\sqrt{2}, so 4.0t=π/44.0t = \pi/4, giving t=π/160.20t = \pi/16 \approx 0.20 s.

Markers reward differentiating to get vv and aa, the vmax=Aωv_{max} = A\omega and amax=Aω2a_{max} = A\omega^2 relations, and the phase positions of the maxima.

AP 2021 (style)1 marksSection I (multiple choice). For an object in simple harmonic motion, where is its speed greatest and its acceleration zero? (A) at the extremes of the motion (B) at the equilibrium position (C) halfway between (D) nowhere; they never coincide. Justify your reasoning.
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A 1-point conceptual MCQ. The answer is (B).

In SHM the acceleration is a=ω2xa = -\omega^2 x, so it is zero exactly where x=0x = 0 (equilibrium). At equilibrium all the energy is kinetic, so the speed is maximum there. At the extremes (x=±Ax = \pm A) the speed is zero and the acceleration is maximum. So the speed peaks and the acceleration vanishes at the same place, equilibrium. The trap is to associate maximum speed with the extremes.

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