Skip to content

Question

A spring-mass system oscillates with \( T = 0.4 \, \text{s} \) when \( m = 0.2 \, \text{kg} \). What is
the spring constant?

Options

Choose one · Correct answer highlighted

Explanation

**Real oscillators** experience damping, amplitude decreasing with time. Critical damping returns to equilibrium fastest without oscillation, overdamping slows return, underdamping shows decaying oscillations, classification based on b relative to 2mω₀, important for practical systems. T = 2π √((m/k)) . 0.4 = 2π √((0.2/k)) ⇒ (0.4/2π) = √((0.2/k)) . (0.0637)² = (0.2/k) ⇒ k = (0.2/0.00406) ≈ 49.26 N/m . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 49.26 N/m follows, reflecting SHM dependence on amplitude A, ω and system parameters.

Discussion

Comments

0 comments

No comments yet. Be the first to start the discussion.