3.6.1.3

The mass-spring system: $T = 2 pi sqrt(m/k)$

Simple Harmonic Motion — AQA A-Level Physics

$$T = 2\pi\sqrt{\frac{m}{k}}$$
  • $T$: time periodThe time taken for one complete oscillation or wave cycle. Measured in seconds (s). (s)
  • $m$: mass (kg)
  • $k$: spring constantThe force per unit extension of a spring. A measure of the stiffness of the spring. Measured in N m⁻¹. (N \(m^{-1}\))
Worked Example
A 2.0 kg mass is attached to a spring of spring constantThe force per unit extension of a spring. A measure of the stiffness of the spring. Measured in N m⁻¹. 0.9 N \(m^{-1}\). Calculate the frequencyThe number of complete oscillations passing a point per unit time. Measured in hertz (Hz)..
Show Solution
1
Calculate time period

$$T = 2\pi\sqrt{\frac{m}{k}} = 2\pi\sqrt{\frac{2.0}{0.9}} = 9.37 \text{ s}$$

2
Calculate frequencyThe number of complete oscillations passing a point per unit time. Measured in hertz (Hz).

$$f = \frac{1}{T} = \frac{1}{9.37} = 0.11 \text{ Hz}$$

Answer
$f = 0.11$ Hz
Examiner Tips and Tricks
  • Hooke's lawThe extension of a spring is directly proportional to the applied force, provided the limit of proportionality is not exceeded. often gets combined with SHM questions.
  • The spring constantThe force per unit extension of a spring. A measure of the stiffness of the spring. Measured in N m⁻¹. k appears in both $F = kx$ and $T = 2 \pi \sqrt{m/k}$.
  • Make sure you are comfortable moving between these.
Simple Harmonic Motion Overview