Unit 14: Simple Harmonic Motion — Numericals

Given amplitude x₀ = 0.05 m, frequency f = 10 Hz.

Angular frequency ω = 2πf = 2π(10) = 62.8 rad s⁻¹

(a) Maximum acceleration (at extreme position):
a = ω²x₀ = (62.8 rad s⁻¹)² × 0.05 m = 3943.4 × 0.05 ≈ 197 m s⁻²

(b) Maximum velocity (at mean position):
v = ωx₀ = 62.8 rad s⁻¹ × 0.05 m ≈ 3.14 m s⁻¹

Result a ≈ 197 m s⁻², v ≈ 3.14 m s⁻¹ — matches the printed answer.

Given m = 0.5 kg, static extension x = 0.098 m, g = 9.8 m s⁻².

At equilibrium the spring force balances the weight: kx = mg
k = mg/x = (0.5 kg × 9.8 m s⁻²)/0.098 m = 4.9/0.098 = 50 N m⁻¹

(i) Time period: T = 2π√(m/k) = 2π√(0.5/50) = 2π√0.01 = 2π(0.1) = 0.628 s

(ii) Frequency: f = 1/T = 1/0.628 ≈ 1.59 Hz

Result T ≈ 0.628 s, f ≈ 1.59 Hz — matches the printed answer.

Given m = 0.5 kg, compression x = 10 cm = 0.10 m, force F = 45 N.

Spring constant k = F/x = 45/0.10 = 450 N m⁻¹

Energy stored (elastic P.E.): P.E. = ½kx² = ½ × 450 × (0.10)² = ½ × 450 × 0.01 = 2.25 J

When the compressing force is suddenly removed, this stored energy converts entirely into kinetic energy as the ball passes through the natural (mean) position:
½mv² = 2.25 J
v² = (2 × 2.25)/0.5 = 9
v = 3 m s⁻¹

Result Energy stored = 2.25 J, maximum velocity = 3 m s⁻¹ — matches the printed answer.

Given T = 1 s, g = 9.8 m s⁻².

From T = 2π√(ℓ/g) ℓ = gT²/4π² = (9.8 × 1²)/(4 × 9.8696) = 9.8/39.478 ≈ 0.248 m ≈ 0.25 m

Frequency f = 1/T = 1/1 = 1 Hz

Result ℓ ≈ 0.25 m, f = 1 Hz — matches the printed answer.

Given m = 5 kg, h = 0.8 m, k = 1960 N m⁻¹, g = 9.8 m s⁻².

By conservation of energy, the gravitational P.E. lost in falling converts into the elastic P.E. stored in the spring at maximum compression x:
mgh = ½kx²
5 × 9.8 × 0.8 = ½ × 1960 × x²
39.2 = 980x²
x² = 39.2/980 = 0.04
x = 0.2 m

Result The spring is compressed by 0.2 m — matches the printed answer.