Dimensional analysis and exponents

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The discussion revolves around finding the powers p and q for dimensional consistency in the equation for the period T of a simple pendulum. The equation is expressed as T = 2 (pi) L^p g^q, where L is the length in meters and g is the acceleration due to gravity in meters per second squared. To achieve dimensional consistency, the dimensions on both sides of the equation must match, leading to the relationships p + q = 0 and -2q = 1. By solving these equations, the values of p and q can be determined. This analysis is crucial for understanding the physics of pendulums and dimensional analysis in general.
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I decided to take physics this year, my senior year in high school, and my teacher has decided he isn't going to explain things before he assigns us homework. Any and all help is appreciated.

1. The period T of a simple pendulum is the amount of time required for it to undergo one complete oscillation. If the length of the pendulum is L and the acceleration of gravity is g, then T is given by:
2. T = 2 (pi) Lp gq
3. T = 2 (pi) Lp (L/T2)q

Find the powers p and q required for dimensional consistency.
 
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Hint. L is in meters, g is in meter/second^2 and period T is in seconds. Now just do the algebra.
 
addedline8 said:
I decided to take physics this year, my senior year in high school, and my teacher has decided he isn't going to explain things before he assigns us homework. Any and all help is appreciated.

1. The period T of a simple pendulum is the amount of time required for it to undergo one complete oscillation. If the length of the pendulum is L and the acceleration of gravity is g, then T is given by:



2. T = 2 (pi) Lp gq



3. T = 2 (pi) Lp (L/T2)q

Find the powers p and q required for dimensional consistency.

keep going the way you have started..

T = 2 (pi) Lp (L/T2)q

T = 2 (pi) Lp Lq/T2q

T = 2 (pi) Lp+q.T -2q

Now compare dimensions left and right.

What must -2q equal?

What must p+q equal?
 
Thanks a lot guys. I appreciate the help.
 
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