Answer :
- The problem provides the length ($L$) and period ($T$) of a pendulum and asks for the acceleration due to gravity ($g$).
- The formula $T = 2\pi\sqrt{\frac{L}{g}}$ is used, and rearranged to solve for $g$: $g = \frac{4\pi^2 L}{T^2}$.
- Substitute $L = 0.369$ m and $T = 1.22$ s into the formula.
- Calculate $g \approx 9.787 \text{ m/s}^2$, so the final answer is $\boxed{9.787 \text{ m/s}^2}$.
### Explanation
1. Understanding the Problem
We are given the length of a pendulum ($L = 36.9$ cm $= 0.369$ m) and its period ($T = 1.22$ s). We need to find the acceleration due to gravity ($g$) at the pendulum's location.
2. Stating the Formula
The period of a simple pendulum is given by the formula: $$T = 2\pi\sqrt{\frac{L}{g}}$$. We need to rearrange this formula to solve for $g$.
3. Rearranging the Formula
Squaring both sides of the equation, we get: $$T^2 = 4\pi^2 \frac{L}{g}$$. Now, we solve for $g$: $$g = \frac{4\pi^2 L}{T^2}$$.
4. Substituting the Values
Now, we substitute the given values of $L$ and $T$ into the equation: $$g = \frac{4\pi^2 (0.369)}{(1.22)^2}$$.
5. Calculating g
Calculating the value of $g$, we find: $$g = \frac{4 \times (3.14159...)^2 \times 0.369}{(1.22)^2} \approx 9.787 \text{ m/s}^2$$.
6. Final Answer
Therefore, the acceleration due to gravity at the pendulum's location is approximately $9.787 \text{ m/s}^2$.
### Examples
Understanding the acceleration due to gravity is crucial in various fields, such as designing clocks, understanding projectile motion, and even in geological surveys to map subsurface densities. For instance, knowing the precise value of 'g' allows engineers to calibrate pendulum clocks accurately, ensuring they keep time correctly. Similarly, in sports, understanding gravity's influence helps athletes and coaches optimize performance in activities like long jump or pole vault.
- The formula $T = 2\pi\sqrt{\frac{L}{g}}$ is used, and rearranged to solve for $g$: $g = \frac{4\pi^2 L}{T^2}$.
- Substitute $L = 0.369$ m and $T = 1.22$ s into the formula.
- Calculate $g \approx 9.787 \text{ m/s}^2$, so the final answer is $\boxed{9.787 \text{ m/s}^2}$.
### Explanation
1. Understanding the Problem
We are given the length of a pendulum ($L = 36.9$ cm $= 0.369$ m) and its period ($T = 1.22$ s). We need to find the acceleration due to gravity ($g$) at the pendulum's location.
2. Stating the Formula
The period of a simple pendulum is given by the formula: $$T = 2\pi\sqrt{\frac{L}{g}}$$. We need to rearrange this formula to solve for $g$.
3. Rearranging the Formula
Squaring both sides of the equation, we get: $$T^2 = 4\pi^2 \frac{L}{g}$$. Now, we solve for $g$: $$g = \frac{4\pi^2 L}{T^2}$$.
4. Substituting the Values
Now, we substitute the given values of $L$ and $T$ into the equation: $$g = \frac{4\pi^2 (0.369)}{(1.22)^2}$$.
5. Calculating g
Calculating the value of $g$, we find: $$g = \frac{4 \times (3.14159...)^2 \times 0.369}{(1.22)^2} \approx 9.787 \text{ m/s}^2$$.
6. Final Answer
Therefore, the acceleration due to gravity at the pendulum's location is approximately $9.787 \text{ m/s}^2$.
### Examples
Understanding the acceleration due to gravity is crucial in various fields, such as designing clocks, understanding projectile motion, and even in geological surveys to map subsurface densities. For instance, knowing the precise value of 'g' allows engineers to calibrate pendulum clocks accurately, ensuring they keep time correctly. Similarly, in sports, understanding gravity's influence helps athletes and coaches optimize performance in activities like long jump or pole vault.