Simple Harmonic Motion

A restoring force proportional to the displacement.

12 problems · about 23 minutes · F-TF.B.5, F-IF.B.4

What this lesson teaches

The student models undamped oscillation and computes period and frequency.

A restoring force proportional to displacement gives y″ + ω²y = 0, with ω² = k/m.
The solution oscillates with period 2π/ω.
The period depends only on stiffness and mass, never on the amplitude.

Straightforward practice. Get the method working first.

5 problems

k = 64 and m = 4. What is ω?
Answer 4
Why 4.
Does the period of simple harmonic motion depend on the amplitude?
Answer No. It depends only on the stiffness and the mass.
Why Amplitude does not affect the period.
k = 16 and m = 1. What is ω?
Answer 4
Why The square root of 16.
k = 45 and m = 5. What is ω?
Answer 3
Why The square root of 9.
ω = 4. What is the period, using 2π as about 6.28?
Answer 1.57
Why 6.28 divided by 4.

The same ideas with more to keep track of.

3 problems

Doubling the amplitude. Does the period change? 1 yes, 0 no.
Answer 0
Why Amplitude and period are independent.
A stiffer spring with the same mass. Does the period get longer or shorter? Enter 1 longer, 2 shorter.
Answer 2
Why A larger k raises ω.
y″ + 25y = 0. What is ω?
Answer 5
Why ω² = 25.

Mixed problems. Work out what kind of question it is before you start.

4 problems

Sort each change by its effect on the period.
Answer Makes the period shorter: A stiffer spring, A lighter mass · No effect on the period: Pulling it further before release, Releasing it at a different moment
Why Only k and m enter ω.
y″ + 49y = 0. What is ω?
Answer 7
Why The square root of 49.
The Spring: k = 100 and m = 4. What is ω?
Answer 5
Why 5.
The Cycle: ω = 2. What is the period, using 2π as about 6.28?
Answer 3.14
Why About 3.14 seconds.