Post C — Tracking Motion Through a Full Cycle: Position, Velocity, Acceleration, and Net Force

Tracking Motion Through a Full Cycle: Position, Velocity, Acceleration, and Net Force

Welcome back! I'm Ahaan Thota, your student teacher. This is the third post in our Simple Harmonic Motion series!

In Post A, we defined Simple Harmonic Motion (SHM) and the key quantities used to describe it: amplitude, period, and frequency. In Post B, we explained why SHM happens: a restoring force always points back toward equilibrium, and in true SHM that restoring force is proportional to displacement.

In this post, we focus on what the motion looks like throughout one full oscillation. Specifically, we will track how position, velocity, acceleration, and net force change and how their directions depend on where the object is in the cycle. This is where SHM becomes predictable: if you know the position, you can figure out the direction (and often the relative magnitude) of velocity, acceleration, and net force.

1) Position in SHM (Mass–Spring and Pendulum)

We define displacement x as the signed distance from equilibrium. For a horizontal spring, choose the right direction as positive. For a pendulum, displacement is often described by angle θ from equilibrium, but for small angles we can treat motion as approximately linear.

  • Equilibrium: x = 0 (or θ = 0)
  • Right of equilibrium: x > 0
  • Left of equilibrium: x < 0
  • Amplitude: maximum displacement, x = ±A

In ideal SHM, position varies smoothly and periodically with time. A standard mathematical model is:

x(t) = A cos(ωt + φ)

where:

  • A is amplitude (m)
  • ω is angular frequency (rad/s)
  • t is time (s)
  • φ is phase constant (rad), which depends on where the object starts

This equation matters because it encodes the repeating nature of SHM. The object does not move randomly — it follows a predictable cycle that repeats every period T.

Period, Frequency, and Angular Frequency

In Post A, we discussed period and frequency. Here are the mathematical relationships that connect them:

  • Frequency: f = 1/T
  • Angular frequency: ω = 2πf = 2π/T

The value of ω tells you how quickly the system cycles through its motion, measured in radians per second.

2) Velocity in SHM: Direction at Multiple Positions

Velocity is the rate of change of position. For SHM modeled by x(t) = A cos(ωt + φ), the velocity function is:

v(t) = dx/dt = -Aω sin(ωt + φ)

You do not need to memorize calculus to use this: the important idea is that velocity is also sinusoidal, but it is shifted in time relative to position. This creates the famous phase relationship between x and v.

Key velocity facts (direction + magnitude)

  • At maximum displacement (x = ±A): v = 0 (turning points)
  • At equilibrium (x = 0): |v| is maximum (speed is greatest)
  • Between equilibrium and endpoints: v has some intermediate value

A very useful relationship that connects velocity and position (without needing time) is:

v² = ω²(A² − x²)

This equation shows:

  • When x = 0, v² = ω²A² → speed is maximum
  • When x = ±A, v² = 0 → speed is zero
  • The farther from equilibrium, the smaller the speed becomes

3) Acceleration in SHM: Direction at Multiple Positions

Acceleration is the rate of change of velocity. Using the SHM velocity function, v(t) = -Aω sin(ωt + φ), the acceleration function is:

a(t) = dv/dt = -Aω² cos(ωt + φ)

Notice that cos(ωt + φ) appears again, which means acceleration is directly related to position. Since x(t) = A cos(ωt + φ), we can substitute and get the most important SHM equation:

a = -ω²x

This equation exactly matches the conceptual rule from Post B: acceleration always points toward equilibrium. The negative sign means the acceleration points in the opposite direction of displacement.

Key acceleration facts (direction + magnitude)

  • At equilibrium (x = 0): a = 0
  • At endpoints (x = ±A): |a| is maximum (largest restoring effect)
  • In between: acceleration has intermediate magnitude, always toward equilibrium

The maximum acceleration magnitude is: amax = ω²A.

4) Net Force in SHM: Direction at Multiple Positions

Net force follows directly from Newton's Second Law:

Fnet = ma

Since a = -ω²x, we get:

Fnet = -mω²x

This means:

  • Net force is zero at equilibrium
  • Net force is maximum at the endpoints
  • Net force always points toward equilibrium

5) Spring System Application (Connecting to Hooke's Law)

For a mass–spring system, the restoring force is given by Hooke's Law:

F = -kx

But we also found that for SHM:

Fnet = -mω²x

Since both expressions describe the same restoring force, we can match them:

-kx = -mω²x

Canceling x (as long as x ≠ 0 during motion), we get:

ω = √(k/m)

This leads to the famous period formula for a mass–spring oscillator:

T = 2π √(m/k)

Interpretation:

  • More mass (m larger) → slower oscillation → larger period
  • Stiffer spring (k larger) → faster oscillation → smaller period

6) Pendulum Application (Small-Angle Approximation)

For a simple pendulum of length L, the restoring influence comes from gravity's tangential component. For small angles, sin(θ) ≈ θ (in radians), which makes the restoring effect approximately linear.

The small-angle SHM model gives:

ω = √(g/L)

and therefore:

T = 2π √(L/g)

Interpretation:

  • Longer pendulum (L larger) → slower oscillation → larger period
  • Greater gravity (g larger) → faster oscillation → smaller period

Note: This formula is most accurate for small angles. At larger angles, the motion is still periodic, but it is not perfectly SHM and the period changes slightly.

7) Phase Relationships Between Position and Velocity

The math confirms the conceptual "out of sync" behavior:

  • x(t) is cosine-based
  • v(t) is sine-based and has a negative sign

This means velocity reaches its maximum when position is zero, and velocity is zero when position is at its maximum. In phase terms, velocity is shifted by one-quarter cycle (90° or π/2 radians) relative to position.

The acceleration is proportional to -x, so acceleration is exactly opposite position. In phase terms, acceleration is 180° (π radians) out of phase with position. Below are three graphs that depict the general functions that correspond to each of the functions x(t), v(t), and a(t). Keep in mind that there are other elements that distort the graphs further, but they generally follow these basic structures.

8) Motion Snapshot Summary (Directions at Key Positions)

Assume the object is moving from the right side toward equilibrium:

  • At x = +A: v = 0, a points left, Fnet points left
  • At x = 0: |v| max, a = 0, Fnet = 0
  • At x = -A: v = 0, a points right, Fnet points right

The key pattern is consistent: velocity tells you where the object is going, but acceleration and net force tell you where the object is being pulled (toward equilibrium).

Conclusion

Post C shows that SHM is mathematically predictable. Position varies sinusoidally, velocity and acceleration follow from that motion, and the defining feature appears clearly in the equation a = -ω²x. This matches the physics: the restoring force pulls toward equilibrium, creating acceleration that continuously reverses the motion.

Next, in Post D, we will track SHM using energy: how kinetic and potential energy trade places throughout the cycle, and how amplitude connects directly to the total mechanical energy of the system.


Practice Problems (Post C)

Answer these without guessing. Your goal is to explain your reasoning using the SHM rules from this post.

Multiple Choice

Question 1

A spring-mass system is at x = +A. Which is correct?

  • A. v is maximum right; a is maximum right
  • B. v is 0; a points left
  • C. v is 0; a is 0
  • D. v is maximum left; a points left

Question 2

At equilibrium in SHM, which statement is true?

  • A. v = 0 and a = 0
  • B. |v| is maximum and a = 0
  • C. |v| is maximum and |a| is maximum
  • D. v = 0 and |a| is maximum

Question 3

Which equation best represents the defining relationship for SHM?

  • A. a = ω²x
  • B. a = -ω²x
  • C. v = ωx
  • D. F = mv

Question 4

For a mass-spring system, the angular frequency is:

  • A. ω = √(m/k)
  • B. ω = √(k/m)
  • C. ω = 2πT
  • D. ω = T/2π

Question 5

For a small-angle pendulum, the period is:

  • A. T = 2π √(g/L)
  • B. T = 2π √(L/g)
  • C. T = 2π √(k/m)
  • D. T = √(L/g)

Answer Key (don't cheat!!!!)

Question Answer
1 B
2 B
3 B
4 B
5 B

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