A mass of 0.90 kg is suspended from the lower end of a light spring of stiffness 80 N m^-1 - AQA - A-Level Physics - Question 25 - 2020 - Paper 1

The angular frequency ($\omega$) of a mass-spring system is given by the formula:

$\omega = \sqrt{\frac{k}{m}}$

where:

• $k = 80 \, \text{N m}^{-1}$ (stiffness of the spring)
• $m = 0.90 \, \text{kg}$ (mass)

Substituting in the values, we get:

$\omega = \sqrt{\frac{80}{0.90}} = \sqrt{88.89} \approx 9.43 \, \text{rad s}^{-1}$