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.