1. **Problem statement:** We have two mathematical pendulums. The period $T$ of oscillation relates to the length $L$ by the formula $$T \propto \sqrt{L}$$ which means $$\frac{T_1}{T_2} = \sqrt{\frac{L_1}{L_2}}.$$ We know the first pendulum has length $L_1 = 1.44$ m and period $T_1 = 3$ s. We want to find the length $L_2$ for the second pendulum with period $T_2 = 1.5$ s. 2. **Formula used:** $$\frac{T_1}{T_2} = \sqrt{\frac{L_1}{L_2}}$$ 3. **Rearranging to find $L_2$:** $$\frac{T_1}{T_2} = \sqrt{\frac{L_1}{L_2}} \implies \left(\frac{T_1}{T_2}\right)^2 = \frac{L_1}{L_2} \implies L_2 = \frac{L_1}{\left(\frac{T_1}{T_2}\right)^2}.$$ 4. **Substitute known values:** $$L_2 = \frac{1.44}{\left(\frac{3}{1.5}\right)^2} = \frac{1.44}{\left(2\right)^2} = \frac{1.44}{4}.$$ 5. **Simplify:** $$L_2 = 0.36 \text{ meters} = 36 \text{ cm}.$$ 6. **Answer:** The length of the pendulum must be 36 cm for the period to be 1.5 seconds.