A spring pulls back in proportion to how far it is stretched. Newton’s second law then gives my″ = −ky.
Step 1: Let's Learn
Read it, or press Listen and follow the words.
The equation
Rearranged, y″ + ω²y = 0 where ω² = k/m. The characteristic roots are ±ωi, so the motion oscillates.
The solution
y = A cos ωt + B sin ωt, which can always be rewritten as a single cosine with an amplitude and a phase shift.
The period
One full cycle takes 2π/ω. A stiffer spring or a lighter mass raises ω and shortens the period.
Amplitude does not matter
The period of simple harmonic motion is independent of how far you pull it. This is why pendulum clocks work.
Energy
With no damping, energy shuttles between kinetic and potential forever. The total never changes.
A restoring force proportional to displacement
A spring pulls back in proportion to how far it is stretched, giving mx″ = −kx. The solution is sinusoidal, and the frequency depends on the mass and stiffness rather than on the amplitude.
The period does not depend on amplitude
A larger swing travels further and moves faster, and the two effects cancel exactly. That independence is what makes pendulum clocks possible and it holds only for small oscillations.
Energy moves between two forms
Kinetic energy is greatest at the centre and potential energy at the extremes, with the total conserved. That exchange is the physical content behind the sinusoidal solution.
The same equation everywhere
Springs, pendulums, LC circuits and molecular vibrations all reduce to this equation. Any system near a stable equilibrium behaves like a harmonic oscillator, which is why it is studied first.
Step 2: Try It Yourself
Tap and try it out.
- Point(0, 2)
Step 3: Watch an Example
One step at a time.
Watch Tomás Find a Period
Tomás has a spring with k = 36 and a mass of 4.
- Step 1
He computes ω² as k over m, which is 36 over 4, giving 9.
Step 4: Your Turn
Practice makes it stick.
The Spring
Problem 1 of 2
k = 100 and m = 4. What is ω?
The Cycle
Problem 2 of 2
ω = 2. What is the period, using 2π as about 6.28?
Oscillate
1 of 8
k = 16 and m = 1. What is ω?
2 of 8
k = 45 and m = 5. What is ω?
3 of 8
ω = 4. What is the period, using 2π as about 6.28?
4 of 8
Doubling the amplitude. Does the period change? 1 yes, 0 no.
5 of 8
A stiffer spring with the same mass. Does the period get longer or shorter? Enter 1 longer, 2 shorter.
6 of 8
y″ + 25y = 0. What is ω?
7 of 8
Sort each change by its effect on the period.
Tap something to move it.
- Empty
- Empty
8 of 8
y″ + 49y = 0. What is ω?
Step 5: Quick Check
Show what you know.
Question 1 of 2
k = 64 and m = 4. What is ω?
Question 2 of 2
Does the period of simple harmonic motion depend on the amplitude?
What You Learned
- 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.