Dimensional Analysis of an oscillation

[Zack K]

Homework Statement

The period of oscillation of a nonlinear oscillator depends on the mass m, with dimensions M; a restoring force constant k with dimensions of ML^-2T^-2, and the Amplitude A, with dimensions L. Use dimensional analysis to show what the period of oscillation would be proportional to.

Homework Equations

N/A

The Attempt at a Solution

So we know that P∝ m^ak^bA^c(a,b and c are numerical exponents to be determined). So rewriting P(period of oscillation) into its dimensions, we get MxML^-2T^-2L which then simplifies to M²L^-1T^-2. That is equal to M^a(M/L²T²)^bL^c. Then multiplying our unknown exponents into our dimensions and setting them equal to the dimension exponents on the right, we should get: a+b=2, -2b+c=-1, -2b=-2. Solving for each variable we get: b=1, a=2, c=1. Then redefining our dimensions into their respective units I got: P∝m²AK. The problem is this is not a possible solution since its a multiple choice question.

 

[Orodruin]

A period is a time.

 

[Zack K]

A period is a time.

Oh wow I feel stupid now. Anyways besides that, I thought to get the dimensions of the left hand side(in this case period) was given by multiplying the dimensions of the variables on your right hand side. Why is this not true?

 

[Orodruin]

Oh wow I feel stupid now. Anyways besides that, I thought to get the dimensions of the left hand side(in this case period) was given by multiplying the dimensions of the variables on your right hand side. Why is this not true?

It is true. That is why you can identify T with the products of the dimension on the RHS. What is not true is, eg, that ##[m^a]=[m]##, which is what you were doing. Your algebra was also off, you should have gotten a=b=c=1 with your (faulty) method, because you essentially assumed that the period could be written proportional to mkA.

You may also want to double check the dimensions you have for the force constant ...