# Algebra - Quartz Tuning Fork Watch

1. Sep 23, 2012

Algebra - "Quartz Tuning Fork" Watch

1. The problem statement, all variables and given/known data
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.

αQuartz = 0.59 × 10-6/°C

2. Relevant 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

3. 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.

2. Sep 23, 2012

### voko

Re: Algebra - "Quartz Tuning Fork" Watch

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

3. Sep 23, 2012

Re: Algebra - "Quartz Tuning Fork" Watch

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?

4. Sep 23, 2012

### voko

Re: Algebra - "Quartz Tuning Fork" Watch

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.