Beats — Question 7

Question 7

Thermal drift turns a tuned pipe into a beat thermometer

A fixed-length ideal pipe is tuned to a stable $440\,\mathrm{Hz}$ reference at $T_0=300\,\mathrm K$. Its frequency obeys $f(T)=440\sqrt{T/300}$. The pipe warms to $303\,\mathrm K$; neglect dimensional expansion.

1. Find the resulting beat frequency.

2. Derive the small-temperature-change relation between beat rate and temperature change. Use it to estimate the positive temperature change corresponding to $1.0\,\mathrm{Hz}$ beats.

Question 7 - Solution

Thermal drift turns a tuned pipe into a beat thermometer

(a) Method and calculation

Warming raises the pipe frequency, so$$f_b=440\left(\sqrt{303/300}-1\right)=2.19453\,\mathrm{Hz}.$$The positive sign is known from the stated warming; the beat count alone would not supply it.

(b) Interpretation and further calculation

Linearizing near $T_0$,$$f(T)\simeq f_0\left(1+\frac{\Delta T}{2T_0}\right),\quad f_b\simeq\frac{f_0|\Delta T|}{2T_0}.$$For warming and $f_b=1.0\,\mathrm{Hz}$, $\Delta T\simeq600/440=1.364\,\mathrm K$. The exact result is $300[(441/440)^2-1]=1.3652\,\mathrm K$.

Physical check

The approximation requires $|\Delta T|\ll T_0$. Gas composition changes or pipe-length changes would confound the thermal inference.