1. **Problem Statement:** A particle moves with Simple Harmonic Motion (SHM) and has speeds 3 m/s and 5 m/s at distances 7 m and 6 m respectively. Show that the period of the SHM is $\frac{1}{2}\pi\sqrt{3}$. 2. **Formula and Concepts:** In SHM, the velocity $v$ at displacement $x$ is given by $$v = \pm \omega \sqrt{A^2 - x^2}$$ where $\omega$ is the angular frequency and $A$ is the amplitude. 3. **Using the given data:** For $x=7$ m, $v=3$ m/s, so $$3 = \omega \sqrt{A^2 - 7^2} = \omega \sqrt{A^2 - 49}$$ For $x=6$ m, $v=5$ m/s, so $$5 = \omega \sqrt{A^2 - 6^2} = \omega \sqrt{A^2 - 36}$$ 4. **Square both equations:** $$9 = \omega^2 (A^2 - 49)$$ $$25 = \omega^2 (A^2 - 36)$$ 5. **Subtract the first from the second:** $$25 - 9 = \omega^2 (A^2 - 36) - \omega^2 (A^2 - 49)$$ $$16 = \omega^2 (A^2 - 36 - A^2 + 49) = \omega^2 (13)$$ 6. **Solve for $\omega^2$:** $$\omega^2 = \frac{16}{13}$$ 7. **Use one equation to find $A^2$:** Using $9 = \omega^2 (A^2 - 49)$, substitute $\omega^2$ $$9 = \frac{16}{13} (A^2 - 49)$$ Multiply both sides by $\frac{13}{16}$: $$\frac{9 \times 13}{16} = A^2 - 49$$ $$\frac{117}{16} = A^2 - 49$$ Add 49 to both sides: $$A^2 = 49 + \frac{117}{16} = \frac{784}{16} + \frac{117}{16} = \frac{901}{16}$$ 8. **Calculate the period $T$:** $$T = \frac{2\pi}{\omega} = 2\pi \times \frac{1}{\sqrt{\omega^2}} = 2\pi \times \frac{1}{\sqrt{\frac{16}{13}}} = 2\pi \times \sqrt{\frac{13}{16}} = 2\pi \times \frac{\sqrt{13}}{4} = \frac{\pi}{2} \sqrt{13}$$ 9. **Simplify the period:** The problem states the period is $\frac{1}{2}\pi\sqrt{3}$, but our calculation shows $\frac{\pi}{2} \sqrt{13}$. Re-examining the problem, the speeds and distances given imply the period is $\frac{\pi}{2} \sqrt{13}$, which is consistent with the data. Possibly the problem's given period is approximate or symbolic. **Final answer:** The period of the SHM is $$T = \frac{\pi}{2} \sqrt{13}$$