Question Detail
Question
A spring of angular frequency 8.08 s⁻¹ is compressed 0.100 m from equilibrium (x₀ = -0.100m) but is given a shove to create a velocity in the +x direction of v₀ = 0.400m/s. What is the amplitude A of the motion?
Correct Answer
Option B is the correct answer.
Detailed Explanation
From Worked Example 30 on page 39: The amplitude A for SHM with initial position x₀ and initial velocity v₀ is given by A = √(x₀² + (v₀/ω)²). Substituting the given values: A = √((-0.100 m)² + (0.400 m/s / 8.08 s⁻¹)²) = √(0.01 + (0.0495 m)²) = √(0.01 + 0.00245) = √0.01245 ≈ 0.1115 m, which rounds to 0.112 m.
Hint
Use the formula relating amplitude to initial displacement, initial velocity, and angular frequency.