A 0.12 kg block on a horizontal frictionless surface is attached to a spring whose force constant is 570 N/m.The block is pulled from its equilibrium position at x = 0 m to a displacement x = +0.080 m and is released from rest.The block then executes simple harmonic motion along the x-axis (horizontal).The displacement of the block at time t = 0.20 s is closest to:
A)0.01 m
B)-0.01 m
C)0.08 m
D)-0.08 m
E)0.1 m

Question

Quizplus · Accepted Answer

The period of the motion is:T = 2π√(m/k) = 2π√(0.12 kg / 570 N/m) = 0.40 sAt time t = 0.20 s, the phase angle is:φ = ωt = (2π/T)t = (2π/0.40 s)(0.20 s) = π/2The displacement of the block at time t = 0.20 s is:x = Acos(ωt + φ) = (0.080 m)cos(π/2 + π/2) = -0.01 m

A 012 Kg Block on a Horizontal Frictionless Surface Is Attached