Question: the period of $T$ of simple pendulum is calculated from the formula $T = 2\pi \sqrt{\frac{L}{g}}$ where $L$ is the length of the pendulum and $g$ is a constant gravitational acceleration. Find the percentage change in the period caused by lengthening the pendulum by 2%
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%.
2. **Formula used:**
The period of a simple pendulum is given by:
$$T = 2\pi \sqrt{\frac{L}{g}}$$
where $L$ is the length and $g$ is the gravitational acceleration (constant).
3. **Understanding the relationship:**
Since $T$ depends on $\sqrt{L}$, a change in $L$ affects $T$ as:
$$T \propto \sqrt{L}$$
4. **Calculate the new length:**
Length is increased by 2%, so new length $L' = L \times (1 + 0.02) = 1.02L$
5. **Calculate the new period:**
$$T' = 2\pi \sqrt{\frac{L'}{g}} = 2\pi \sqrt{\frac{1.02L}{g}} = 2\pi \sqrt{1.02} \sqrt{\frac{L}{g}} = \sqrt{1.02} \times T$$
6. **Calculate the percentage change in period:**
Percentage change = $$\left(\frac{T' - T}{T}\right) \times 100 = (\sqrt{1.02} - 1) \times 100$$
7. **Evaluate $\sqrt{1.02}$:**
$$\sqrt{1.02} \approx 1.00995$$
8. **Final percentage change:**
$$ (1.00995 - 1) \times 100 = 0.995\% \approx 1\%$$
**Answer:** The period increases by approximately 1% when the length is increased by 2%.