What are the powers required for dimensional consistency in a simple pendulum?

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The discussion focuses on determining the powers p and q for dimensional consistency in the equation for the period T of a simple pendulum, expressed as T=2(3.14)L^(P)g^(q). To achieve dimensional consistency, the units of time on the left side must match the right side, which involves manipulating the units of length and gravity. The correct values found are p=1/2 and q=-1/2, leading to the conclusion that T simplifies to T, confirming the dimensional analysis. This illustrates the importance of dimensional consistency in physics problems. The conversation highlights the value of understanding the underlying principles rather than just solving for the variables.
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The period T of a simple pendulm is he amount of time required for it to undergo one complete oscillation. If the length of the pendulm is L and the acceleration of gravity is g, then T is given by T=2(3.14)L^(P)g^q Find the powers p and q required for dimensional consistency.
 
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Basically, they're asking you to make the units on the right side of the equation match the units on the left. You have units of TIME on the left side of the equation, so you need to end up with the same thing on the right. On the right, you have units of LENGTH (L) raised to some power {p} and units of LENGTH / TIME2 (g) raised to some other power {q} You just need to find a value for {p} and {q} which leaves you with TIME when you're done.
 
A good problem aimed to illustrate the power of dimensional analysis in physics... :approve: It would be a shame if you eventually solved but didin't catch the idea behind it... :rolleyes:

Daniel.
 
So then it would be T=(L)^1/2(L/T^2)^(-1/2)
 
U should present us with the answer
p=...
q=...

Only then we can be sure you solved the problem...

Daniel.
 
T= 2(3.14) L^(P) g^(q)
T=(L)^p (L/T^2)^q
T=(L)^(1/2)(L/T^2)^(-1/2)
T=(L)^(1/2)(T^2/L)^(1/2)
T=(T^2)^(1/2)
T=T
 
thank you very much
 

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