Practice question
Question
A particle’s x-projection from circular motion is \( x = 5 \cos (4t) \) (in m). What is its maximum
speed?
Explanation
**Effect of damping** is gradual amplitude reduction while period remains nearly constant for light damping. Mechanical energy decreases as work done against damping force, E(t) = ½ k A(t)² decaying exponentially, and motion ceases without external energy input, distinguishing from ideal undamped SHM. Maximum speed: vₘₐₓ = ω A . A = 5 m, ω = 4 s⁻¹ . vₘₐₓ = 4 × 5 = 20 m/s . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 20 m/s follows, reflecting SHM dependence on amplitude A, ω and system parameters.
Discussion
Comments
Please log in to join the discussion.
Login to commentNo comments yet. Be the first to start the discussion.