FULL REVIEW

Full Review: Simple Harmonic Motion — Calculus-Based

Review the essential ideas, relationships, and problem-solving tools for Simple Harmonic Motion.

TIME

45–60 minutes

BEST FOR

A complete topic review

FINISH WITH

A readiness check

After this full review, you'll be able to...

recall the essential ideas, apply them to representative problems, and determine what to study next.

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Topic Alignment

This bundle is aligned to the approved Physics Sensei topic specification below. Use it to recover the topic structure, reinforce key decisions, and confirm readiness for the next study task.

RESOURCE: Independent Physics Sensei Unit Review

UNIT: Mechanics • Unit MEC-U09

TOPIC: Simple Harmonic Motion

COURSE LEVEL: Calculus-Based

BEST USED

✓ After reading the chapter

✓ Before starting homework

✓ Before a quiz or exam

Your Review Plan

Complete these six stages in order. Each stage builds on the previous one and prepares you for the final readiness check.

6 Stages • Approximately 45-60 minutes.

Warm-Up Check

Activate prior knowledge.

Core Concepts

Review the essential ideas.

Guided Practice

Apply what you learned.

Confidence Check

Confirm your understanding.

Summary

Review the key ideas.

Next Step

Continue your learning.

Warm-Up Check

Before you begin, take a moment to see what you already remember. Do not worry about getting everything right. This is only a starting point.

ACTIVITY 1

Recall Activity 1

CALCULUS-BASED
Connect the force law to the equation of motion.

Starting from F = −kx and Newton’s second law, write the differential equation for a mass m attached to an ideal spring.
Reveal Answer
m d²x/dt² = −kx, or d²x/dt² + (k/m)x = 0.

Why it works: The defining linear SHM equation has acceleration proportional to −x, with angular frequency squared equal to k/m.

ACTIVITY 2

Recall Activity 2

CALCULUS-BASED
Differentiate the position function carefully.

If x(t) = A cos(ωt + φ), write v(t) and a(t).
Reveal Answer
v = −Aω sin(ωt + φ); a = −Aω² cos(ωt + φ) = −ω²x.

Why it works: Velocity is one quarter-cycle out of phase with displacement; acceleration is exactly opposite displacement.

ACTIVITY 3

Recall Activity 3

CALCULUS-BASED
Use the potential-energy picture near stable equilibrium.

For a conservative system with a stable equilibrium at x = 0, what feature of U(x) makes small oscillations approximately simple harmonic?
Reveal Answer
U has a local minimum and is approximately quadratic: U ≈ U₀ + ½kₑ x², so F ≈ −kₑ x.

Why it works: A smooth potential near a stable minimum has a leading quadratic term, producing a linear restoring force.

Ready to strengthen your understanding?

You've refreshed what you already know. Next, you'll reinforce the essential concepts that will help you solve problems with confidence. Need to see the learning path again?

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Core Concepts

Let's rebuild the key ideas one step at a time. Focus on understanding the relationships before worrying about solving problems.

KEY CONCEPT 1

Restoring force and the SHM model

SHM is defined by d²x/dt² = −ω²x. For a spring-mass system, ω² = k/m. More generally, linearization about a stable equilibrium gives the same form when the leading restoring term is proportional to displacement.

d²x/dt² + ω²x = 0; spring: ω² = k/m; near a stable minimum, kₑ = d²U/dx² evaluated at equilibrium

Example: If U(x) ≈ U₀ + ½(18 N/m)x² near equilibrium, then small motion has ω = √(18/m).

Sensei Note: Do not call every back-and-forth motion SHM. The key test is whether the restoring force is approximately linear in displacement over the motion being modeled.

KEY CONCEPT 2

Amplitude, period, phase, displacement, velocity, and acceleration

Differentiating x(t) gives v(t) and a(t). Displacement and acceleration differ by π radians; velocity differs from displacement by π/2 radians. Phase-space and x-v-a graphs encode these relations without treating the three quantities as simultaneous maxima.

v = dx/dt = −Aω sin(ωt + φ); a = d²x/dt² = −Aω² cos(ωt + φ) = −ω²x

Example: For x = 0.040 cos(6t + π/3) m, v = −0.240 sin(6t + π/3) m/s and a = −1.44 cos(6t + π/3) m/s².

Sensei Note: Do not confuse amplitude with distance traveled in a cycle. One full cycle covers a path length 4A, while the amplitude is only A.

KEY CONCEPT 3

Energy, spring oscillators, and the simple pendulum

Energy conservation follows directly from the SHM equation. For a pendulum, sin θ ≈ θ converts θ¨ = −(g/L)sin θ into θ¨ + (g/L)θ = 0. This approximation is the mathematical reason the small-angle pendulum is simple harmonic.

d/dt[½m(dx/dt)² + ½kx²] = 0; pendulum: θ¨ + (g/L)θ ≈ 0 for |θ| ≪ 1 rad

Example: For m = 0.50 kg, k = 200 N/m, A = 0.080 m, at x = 0.040 m: U = 0.16 J, K = 0.48 J, and |v| ≈ 1.39 m/s.

Sensei Note: The simple-pendulum formula is a small-angle result. At large amplitude, the motion is periodic but not exactly simple harmonic and the period increases with amplitude.

Ready to apply these ideas?

You've reinforced the essential concepts. Now it's time to put them into practice by working through guided examples and building your problem-solving confidence. Need a quick reminder?

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Guided Practice

Now it's time to apply what you've reviewed.

Work through each activity in order. The examples become gradually more challenging, and each one prepares you for the final readiness check.

PRACTICE 1

Worked Example

Identify equilibrium, choose the positive direction, and apply the restoring-force model.

A particle satisfies d²x/dt² = −36x. Identify ω and write the general sinusoidal form for x(t).
Reveal Answer
ω = 6 rad/s; x(t) = A cos(6t + φ), equivalently a sine-cosine combination.

Why it works: Compare the equation directly with x¨ = −ω²x.

PRACTICE 2

Guided Problem

Use the cycle relationships among x, v, a, T, f, and ω.

For x(t) = 0.030 cos(4t − π/6) m, find v(0) and a(0).
Reveal Answer
v(0) = −Aω sin(−π/6) = +0.060 m/s; a(0) = −Aω² cos(−π/6) ≈ −0.416 m/s².

Why it works: Differentiate the position function first, then substitute t = 0 and preserve the phase sign.

PRACTICE 3

Independent Problem

Use energy or the pendulum model, and state the assumptions that make the model valid.

Derive the small-angle pendulum angular frequency from θ¨ = −(g/L)sin θ.
Reveal Answer
For small θ in radians, sin θ ≈ θ, so θ¨ + (g/L)θ = 0 and ω = √(g/L).

Why it works: The pendulum is approximately SHM only after linearizing sin θ about θ = 0.

Ready to check your understanding?

You've practiced the essential skills with guidance. Now it's time to solve a few short problems on your own and confirm you're ready to move forward. Need a quick reminder?

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Confidence Check

You've rebuilt the key ideas and practiced them with guidance. Now try these short questions on your own to check your understanding before moving on.

QUICK CHECK 1

Restoring-force model check

Answer and justify in one sentence.

What mathematical relation between acceleration and displacement defines SHM?
Reveal Answer
a = −ω²x, equivalently x¨ + ω²x = 0.

Why it works: The proportionality to −x produces sinusoidal motion with constant angular frequency.

QUICK CHECK 2

Cycle and phase check

Use the cycle, not memorized slogans.

For x = A cos(ωt), at what phase is v most negative?
Reveal Answer
At ωt = π/2 modulo 2π, because v = −Aω sin(ωt).

Why it works: Maximum negative velocity occurs when sin(ωt) = 1.

QUICK CHECK 3

Energy and pendulum model check

State the model assumption with the answer.

Why does the exact pendulum equation fail to be linear SHM at large angles?
Reveal Answer
Because sin θ is not proportional to θ at large angle, so the restoring torque is nonlinear.

Why it works: Only the small-angle approximation sin θ ≈ θ gives a constant-coefficient linear SHM equation.

How did it go?

You've checked your understanding. Take one final look at the essential ideas before deciding what to do next. Need a quick reminder?

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Summary

Before moving on, take one final look at the most important ideas from this review.

KEY TAKEAWAY 1

Test the restoring law first

SHM requires a stable equilibrium and acceleration proportional to −displacement. For a spring, F = −kx and ω = √(k/m).

KEY TAKEAWAY 2

Read the cycle through phase relationships

Use ω = 2πf, x = A cos(ωt + φ), v = dx/dt, and a = −ω²x. At equilibrium speed is maximum; at turning points speed is zero.

KEY TAKEAWAY 3

Use energy and respect the small-angle limit

For an ideal spring, E = ½kA² is constant. A simple pendulum is approximately SHM only when sin θ ≈ θ is valid.

Ready for your next step?

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