A particle at the end of a spring executes simple harmonic motion with a period $$t_1$$, while the corresponding period for another spring is $$t_2$$. If the period of oscillation with the two springs in series is $$t$$, then

The time period of a mass $m$ attached to a spring with constant $$k$$ is $$t = 2\pi\sqrt{\frac{m}{k}}$$.

Squaring both sides gives $$t^2 = \frac{4\pi^2 m}{k}$$, which can be rearranged to find the spring constant:

$$ \frac{1}{k} = \frac{t^2}{4\pi^2 m} $$

For the first spring:

$$ \frac{1}{k_1} = \frac{t_1^2}{4\pi^2 m} $$

For the second spring:

$$ \frac{1}{k_2} = \frac{t_2^2}{4\pi^2 m} $$

When two springs are connected in series, their equivalent spring constant $$k_{\text{eq}}$$ is given by:

$$ \frac{1}{k_{\text{eq}}} = \frac{1}{k_1} + \frac{1}{k_2} $$

Let $$T$$ be the period of the series combination. Using the same logic as above:

$$ \frac{1}{k_{\text{eq}}} = \frac{T^2}{4\pi^2 m} $$

Substitute the period equations into the series equivalent equation:

$$ \frac{T^2}{4\pi^2 m} = \frac{t_1^2}{4\pi^2 m} + \frac{t_2^2}{4\pi^2 m} $$

Multiply the entire equation by $$4\pi^2 m$$ to simplify:

$$ T^2 = t_1^2 + t_2^2 $$