The force law for SHM
From a = −ω²x to F = −kx
We found that in SHM the acceleration is a equals minus ω squared x. Now bring in Newton’s second law: the force on the particle is F equals m a.
So the force is F equals minus m ω squared x. Since m and ω are constants, we write k equals m ω squared, and the force law becomes F equals minus k x.
The force is proportional to the displacement and always directed towards the mean position. Watch the red arrow: it shrinks to zero at the centre and is largest at the ends. This is a restoring force.
The idea also works backwards. Whenever the net force on a body is F equals minus k x, the body performs simple harmonic motion, with angular frequency ω equals the square root of k over m.
Because the force varies linearly with x, such a system is called a linear harmonic oscillator. k is called the force constant.
The best example is a block on a spring, resting on a smooth floor. For small stretches, the spring obeys Hooke’s law: F equals minus k x, where k is the spring constant.
Pull the block out and let go. It oscillates in SHM with ω equals the square root of k over m, so its period is T equals 2π times the square root of m over k.
A heavier block responds more slowly, so the period grows. Make the mass four times bigger and the period doubles. A stiffer spring pulls harder, so the period shrinks.
And notice what is missing from the formula: the amplitude. Pull the block out 5 cm or 15 cm, and it takes the same time for each oscillation.
Worked examples
A 5 kg collar slides without friction on a horizontal rod. It is attached to a spring with spring constant 500 newtons per metre. It is pulled 10 cm from equilibrium and released.
First, the period. ω is the square root of k over m: the square root of 500 over 5, which is 10 radians per second. So T equals 2π over 10, about 0.63 seconds.
The maximum speed is ωA: 10 times 0.1, which is 1 metre per second. It happens as the collar shoots through the equilibrium position.
The maximum acceleration is ω squared A: 100 times 0.1, which is 10 metres per second squared. It happens at the two ends, where the spring is stretched or squeezed the most.
Now a block of mass m sits between two identical springs, each of spring constant k, fixed to walls on either side. Push the block a distance x to the right.
The left spring is stretched by x and pulls it back with k x. The right spring is compressed by x and pushes it back with another k x. The total force is minus 2 k x.
So the system behaves like a single spring of constant 2k. The period is 2π times the square root of m over 2k: smaller than with one spring, by a factor of root 2.
The method is always the same: find the net restoring force, write it as minus some constant times x, and then ω equals the square root of that constant over the mass.
Try it: what sets the period?
- Drag the block sideways and let go, or press Release (Space).
- The amber bracket on the graph marks one period T.
- Change m or k after a run: the last run stays as a dashed ghost so you can compare.
- Change only the amplitude. Does T change?
Try it: the force law
- The F–x graph lines up with the block: the dot is always right below it.
- The graph is a straight line through the origin with slope −k. Change k and watch it tilt.
- The force (red arrow) always points back towards x = 0.
Four times the mass
A block on a spring oscillates with a period of 1 s. The block is replaced by one with four times the mass, on the same spring. What is the new period?
- 0.25 s
- 0.5 s
- 2 s
- 4 s
Answer
C. T = 2π√(m/k), so T ∝ √m. Four times the mass gives √4 = 2 times the period: 2 s. A heavier block accelerates less for the same spring force, so it oscillates more slowly.
Between two springs
A 0.5 kg block on a smooth floor is attached between two identical springs, each with k = 100 N/m, fixed to walls on both sides. What is the period of its oscillation?
- 0.22 s
- 0.31 s
- 0.44 s
- 0.63 s
Answer
B. Displaced by x, both springs push the block back: F = −2kx, so the effective constant is 2k = 200 N/m. Then T = 2π√(m/2k) = 2π√(0.5/200) = 2π × 0.05 = 0.31 s. Using a single spring (100 N/m) gives 0.44 s.
Summary
- Force law for SHM:
F = −kx - Force constant:
k = mω² - Angular frequency:
ω = √(k/m) - Period of a spring–block system:
T = 2π√(m/k)
- A body moves in SHM when the net force on it is a restoring force proportional to the displacement: F = −kx.
- For a block on an ideal spring, k is the spring constant from Hooke’s law.
- A heavier block or a softer spring gives a longer period. The period does not depend on the amplitude.
- For two springs pulling the same way on a block, the constants add: a block between two springs of constant k has T = 2π√(m/2k).