]> Exercise 4

# Presentation, Exercise 4

## Question

The force of a non-linear spring is:

where x is the distance from the origin and  q>0 is the constant of the spring. After attaching a mass m to the spring and stretching it to a length length  l  from the origin, the mass starts a periodic motion with period of time T.
Find out how  T  depends on  l, q and m  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

The dimensions of q can be expressed by the dimensions of the force F and the elongation l according to the above given relation:

$\left[q\right]\text{\hspace{0.17em}}=\text{\hspace{0.17em}}\left[F\right]{\left[l\right]}^{-3}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}\left(Μ\Lambda {\Theta }^{-2}\right)\left({\Lambda }^{-3}\right)\text{\hspace{0.17em}}=\text{\hspace{0.17em}}Μ{\Lambda }^{-2}{\Theta }^{-2}$

The rest of the variables have obviously the dimensions:

• [T]=Θ
• [m]=Μ
• [l]=Λ

## Part 2

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

$T=c{m}^{\alpha }{l}^{\beta }{q}^{\gamma }$ , where c is a dimensionless constant.

## Part 3

From part 1 and part 2 one obtains

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

## Part 4

The comparison of the powers of part 3 gives:

• α+γ=0
• β−2γ=0
• −2γ=1

## Part 5

The solution of part 4 is

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

and finally

$T=\frac{c}{l}\sqrt{\frac{m}{q}}$ .

## Score

Part 1 is worth 3 points.

Parts 2 is worth 1 point.
Parts 3,4 and 5 are worth 2 points each.