Problem Solution

The car oscillates at approximately 2.75 Hz when loaded with passengers. This result comes from applying the spring-mass oscillation formula to the increased total mass.

A car suspension with four identical springs behaves like a mass-spring system where the springs act in parallel, giving an equivalent spring constant k_eq = 4k. The oscillation frequency follows f = (1/(2π)) √(k_eq/m), where m is the total mass.

For the empty car (m = 1500 kg, f = 3 Hz), the angular frequency ω = 2πf = 2π × 3. Thus, k_eq = ω²m = (2π × 3)² × 1500 ≈ 532959 N/m.

With four 75 kg passengers, total mass m’ = 1500 + 300 = 1800 kg. The new frequency is f’ = (1/(2π)) √(k_eq/m’) ≈ 2.74 Hz, closest to 2.75 Hz (option c).

Option Analysis

• a. 0.75 Hz: This assumes frequency scales inversely with mass (f’ = f × (m/m’)), but ignores the square root, yielding 1500/1800 × 3 ≈ 2.74 Hz, not 0.75 Hz.
• b. 1.75 Hz: Incorrect; might stem from linear mass scaling without square root or confusing passenger distribution.
• c. 2.75 Hz: Correct, as f ∝ 1/√m, so f’/f = √(1500/1800) ≈ 0.9129, and 3 × 0.9129 ≈ 2.74 Hz.
• d. 3.75 Hz: Wrong; higher frequency would require stiffer springs or less mass, opposite to added load.

Physics Behind Suspension Frequency

The car suspension springs oscillation frequency passengers mass problem models vehicle vertical motion as simple harmonic oscillation. A 1500 kg car on four springs vibrates at 3 Hz empty, dropping to about 2.75 Hz with 300 kg added passenger mass due to inverse square root mass dependence.

Suspension systems use springs in parallel, so f = (1/(2π)) √(4k/m). Added mass lowers frequency for ride comfort, matching real cars at 1-3 Hz unloaded.

Step-by-Step Calculation

1. Compute k_eq from empty car data.
2. Add passenger mass (4 × 75 kg).
3. Recalculate f’ using same k_eq.