The student connects oscillatory motion to uniform circular motion by mapping correspondences between variables, applies Hookeās law and basic energy relationships, and creates analogies between springāmass and pendulum systems to explain simple harmonic motion.
A spring stretches 0.08 m when a force of 12 N is applied.
Find k.
A spring has k=250 N/m.
Find the force at x=0.06 m.
A massāspring system has k=200 N/m and amplitude A=0.10 m.
Find total mechanical energy E=(1/2) k A^2.
If amplitude doubles, energy becomes:
2E / 4E / 8E
Match the Simple Harmonic Motion concept to a Uniform Circular Motion idea:
amplitudeĀ A
angular frequency Ļ
equilibrium position
restoring force
Choices (use each once):
radius r
constant angular speed
center point
inward net force
A mass m=0.50 kg oscillates on a spring with k=80 N/m.
Find angular frequency Ļ^2 = k/mā.
If amplitude is 0.15 m, find max speed vmaxā = Ļ A .