Can Someone Explain Proportional Reasoning to Me?

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Proportional reasoning involves understanding the relationship between quantities and how changing one affects the other. The discussion centers on finding the square root of a variable L, specifically when L equals 2, which results in √L being equal to √2. Participants suggest replacing L with 2L to explore how this change impacts the original equation. The emphasis is on simplifying the problem and expressing the mathematical relationship in clear terms. Overall, the conversation highlights the importance of clarity and straightforwardness in approaching proportional reasoning.
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Homework Statement
The question is asking what would happen to T if I double L. The answer has to be in words (i.e: “T is cut in half”)
Relevant Equations
T = 2𝝅√L/g
I already tried plugging in values I saw from example problems and tried to find the correlation between the two solutions I would get, but thus far haven’t been able to find any. Maybe I’m overcomplicating this or looking at this the wrong way?
 
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If L=1, what is √L ?
 
Replace L with 2L and compare the result to your original equation. Then find a way to express the mathematical relationship between those values in words.

Don’t overthink it.
 
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berkeman said:
If L=1, what is √L ?
it should still be 1
 
Oh sorry, I apologize. It's been a long day. Let me rephrase...

If L=2, what is √L ?

And the suggestion by @Emmo Amaranth is equivalent to my corrected question, BTW. :smile:
 
berkeman said:
Oh sorry, I apologize. It's been a long day. Let me rephrase...

If L=2, what is √L ?

And the suggestion by @Emmo Amaranth is equivalent to my corrected question, BTW. :smile:
no worries, that’s understandable! i think i’m catching your drift. thank you for your quick response and your help :smile:
 
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Emmo Amaranth said:
Replace L with 2L and compare the result to your original equation. Then find a way to express the mathematical relationship between those values in words.

Don’t overthink it.
thank you!
 
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You're quite welcome! Glad to help.
 
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