# I don't get this am confused

1. Jan 21, 2005

### dg_5021

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.

2. Jan 21, 2005

### Grogs

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.

3. Jan 21, 2005

### dextercioby

A good problem aimed to illustrate the power of dimensional analysis in physics... It would be a shame if you eventually solved but didin't catch the idea behind it...

Daniel.

4. Jan 23, 2005

### dg_5021

So then it would be T=(L)^1/2(L/T^2)^(-1/2)

5. Jan 23, 2005

### dextercioby

U should present us with the answer
$$p=...$$
$$q=...$$

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

Daniel.

6. Jan 23, 2005

### dg_5021

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

7. Jan 23, 2005

### dextercioby

Staggering...Congratulations!!!

Daniel.

8. Jan 23, 2005

### dg_5021

thank you very much