Pendulum Period Change Bb5A57

1. **State the problem:** We need to find the percentage change in the period $T$ of a simple pendulum when its length $L$ is increased by 2%, using the formula: $$T = 2\pi \sqrt{\frac{L}{g}}$$ where $g$ is constant. 2. **Formula and rule:** The period $T$ depends on the square root of $L$. For small percentage changes, if $L$ changes by $\Delta L\%$, then the percentage change in $T$ is approximately half of that because of the square root. 3. **Calculate the percentage change:** Let the original length be $L$. The new length is $L' = L \times (1 + 0.02) = 1.02L$. The new period is: $$T' = 2\pi \sqrt{\frac{L'}{g}} = 2\pi \sqrt{\frac{1.02L}{g}} = 2\pi \sqrt{1.02} \sqrt{\frac{L}{g}} = T \times \sqrt{1.02}$$ 4. **Approximate $\sqrt{1.02}$ using binomial expansion for small $x$:** $$\sqrt{1+x} \approx 1 + \frac{x}{2}$$ Here, $x = 0.02$, so: $$\sqrt{1.02} \approx 1 + \frac{0.02}{2} = 1 + 0.01 = 1.01$$ 5. **Percentage change in $T$:** $$\frac{T' - T}{T} \times 100\% = (1.01 - 1) \times 100\% = 1\%$$ **Final answer:** The period $T$ increases by approximately 1% when the length $L$ is increased by 2%.