Mathematical Introduction for Physics and Engineering

by Samuel Dagan (Copyright © 2007-2020)

**
A simple pendulum is swinging. In principle the time period of a swing T could depend on the following parameters: the swinging mass m, the length of the pendulum l and the constant acceleration on the surface of the earth g.**

We'll follow the notation for the 3 basic physical dimensions already used.

*Μ*denotes the dimension of a mass*m*: [*m*]*Θ*denotes the dimension of time*t*: [*t*]*Λ*denotes the dimension of length*l*: [*l*]

By use of this notation, the dimensions of the variables from this exercise are

- [
*T*]=*Θ* - [
*m*]=*Μ* - [
*l*]=*Λ* - [
*g*]=*Λ**Θ*^{−2}

We assume that the period *T* is expressed by the other variables by

$T=c{m}^{\alpha}{l}^{\beta}{g}^{\gamma}$ ,
where *c* is a dimensionless constant.
## Part 3

From part 1 and part 2 one obtains

$\Theta ={{\rm M}}^{\alpha}{\Lambda}^{\beta}{\left(\Lambda {\Theta}^{-2}\right)}^{\gamma}$ .
## Part 4

## Part 5

## Note

## Score

The comparison of the powers of part 3 gives:

*α*=0*β*+*γ*=0- −2
*γ*=1

The solution of part 4 is

- $\alpha =0$
- $\beta ={\scriptstyle \frac{1}{2}}$
- $\gamma =-{\scriptstyle \frac{1}{2}}$

and finally

**$T=c\sqrt{\frac{l}{g}}$**

For oscilations with small amplitude $c=2\pi $ .

Parts 1 and 2 are worth 1 point each.

Parts 3 and 5 are worth 3 points each.

Part 4 is worth 2 points.