Algebra - Quartz Tuning Fork Watch

[CallMeShady]

Algebra - "Quartz Tuning Fork" Watch

Homework Statement

Many modern watches operate based on a small quartz tuning fork which transduces a mechanical oscillation into an electrical signal. The frequency of the tuning fork is
inversely proportional to sqrt(l), with l being the length of the fork. If the watch keeps perfect time at 20°C, what is the fractional gain or less in time for a quartz tuning fork that is 6mm long at:
(a) 0°C
(b) 30°C
Hint: try working this out algebraically. The changes are small and prone to rounding issues.

α_Quartz = 0.59 × 10^-6/°C

Homework Equations

Not sure if the following equations are useful, but my topic of study right now is based on thermal expansion, as well as calculations based on heat energy (Q = mcΔT).
ΔL = αL₀ΔT (linear thermal expansion)

If a body has length L₀ at temperature T₀, then its length L at a temperature T = T₀ + ΔT is:
L = L₀ + ΔL = L₀ + αL₀ΔT = L₀(1 + αΔT)

Q = mcΔT

The Attempt at a Solution

This question appears to me on a topic that I haven't touched so far in my Physics class; however, I know that the topics I am studying is interconnected with this question somehow. From what I read from the question, I concluded that frequency = 1/√length but I don't know how to incorporate that into an equation in which I can find a "ratio." I am stuck here and do not know how to further approach this problem.

 

[voko]

What is the frequency at the higher temperature? What is it ratio with that of the "perfect" frequency?

 

[CallMeShady]

I understand that I need to compare the ratio of the frequencies at 0°C and 30°C with that of the "perfect" frequency at 20°C, however, I don't know how to express both frequencies in equations that I can compare them to each other. Can you shine some light on that please?

 

[voko]

The frequency is said to be inversely proportional to $\sqrt {l}$, which means it is a product of some constant $c$ and $\frac 1 {\sqrt {l} }$, i.e., $f = \frac {c} {\sqrt {l} }$. $l$ changes with temperature, $c$ does not.