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.

 

[Nickie]

I would first like to clarify that the topic being discussed here is not algebra, but rather the concept of thermal expansion and its effect on the frequency of a quartz tuning fork used in watches. Algebra is simply the mathematical tool used to solve the problem at hand.

To solve this problem, we first need to understand the relationship between the length of the tuning fork and its frequency. As stated in the problem, the frequency is inversely proportional to the square root of the length. This can be expressed as f ∝ 1/√l, where f represents frequency and l represents length.

Next, we need to take into account the effect of temperature on the length of the tuning fork. As the temperature increases, the length of the tuning fork will also increase due to thermal expansion. This can be expressed as Δl = αlΔT, where Δl represents the change in length, α represents the coefficient of linear thermal expansion for quartz, and ΔT represents the change in temperature.

Now, we can combine these two equations to find the relationship between frequency and temperature. We get f ∝ 1/√(l + αlΔT), which can be simplified to f ∝ 1/√l(1 + αΔT). Since the watch keeps perfect time at 20°C, we can set the frequency at this temperature as the reference frequency, f0. This gives us the equation f0 ∝ 1/√l0, where l0 represents the length at 20°C.

To find the fractional gain or loss in time at different temperatures, we can use the equation Δf/f0 = (f-f0)/f0. Plugging in the equation for f0, we get Δf/f0 = (√l0 - √l)/√l0. Substituting the given values, we get Δf/f0 = (√6 - √6.36)/√6. At 0°C, ΔT = -20°C, so we get Δf/f0 = (√6 - √6.36)/√6 = -0.02%. Similarly, at 30°C, ΔT = 10°C, so we get Δf/f0 = (√6 - √6.18)/√6 = -0.004%.

In conclusion, at 0°C, the fractional gain