Representing and Analyzing SHM: Study Guide (Physics C: Mechanics 7.3)

How this differs from the Physics 1 version. Physics C: Mechanics and Physics 1 are separate courses that both have a Topic 7.3 called Representing and Analyzing SHM. This guide is the calculus-based one. It builds the equation of motion from Newton’s second law, uses a phase constant, and finds velocity and acceleration by differentiation. The algebra-based treatment is in the Physics 1 study guide; do not mix the two when you revise.

From Newton’s second law to the SHM equation

Take a block of mass m on a frictionless horizontal surface, attached to an ideal spring of constant k. Measure x from the equilibrium position, where the net force is zero, with +x to the right. When the block is at x, the spring exerts the restoring force F_x = −kx. Newton’s second law gives:

m d²x/dt² = −kx, so d²x/dt² = −(k/m)x

Write the positive constant k/m as ω². Then:

d²x/dt² = −ω²x, with ω = √(k/m)

This is a second-order differential equation: it links x to its own second derivative. Read it in words: the acceleration is always proportional to the displacement and points the other way.

This gives you a test for SHM in any system. Apply Newton’s second law (or its rotational form), write the result for the displacement from equilibrium, and see whether it takes the form d²x/dt² = −(positive constant) × x. If it does, the motion is simple harmonic and ω is the square root of that constant. If the constant depends on x, or the force is not proportional to x, the motion may be periodic but it is not SHM.

The solution: x = A cos(ωt + φ₀)

The function that satisfies d²x/dt² = −ω²x is a sinusoid:

x(t) = A cos(ωt + φ₀)

• A is the amplitude, the largest displacement from equilibrium (in m, always positive).
• ω is the angular frequency (rad/s). It is set by the system, here √(k/m). Then T = 2π/ω and f = ω/2π (Topic 7.2).
• φ₀ is the phase constant (rad). It says where in the cycle the object is at t = 0.

The sine form x = A sin(ωt + φ₁) describes the same motion with φ₁ = φ₀ + π/2. Use whichever makes the starting condition simplest:

• Released from rest at x = +A at t = 0: x = A cos(ωt) (φ₀ = 0).
• Passing through x = 0 moving in +x at t = 0: x = A sin(ωt).

You do not have to prove that this solves the equation. A quick check is still useful: differentiate twice and you get −ω² times the original function.

Finding A and φ₀ in general. Suppose that at t = 0 the object is at x₀ with velocity v_x0. From x(0) = A cos φ₀ and v_x(0) = −Aω sin φ₀:

A = √(x₀² + (v_x0/ω)²), then cos φ₀ = x₀/A and sin φ₀ = −v_x0/(Aω).

Use both the cosine and the sine to choose the right quadrant for φ₀. Using tan φ₀ on its own can give an angle that is π out.

Velocity and acceleration by differentiation

Differentiate x(t) once and then again:

v_x = dx/dt = −Aω sin(ωt + φ₀)

a_x = dv_x/dt = −Aω² cos(ωt + φ₀) = −ω²x

Three results follow directly:

• a_x = −ω²x. The acceleration is proportional to the displacement and opposite to it. This is the differential equation again, which is a good check.
• v_max = Aω, because the largest value of |sin| is 1.
• a_max = Aω², because the largest value of |cos| is 1.

Using sin² + cos² = 1 you can also write the speed at any position: |v_x| = ω√(A² − x²). This is zero at x = ±A and equal to Aω at x = 0.

Zeros and extremes

The three graphs in Figure 1 are for x = A cos(ωt). Each is a sinusoid with the same period. The velocity graph is a quarter-cycle ahead of the displacement graph, and the acceleration graph is the displacement graph turned upside down and scaled by ω².

Where the object is	Displacement	Velocity	Acceleration
Turning point x = +A	maximum, +A	zero	−Aω² (most negative)
Equilibrium, moving in −x	zero	−Aω (most negative)	zero
Turning point x = −A	minimum, −A	zero	+Aω² (most positive)
Equilibrium, moving in +x	zero	+Aω (most positive)	zero

Two features are worth knowing without thinking. Wherever v_x = 0, the object is at a turning point and |a_x| is largest. Wherever a_x = 0, the object is at equilibrium and the speed is largest. On the x–t graph, velocity is the tangent slope, so the slope is zero at the peaks and steepest where the curve crosses the axis.

Amplitude does not change the period

In d²x/dt² = −ω²x, ω depends only on the system (k and m for a spring). A does not appear in it. Double the amplitude and the object has twice as far to go, but the restoring force at each point is also twice as large, so it moves twice as fast at each matching point of the cycle. The two effects cancel exactly: the period stays the same. What does change: v_max = Aω and a_max = Aω² both double.

This is true for ideal SHM, where the restoring force is exactly proportional to the displacement. For a pendulum it holds only for small angles (Topic 7.5).

Reading SHM from graphs

• From an x–t graph: the amplitude is the height of a peak above the equilibrium line; the period is the time between matching points; φ₀ comes from where the cycle is at t = 0. The tangent slope at any instant is v_x.
• From a v_x–t graph: the peak gives v_max = Aω. The signed area under the graph over an interval is the displacement (Topic 1.2), so each half-loop has area of size 2A.
• From an a_x–x graph: SHM gives a straight line through the origin with negative slope. The slope is −ω². A curved line, or one that misses the origin, means the motion is not SHM about that origin.

Resonance

The natural frequency of a system is the frequency it oscillates at when you displace it from equilibrium and let go: f₀ = ω/2π. Now push it with an external force that varies sinusoidally in time at a driving frequency f. If f equals the natural frequency, each push arrives at the same point in every cycle and is in the direction the object is already moving. Energy is added every cycle, so the amplitude grows. This is resonance.

Driving at other frequencies gives a smaller amplitude, because the pushes are only sometimes in step with the motion. Figure 2 sketches the pattern.

Worked example 1: from starting conditions to x(t)

Question. Take +x to the right, origin at equilibrium. A 0.50 kg cart on a level, frictionless track is held between springs that act together like one spring with k = 32 N/m. At t = 0 the cart is at x₀ = +0.030 m moving with v_x0 = +0.32 m/s. Find ω, the amplitude, the phase constant for the cosine form, v_max, a_max, and the first time the cart passes through equilibrium.

1. Angular frequency: ω = √(k/m) = √(32/0.50) = 8.0 rad/s. So T = 2π/8.0 = 0.785 s.
2. Amplitude: v_x0/ω = 0.32/8.0 = 0.040 m. A = √(0.030² + 0.040²) = 0.050 m.
3. Phase constant: cos φ₀ = 0.030/0.050 = 0.60 and sin φ₀ = −0.32/(0.050 × 8.0) = −0.80. Cosine positive and sine negative put φ₀ in the fourth quadrant: φ₀ = −0.93 rad. So x(t) = (0.050 m) cos(8.0t − 0.93).
4. Maxima: v_max = Aω = 0.050 × 8.0 = 0.40 m/s; a_max = Aω² = 0.050 × 64 = 3.2 m/s².
5. First passage through equilibrium: the cart is moving right, so it first reaches x = +A when 8.0t − 0.93 = 0, at t = 0.116 s. It then returns, and x = 0 when 8.0t − 0.927 = π/2: t = (1.571 + 0.927)/8.0 = 0.31 s, moving in −x at 0.40 m/s.

Check. At t = 0, a_x = −ω²x₀ = −64 × 0.030 = −1.92 m/s²: negative while x is positive, as it must be. And |v_x| = ω√(A² − x₀²) = 8.0 × 0.040 = 0.32 m/s, the speed we started with.

Worked example 2: working from a velocity–time graph

Question. Take +x to the right, origin at equilibrium. A glider oscillates in SHM. Its velocity–time graph (Figure 3) is a sine curve with v_x = 0 at t = 0, a peak of +0.40 m/s, and a period of 0.628 s. Find ω, A, the glider’s position at t = 0, its acceleration at t = 0, and the distance it covers in the first half-period.

1. Angular frequency: ω = 2π/T = 2π/0.628 = 10 rad/s.
2. Amplitude: v_max = Aω, so A = 0.40/10 = 0.040 m.
3. Velocity function: the graph is v_x = (0.40 m/s) sin(10t).
4. Position by integration: x = ∫v_x dt = −(0.040 m) cos(10t) + C. The motion is centred on the origin (equilibrium), so C = 0. At t = 0: x = −0.040 m = −A. The glider starts at the left turning point, which fits v_x = 0 there.
5. Acceleration by differentiation: a_x = dv_x/dt = (4.0 m/s²) cos(10t). At t = 0, a_x = +4.0 m/s², the largest value. Check with a_x = −ω²x = −100 × (−0.040) = +4.0 m/s².
6. Distance in the first half-period: v_x is positive throughout, so distance = area of the first loop = ∫₀^0.314 0.40 sin(10t) dt = 0.080 m. That is 2A: from −A to +A.

Interpretation. A velocity graph that starts at zero and rises means the object starts at rest and is pushed in +x. The push in +x at a turning point only happens at x = −A.

Common misconceptions

• “The speed is greatest at the ends.” The object is momentarily at rest at x = ±A. The speed is greatest at equilibrium.
• “a_x = 0 means v_x = 0.” At equilibrium a_x = 0 but the speed is v_max.
• “Bigger amplitude, longer period.” For ideal SHM the period is independent of A. Only v_max and a_max scale with A.
• Mixing up ω and f. ω is in rad/s and f in Hz; ω = 2πf. v_max = Aω, not Af.
• Calculator in degrees. ωt + φ₀ is in radians.
• Choosing φ₀ from tan alone. Check the signs of cos φ₀ and sin φ₀ to get the right quadrant.
• “Any repeating motion is SHM.” A ball bouncing on the floor repeats, but the force on it is not proportional to its displacement, so it is periodic but not simple harmonic.
• “Any regular push makes a big amplitude.” Only pushes at (or very close to) the natural frequency produce resonance.

Where this leads

Topic 7.4 (Energy of Simple Harmonic Oscillators) uses x(t) and v_x(t) from this guide to show that kinetic and potential energy trade back and forth while their sum stays constant: read the Topic 7.4 study guide. Topic 7.5 applies the same differential-equation test to pendulums. Now try the practice questions, then use the revision notes and the checklist. You can also return to the course roadmap.