Problem — Period and frequency of a mass-spring system

Problem

A mass $m = 0{,}40\;\text{kg}$ is attached to a spring of elastic constant $K = 40\;\text{N/m}$. Calculate the period and frequency of the small oscillations.

Solution

The angular frequency of the mass-spring system is:

$\omega = \sqrt{\frac{K}{m}} = \sqrt{\frac{40}{0{,}40}} = \sqrt{100} = 10\;\text{rad/s}$

The period is $T = 2\pi/\omega$:

$T = \frac{2\pi}{10} \approx 0{,}628\;\text{s}$

$\ev{T \approx 0{,}628\;\text{s}}$

The frequency is the inverse of the period:

$f = \frac{1}{T} = \frac{1}{0{,}628}\;\text{Hz} \approx 1{,}59\;\text{Hz}$

$\ev{f \approx 1{,}59\;\text{Hz}}$

Links

Topics: Oscillations and harmonic motion Concepts: Simple harmonic motion · Period and frequency Objects: Spring