How Does Temperature Affect the Resonant Frequency of a Hydrometer?

[capslock]

A glass hydrometer of mass m floats in water of density p. The area of the neck of the hydrometer is A. The resonant frequancy for the vertical oscillations of the hydrometer is given by

f = (1/2\pi) (Apg/m)^(1/2)

Calculuate the fractional change in the resonant frequancy when the temprature changes from 20 to 30 degrees C.

My effort:

(2f\pi)^2 = Apg/m
2ln(2f\pi) = ln(Apg/m)
2ln(2) + 2ln(\pi) + 2ln(f) = ln(A) + ln(P) + ln(g) - ln(m)

I tried to then differentiate but I got in a total kefuffel.

In the question it gives the 'coefficient of linear expansion' for glass and the 'coefficient of volume expansion for water.

Thanks in advance for any kind help you can offer. Best Regards, James.

 

[mezarashi]

First off, I'm not familiar with hydrometers, but as for rates of change, the question is asking you to find something of the form

$$\frac{df}{dT} = g(f, T)$$

and of course hope that the equation turns out to be separable and thus easily solvable by integration over the range you need (i.e. T=20, T=30).

The information given to you, change in volume over time and change in length over time are: dV/dT and dx/dT.

These two variables V, and x are not in the original equation. You need to successively find relationships, for example:

$$\rho = \frac{m}{V}$$
$$\frac{d\rho}{dT} = -\frac{m}{V^2}\frac{dV}{dT}$$

Once you have the right relationships you can plug in.

 

[quicknote]

Dear James,

Thank you for your question. I can definitely help you with computing differential changes in this scenario. First, let's define the variables in the equation provided:

f = resonant frequency for vertical oscillations of the hydrometer
A = area of the neck of the hydrometer
p = density of water
g = acceleration due to gravity
m = mass of the hydrometer

To calculate the fractional change in the resonant frequency, we need to first find the initial value of the frequency at 20 degrees Celsius. We can do this by substituting the given values into the equation:

f1 = (1/2π)(Ap(20)g/m)^(1/2)

Next, we need to find the final value of the frequency at 30 degrees Celsius. To do this, we can use the coefficient of linear expansion for glass to calculate the change in the area of the neck of the hydrometer. This will give us the new value for A at 30 degrees Celsius. We can then substitute this new value into the equation to find the final frequency:

f2 = (1/2π)(A(1+αΔT)p(30)g/m)^(1/2)

Where α is the coefficient of linear expansion for glass and ΔT is the change in temperature (30-20=10 degrees Celsius).

To calculate the fractional change in the resonant frequency, we can use the formula:

Δf/f = (f2-f1)/f1

Substituting the values we calculated, we get:

Δf/f = [(1/2π)(A(1+αΔT)p(30)g/m)^(1/2) - (1/2π)(Ap(20)g/m)^(1/2)] / (1/2π)(Ap(20)g/m)^(1/2)

Simplifying this equation will give us the fractional change in the resonant frequency when the temperature changes from 20 to 30 degrees Celsius. I hope this helps clarify the process of computing differential changes in this scenario. If you have any further questions, please do not hesitate to ask. Best regards,