IDNLearn.com is designed to help you find the answers you need quickly and easily. Join our Q&A platform to access reliable and detailed answers from experts in various fields.
Sagot :
To analyze the relationship between the time period [tex]\( T \)[/tex] of a simple pendulum and the gravitational field [tex]\( g \)[/tex], let's start by looking at the given formula:
[tex]\[ T = 2 \pi \sqrt{\frac{L}{g}} \][/tex]
We will break this down step by step:
1. Understanding the Formula: The formula [tex]\( T = 2 \pi \sqrt{\frac{L}{g}} \)[/tex] indicates that the time period [tex]\( T \)[/tex] depends on both the length of the pendulum [tex]\( L \)[/tex] and the gravitational field [tex]\( g \)[/tex].
2. Rewriting the Formula: To analyze how [tex]\( T \)[/tex] is related to [tex]\( g \)[/tex], we can rewrite the formula to better see the dependency:
[tex]\[ T = 2 \pi \sqrt{L} \cdot \sqrt{\frac{1}{g}} \][/tex]
3. Isolating the Dependency on [tex]\( g \)[/tex]: From the rewritten formula, it’s clear that [tex]\( T \)[/tex] involves [tex]\( g \)[/tex] in the form of [tex]\( \sqrt{\frac{1}{g}} \)[/tex]. Thus, we can represent this as:
[tex]\[ T ∝ \sqrt{\frac{1}{g}} \][/tex]
4. Interpreting the Proportional Relationship: This means that [tex]\( T \)[/tex] is directly proportional to the inverse of the square root of [tex]\( g \)[/tex]. In mathematical terms:
[tex]\[ T \propto \frac{1}{\sqrt{g}} \][/tex]
5. Conclusion: Therefore, the statement [tex]\( T \propto g \)[/tex] is incorrect. From our analysis, the correct proportionality relationship is:
[tex]\[ T \propto \frac{1}{\sqrt{g}} \][/tex]
In summary, the time period [tex]\( T \)[/tex] of a simple pendulum is proportional to the inverse of the square root of the gravitational field [tex]\( g \)[/tex], and not directly proportional to [tex]\( g \)[/tex].
[tex]\[ T = 2 \pi \sqrt{\frac{L}{g}} \][/tex]
We will break this down step by step:
1. Understanding the Formula: The formula [tex]\( T = 2 \pi \sqrt{\frac{L}{g}} \)[/tex] indicates that the time period [tex]\( T \)[/tex] depends on both the length of the pendulum [tex]\( L \)[/tex] and the gravitational field [tex]\( g \)[/tex].
2. Rewriting the Formula: To analyze how [tex]\( T \)[/tex] is related to [tex]\( g \)[/tex], we can rewrite the formula to better see the dependency:
[tex]\[ T = 2 \pi \sqrt{L} \cdot \sqrt{\frac{1}{g}} \][/tex]
3. Isolating the Dependency on [tex]\( g \)[/tex]: From the rewritten formula, it’s clear that [tex]\( T \)[/tex] involves [tex]\( g \)[/tex] in the form of [tex]\( \sqrt{\frac{1}{g}} \)[/tex]. Thus, we can represent this as:
[tex]\[ T ∝ \sqrt{\frac{1}{g}} \][/tex]
4. Interpreting the Proportional Relationship: This means that [tex]\( T \)[/tex] is directly proportional to the inverse of the square root of [tex]\( g \)[/tex]. In mathematical terms:
[tex]\[ T \propto \frac{1}{\sqrt{g}} \][/tex]
5. Conclusion: Therefore, the statement [tex]\( T \propto g \)[/tex] is incorrect. From our analysis, the correct proportionality relationship is:
[tex]\[ T \propto \frac{1}{\sqrt{g}} \][/tex]
In summary, the time period [tex]\( T \)[/tex] of a simple pendulum is proportional to the inverse of the square root of the gravitational field [tex]\( g \)[/tex], and not directly proportional to [tex]\( g \)[/tex].
Your participation means a lot to us. Keep sharing information and solutions. This community grows thanks to the amazing contributions from members like you. Thank you for trusting IDNLearn.com with your questions. Visit us again for clear, concise, and accurate answers.