A mass on a spring is the model every oscillator in this course is a version of. Newton's law says mass times acceleration is the total force, Hooke's law says the spring pulls back in proportion to the stretch, and the result is m·x″ = −k·x when nothing else acts. Its solutions swing at the natural frequency ω₀, where ω₀² = k/m, so a stiffer spring swings faster and a heavier mass slower.
Core ideas
Three ideas, in the order they are used: where the equation comes from, what its frequency is, and what the period is made of.
Let x be how far the mass is from where the spring is relaxed, positive to the right. Newton's law says m·x″ is the sum of the forces. Hooke's law gives the spring's force as −k·x: proportional to the stretch and pointing back toward rest, which is what the minus sign records. A damper gives −c·x′, proportional to the speed and opposing it. Moving everything to the left gives m·x″ + c·x′ + k·x = 0.
With no damper, c = 0, divide by m to get x″ = −(k/m)·x. Call k/m the number ω₀². The previous course showed that x″ = −ω₀²·x has the solutions C₁cos(ω₀t) + C₂sin(ω₀t). A bigger k makes the pull stronger and ω₀ larger; a bigger m is harder to accelerate with the same pull, so ω₀ is smaller.
Cosine and sine repeat after 2π in their argument, so the motion repeats when ω₀·t has grown by 2π: the period is T = 2π/ω₀. When ω₀ is rational the period is a rational multiple of π; when it is a surd the period is a rational multiple of π over a square root. In either case it is irrational, and the lab prints it with ≈.
The material
The forces, the equation they give, the frequency in the undamped case, and a proof that the period does not depend on how far the mass is pulled.
Hang a mass m on a spring and let it move along a line. Call x its position measured from the spot where the spring is neither stretched nor squeezed. Units are left out throughout: any consistent set of units will do, and the numbers in the lab are pure.
The force a spring exerts on the mass is −k·x, where k > 0 is the spring constant. The force is proportional to the displacement and points the other way.
A stiffer spring has a larger k. The law is an approximation that real springs satisfy for small stretches, and in this course it is an assumption.
Newton's law says that mass times acceleration is the total force. Acceleration is x″, so m·x″ equals the spring's force plus whatever else acts. A damper, such as a plunger in oil, opposes the motion with a force proportional to the speed x′, which is −c·x′ for a damping constant c ≥ 0.
This lesson takes c = 0; the damper arrives in the lessons after next. With no damper the equation is x″ + ω₀²·x = 0 where ω₀² = k/m. That is the equation of “Complex Roots and Oscillation” in the previous course, with ω₀ in the place of β, and its general solution is x = C₁cos(ω₀t) + C₂sin(ω₀t). Two constants, fixed by a start.
Every solution of x″ + ω₀²·x = 0 satisfies x(t + T) = x(t) for T = 2π/ω₀, whatever the constants C₁ and C₂.
Replace t by t + T in cos(ω₀t). The argument becomes ω₀t + ω₀T = ω₀t + 2π, and cosine and sine repeat after 2π.
So both terms, and any combination of them, take the same value at t + T as at t. The constants C₁ and C₂ come out as multipliers and never enter. This is why a larger start does not make a slower swing.
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Take m = 1 and k = 4. Then x″ = −4x, ω₀² = 4, ω₀ = 2, and the solutions are C₁cos(2t) + C₂sin(2t). The period is 2π/2 = π, which is about 3.14159 and is rounded.
Check by substituting x = cos(2t): x″ = −4cos(2t) = −4x. The residual is zero, so it is a solution, and the lab's tiles for the first preset show ω₀² = 4 and the period π.
Now change one number at a time and see which way it moves. Doubling the mass, with the same spring, divides ω₀² by 2. The new frequency is ω₀ = √2, the period is 2π/√2, and the period has been multiplied by √2: the heavier mass is slower. Stiffening the spring to k = 9 with m = 1 gives ω₀ = 3 and the period 2π/3, shorter than π.
Take m = 2 and k = 5. Then ω₀² = 5/2, which has no rational square root. The frequency is ω₀ = √(5/2), written √10/2 after clearing the fraction under the root, and the period is the irrational number the lab prints with ≈.
The lab keeps ω₀² = 5/2 exact and prints ω₀ as a surd with its rounded value beside it. Only the second is rounded.
Interactive
Free, damped and forced motion, with every exact quantity
Method
Four moves, then a check. The check is the one that catches a mass and a spring constant entered the wrong way round.
Take positive to the right, so a stretch is x > 0. The spring's force is −k·x and a damper's is −c·x′. Each force opposes what the mass is doing, which is why each has a minus sign.
m·x″ equals the sum of the forces. Move every term to the left to get m·x″ + c·x′ + k·x = 0. For this lesson c = 0.
x″ = −(k/m)·x, and ω₀² = k/m is a fraction you compute exactly. Keep it as a fraction. Take its root only when you need ω₀, and keep that as a surd if the root is not rational.
T = 2π/ω₀. Leave π and the root in the symbol, and round only once, at the end, with ≈. The lab prints both.
A larger k or a smaller m must make ω₀ larger and the period shorter. If your answer moves the other way, m and k have been swapped.
Start with m = 1 and k = 4, the first preset. Then double the mass and compare.
The first period is exact as π and the lab prints it with ≈. The second is the lab's double preset. Its period is √2 times larger because ω₀ is √2 times smaller, and no other figure in the lesson needs rounding.
You should be able to write m·x″ + c·x′ + k·x = 0 from Newton's law and Hooke's law, compute ω₀² = k/m exactly, give ω₀ as an integer or a surd, and give the period 2π/ω₀ as a symbol and as a rounded number marked ≈.
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