SHM and Uniform Circular Motion, and Phase: PECTAA Class 12 Physics, Punjab Board

Shine a light on a ball fixed to the rim of a turning table and watch its shadow on a wall. The ball goes round in a circle, but the shadow only slides back and forth, fastest in the middle and pausing at the ends. That shadow is doing simple harmonic motion, and the circle gives us its equations.

Chapter 14 of the PECTAA Physics 12 book is Simple Harmonic Motion. After the simple pendulum, the book links SHM with uniform circular motion and then defines phase (section 14.5). This note is for FSc Part 2 (2nd year, Class 12) students in Lahore, Faisalabad, Sargodha, Rawalpindi, Multan, Gujranwala and the other Punjab boards. Gujrat district comes under the Gujranwala board. PECTAA writes the books. Your own board sets your paper.

What to know

The turntable picture (Fig. 14.4). A ball on the rim of a turntable of radius r turns with constant angular speed ω. A beam of light throws its shadow on a screen. The shadow oscillates like a body on a spring. In general, the projection of a particle in uniform circular motion on a diameter executes SHM.

Set-up for the analysis. A particle P moves round a circle of radius x₀ with uniform angular velocity ω. Its linear speed along the tangent is vₚ = x₀ω. Q is the projection of P on the diameter, and O is the centre, the mean position. Starting from A, the radius sweeps angle θ = ωt in time t.

Instantaneous displacement (Eq. 14.11). From triangle POQ, cos θ = x/x₀, so

x = x₀ cos ωt.

Instantaneous velocity. Q's velocity is the horizontal component of P's velocity: v = vₚ cos(90° − θ) = x₀ω sin θ. Using sin θ = √(1 − cos²θ) and cos θ = x/x₀:

v = ω√(x₀² − x²).

At the mean position (x = 0) the speed is greatest, x₀ω. At the extremes (x = ±x₀) it is zero.

Instantaneous acceleration. P's centripetal acceleration x₀ω² points to the centre. Its horizontal component is Q's acceleration: a = −x₀ω²(x/x₀), so

a = −ω²x.

Acceleration is proportional to displacement and directed towards the mean position, which is the condition for SHM. This confirms that the projection moves with SHM.

Phase (section 14.5). The general form is x = x₀ cos(ωt + φ).

• Phase: the time-varying angle ωt + φ. It fixes both the displacement and the direction of motion at a given time.
• Phase constant φ: set by the displacement and velocity at t = 0.
• φ = 0°: at t = 0, T/2 and T the point is at an extreme; at T/4 and 3T/4 it is at the mean position.
• φ = 90°: the reverse: at t = 0, T/2 and T it is at the mean position; at T/4 and 3T/4 at an extreme.
• Phase difference 0° or 360°: in phase. 180°: out of phase.

Key points

• SHM is the shadow of uniform circular motion on a diameter.
• x = x₀ cos ωt, v = ω√(x₀² − x²), a = −ω²x.
• Greatest speed at the mean position; greatest acceleration at the extremes.
• For a mass on a spring, ω = √(k/m).
• Angles in ωt are in radians when you use a calculator.

Worked examples

Example 1 (book example, first spring). A 0.2 kg mass on a spring of constant 10 N m⁻¹ is pulled 0.1 m from equilibrium and released. Find (a) the acceleration at x = 0.05 m and (b) the velocity at x = 0.06 m.

1. ω = √(k/m) = √(10/0.2) = √50 ≈ 7.07 rad s⁻¹, so ω² = 50.
2. (a) a = −ω²x = −50 × 0.05 = −2.5 m s⁻².
3. (b) v = ω√(x₀² − x²) = 7.07 × √(0.01 − 0.0036) = 7.07 × 0.08 ≈ 0.566 m s⁻¹.
4. The book's working puts x = 0.05 m into part (b) and prints 0.612 m s⁻¹; that is the speed at 0.05 m, not at the 0.06 m asked.

Example 2 (book example, second spring). A 0.5 kg mass on a spring of constant 200 N m⁻¹ is pulled 0.2 m and released. Find displacement, velocity and acceleration at t = 0.5 s.

1. ω = √(200/0.5) = 20 rad s⁻¹, so ωt = 10 rad.
2. In radians, cos 10 ≈ −0.839 and sin 10 ≈ −0.544.
3. Displacement: x = 0.2 × cos 10 ≈ −0.168 m (0.168 m on the far side of the mean position).
4. Speed: x₀ω|sin 10| = 4 × 0.544 ≈ 2.18 m s⁻¹.
5. Acceleration: a = −ω²x = −400 × (−0.168) ≈ +67.1 m s⁻², pointing back towards the mean position.
6. The book prints x = 0.168 m and a = −67.12 m s⁻², treating cos 10 as positive; the sizes are right but both signs should be reversed.

Common mistakes in the exam

1. Using degrees for ωt. ω is in rad s⁻¹, so ωt is in radians.
2. Putting the speed maximum at the extremes. Speed is zero there; acceleration is largest.
3. Dropping the minus sign in a = −ω²x. It shows the pull is towards the mean position.
4. Calling 180° "in phase". 180° is out of phase.
5. Writing v = ω(x₀ − x). The formula has a square root of x₀² − x².

Exam technique

• For the derivation, draw the circle, P, Q and the angle θ, and give one line per step.
• Find ω first in every spring question.
• State the sign meaning of each answer in words.
• Check your board's notice for the sections in your paper.

Practice questions

1. State the condition for SHM shown by a = −ω²x. Answer: Acceleration is proportional to displacement and directed towards the mean position.
2. x₀ = 0.1 m and ω = 10 rad s⁻¹. Find the greatest speed. Answer: x₀ω = 1 m s⁻¹.
3. With the same values, find the speed at x = 0.06 m. Answer: 10 × √(0.01 − 0.0036) = 0.8 m s⁻¹.
4. Find ω for m = 2 kg and k = 50 N m⁻¹. Answer: √25 = 5 rad s⁻¹.
5. Two pendulums have a phase difference of 360°. Describe them. Answer: They oscillate in phase.

Ready to practise this? Continue in 2nd Year Physics.