Spring pendulum – Scitura

What is it about?

A body of mass $m$ hangs on a coil spring with spring constant $D$ . At the equilibrium position the spring is stretched by $s_0 = \frac{m g}{D}$ , so that the spring force balances the weight. If the body is displaced by $y$ (here: upwards positive), the resulting force is the restoring force $F = -D \cdot y$ . This linear force law leads to a harmonic oscillation

y(t) = \hat y \sin(\omega t + \varphi_0), \qquad T = 2\pi\sqrt{\frac{m}{D}} .

The period depends neither on the amplitude nor on $g$ . In the phasor diagram a pointer of length $\hat y$ rotates with angular frequency $\omega$ ; its vertical projection is the displacement. Velocity and acceleration are $v = \hat y\,\omega\cos(\omega t + \varphi_0)$ and $a = -\omega^2 y$ . Without friction $\tfrac12 D y^2 + \tfrac12 m v^2 = \tfrac12 D \hat y^2$ stays constant; with damping the amplitude decreases as $e^{-\delta t}$ .

Try it

Pull the body down and let go. Watch the phasor and the graphs.
Pause and tap a graph to set the time marker, e.g. at a maximum of $y(t)$ . What are $v$ and $a$ there?
Time 10 full turns of the phasor with the stopwatch and compare $t/10$ with the formula.
Compare two pendulums: four times the mass or four times as stiff.
Look at the forces and all three forms of energy. Forces and energy

Tasks

Task 1: Period

Calculate the period for $m = 0.5\,\text{kg}$ and $D = 20\,\text{N/m}$ . Load task

Show solution

$T = 2\pi\sqrt{0.5/20}\,\text{s} \approx 0.99\,\text{s}$ , $f \approx 1.01\,\text{Hz}$ .

Task 2: Doubling the period

How must the mass be changed to double the period?

Show solution

Since $T \sim \sqrt{m}$ , the mass must be multiplied by four. Show

Task 3: Maximum values

For $m = 0.5\,\text{kg}$ , $D = 20\,\text{N/m}$ and $\hat y = 8\,\text{cm}$ find the maximum speed and acceleration.

Show solution

$\omega \approx 6.32\,\text{s}^{-1}$ , $\hat v = \hat y\,\omega \approx 0.51\,\text{m/s}$ (at the equilibrium position), $\hat a = \hat y\,\omega^2 = 3.2\,\text{m/s}^2$ (at the turning points).

Ideas for the classroom

Model: ideal massless spring obeying Hooke’s law, displacement positive upwards, velocity-proportional damping (with damping the body starts with a small initial velocity so that $y = \hat y\,e^{-\delta t}\sin(\omega_d t + \varphi_0)$ holds exactly), Runge–Kutta integration with a step of $1\,\text{ms}$ .
The two pendulums in comparison mode are suspended so that their equilibrium positions are at the same height.
The stopwatch counts full turns of the phasor since the start (green mark).
Common stumbling blocks: “A larger amplitude means a longer period.” – “At the turning point the acceleration is zero.” – “On the Moon a spring pendulum oscillates more slowly.”