## Essential University Physics: Volume 1 (4th Edition)

$\sqrt{\frac{k_1+k_2}{m}}$
We use Newton's second law to find: $m \frac{d^2x}{dt^2}=F \\ m \frac{d^2x}{dt^2}= -k_1x-k_2x \\ m \frac{d^2x}{dt^2}= -x(k_1+k_2)$ We see from this equation that k is equal to the sum of $k_1$ and $k_2$, so it follows: $\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{k_1+k_2}{m}}$