Simple Harmonic Motion Notes

Simple Harmonic Motion - Practice Exam 3

Problem Setup

  • A block-spring system is initially at rest on a horizontal surface.
  • Equilibrium position is defined as x=0x = 0.
  • The spring is compressed to x=−0.5 mx = -0.5 \text{ m}.
  • The block is released and oscillates in simple harmonic motion (SHM).
  • The maximum speed of the block during oscillation is vmaxv_{max}.

Question

  • At what location (x-position) will the block's speed be vmax2\frac{v_{max}}{2}?

Given Information

  • Amplitude of oscillation: A=0.5 mA = 0.5 \text{ m}
  • Equilibrium position: x=0x = 0
  • Initial compression: x=−0.5 mx = -0.5 \text{ m}

Possible Answers

  • -0.25 m
  • -0.43 m
  • -0.50 m