Can u help me sketch a graph

Here is a sketch of the graph of $m$ (on the vertical axis) against $T^2$ (on the horizontal axis) for both the single spring and the parallel springs.

• The relationship for a single spring is $m = \left(\frac{k_1}{4\pi^2}\right) T^2$.
• The relationship for two identical springs in parallel is $m = \left(\frac{2k_1}{4\pi^2}\right) T_0^2$.

Both equations represent straight lines passing through the origin. The slope for the parallel springs ($s_2 = \frac{2k_1}{4\pi^2}$) is twice the slope for the single spring ($s_1 = \frac{k_1}{4\pi^2}$), meaning the line for the parallel springs will be steeper.

\begin{tikzpicture}[scale=0.8] % Axes \draw[->] (0,0) -- (7,0) node[below] {$T^2 \text{ (s}^2\text{)}$}; \draw[->] (0,0) -- (0,7) node[left] {$m \text{ (kg)}$}; % Origin label \node at (-0.3,-0.3) {0}; % Line for single spring (less steep) \draw[blue, thick] (0,0) -- (6,3) node[above right, blue] {Single Spring ($m$ vs $T^2$)}; % Line for parallel springs (steeper) \draw[red, thick] (0,0) -- (3,6) node[above right, red] {Parallel Springs ($m$ vs $T_0^2$)}; % Example ticks (optional, but good for a sketch) \foreach \x in {1,2,3,4,5,6} \draw (\x,0.1) -- (\x,-0.1) node[below] {\x}; \foreach \y in {1,2,3,4,5,6} \draw (0.1,\y) -- (-0.1,\y) node[left] {\y}; \end{tikzpicture}