1. **Problem statement:** An object performs simple harmonic motion (SHM) with frequency $f$. The distance between the extreme positions (maximum displacements) is $b$. We need to find the speed of the object when it is halfway between its equilibrium position $O$ and an extreme position. 2. **Understanding the problem:** The distance between extremes is $b$, so the amplitude $A$ (maximum displacement from equilibrium) is half of $b$: $$A = \frac{b}{2}$$ 3. **Formula for velocity in SHM:** The velocity $v$ at displacement $x$ from equilibrium is given by: $$v = \omega \sqrt{A^2 - x^2}$$ where angular frequency $\omega = 2 \pi f$. 4. **Halfway position:** Halfway between equilibrium ($x=0$) and extreme ($x=A$) means: $$x = \frac{A}{2} = \frac{b}{4}$$ 5. **Calculate velocity at $x = \frac{b}{4}$:** $$v = 2 \pi f \sqrt{\left(\frac{b}{2}\right)^2 - \left(\frac{b}{4}\right)^2} = 2 \pi f \sqrt{\frac{b^2}{4} - \frac{b^2}{16}}$$ 6. **Simplify inside the square root:** $$\sqrt{\frac{b^2}{4} - \frac{b^2}{16}} = \sqrt{\frac{4b^2}{16} - \frac{b^2}{16}} = \sqrt{\frac{3b^2}{16}} = \frac{b}{4} \sqrt{3}$$ 7. **Substitute back:** $$v = 2 \pi f \times \frac{b}{4} \sqrt{3} = \frac{\sqrt{3}}{2} \pi f b$$ 8. **Final answer:** $$v = \frac{\sqrt{3}}{2} \pi f b$$ This corresponds to option D.