]> Exercise 3

# Presentation, Exercise 3

## Question

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.

Find how T depends from the parameters by using physical dimensions!

## Notation

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]

## Part 1

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

• [T]=Θ
• [m]=Μ
• [l]=Λ
• [g]=ΛΘ−2

## Part 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

$\Theta ={Μ}^{\alpha }{\Lambda }^{\beta }{\left(\Lambda {\Theta }^{-2}\right)}^{\gamma }$ .

## Part 4

The comparison of the powers of part 3 gives:

• α=0
• β+γ=0
• −2γ=1

## Part 5

The solution of part 4 is

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

and finally

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

## Note

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

## Score

Parts 1 and 2 are worth 1 point each.

Parts 3 and 5 are worth 3 points each.
Part 4 is worth 2 points.