]> Exercise 4

Math Animated™
Mathematical Introduction for Physics and Engineering
by Samuel Dagan (Copyright © 2007-2020)

Chapter 1: Differentiation; Section 1: Real Numbers; page 4

Presentation, Exercise 4


The force of a non-linear spring is:

F=q x 3    
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.


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

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:

[ q ]=[ F ] [ l ] 3 =( ΜΛ Θ 2 )( Λ 3 )=Μ Λ 2 Θ 2

The rest of the variables have obviously the dimensions:

Part 2

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

T=c m α l β q γ , where c is a dimensionless constant.

Part 3

From part 1 and part 2 one obtains

Θ= Μ α Λ β ( Μ Λ 2 Θ 2 ) γ .

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

  • α= 1 2
  • β=1
  • γ= 1 2

and finally

T= c l m q .


Part 1 is worth 3 points.

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