1. Let's start by stating the problem: We want to prove the approximate formula for the wave speed $v$ of a sinusoidal wave on a string, given by $$v \approx \sqrt{\frac{gL}{2\pi}}.$$ Here, $g$ is the acceleration due to gravity and $L$ is the wavelength. 2. The formula relates to the speed of a wave traveling on a string or surface under tension influenced by gravity, such as water waves or waves on a string. 3. The wave speed $v$ is generally related to the wavelength $L$ and the period $T$ by the formula $$v = \frac{L}{T}.$$ The period $T$ is the time it takes for one full wave cycle. 4. For waves influenced by gravity, the period $T$ can be approximated by the formula $$T \approx 2\pi \sqrt{\frac{L}{g}}.$$ This comes from the physics of pendulum-like motion or gravity waves. 5. Substitute the expression for $T$ into the wave speed formula: $$v = \frac{L}{T} = \frac{L}{2\pi \sqrt{\frac{L}{g}}} = \frac{L}{2\pi} \cdot \frac{1}{\sqrt{\frac{L}{g}}}.$$ 6. Simplify the expression inside the square root: $$\frac{1}{\sqrt{\frac{L}{g}}} = \sqrt{\frac{g}{L}}.$$ 7. So the wave speed becomes: $$v = \frac{L}{2\pi} \cdot \sqrt{\frac{g}{L}} = \frac{L}{2\pi} \cdot \frac{\sqrt{g}}{\sqrt{L}} = \frac{\sqrt{g} \cdot L}{2\pi \sqrt{L}} = \frac{\sqrt{g} \cdot \sqrt{L} \cdot \sqrt{L}}{2\pi \sqrt{L}}.$$ 8. Cancel one $\sqrt{L}$ from numerator and denominator: $$v = \frac{\sqrt{g} \cdot \sqrt{L}}{2\pi}.$$ 9. Rearranging gives the approximate formula: $$v \approx \sqrt{\frac{gL}{2\pi}}.$$ 10. This shows that the wave speed depends on the square root of the product of gravitational acceleration and wavelength divided by $2\pi$. This derivation uses the relationship between period and wavelength for gravity-influenced waves and basic algebraic manipulation to arrive at the formula.