Question

A force of 500N is used to stretch a spring with a 0.5kg mass attached to it by 0.35m (a)

what is the value of the spring constant (b) calculate the f of the oscillator.

Answer 1

(a)

$$F_{stretch}=kx,$$ $$k=\dfrac{F_{stretch}}{x}=\dfrac{500\ N}{0.35\ m}=1428.6\ \dfrac{N}{m}.$$

(b)

$$f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}},$$ $$f=\dfrac{1}{2\pi}\sqrt{\dfrac{1428.6\ \dfrac{N}{m}}{0.5\ kg}}=8.5\ Hz.$$