Beats — Question 3

Question 3

A controlled tuning change resolves the sign of a beat offset

An unknown tone produces $6\,\mathrm{Hz}$ beats with a $440\,\mathrm{Hz}$ reference. Increasing the unknown frequency by exactly $2\,\mathrm{Hz}$ changes the beat rate to $4\,\mathrm{Hz}$. The reference remains fixed.

1. Find the original unknown frequency and its value after the change.

2. If it comes from a fixed-length string with $f\propto\sqrt F$, find the tension multiplier required to tune the original tone exactly to the reference.

Question 3 - Solution

A controlled tuning change resolves the sign of a beat offset

(a) Method and calculation

The initial observation permits $f=440\pm6$, or $434$ and $446\,\mathrm{Hz}$. After the known change these would become $436$ and $448\,\mathrm{Hz}$, giving beat rates $4$ and $8\,\mathrm{Hz}$ respectively. Thus$$f_{\mathrm{initial}}=434\,\mathrm{Hz},\qquad f_{\mathrm{changed}}=436\,\mathrm{Hz}.$$

(b) Interpretation and further calculation

The target tension follows from the squared frequency ratio:$$\frac{F_{\mathrm{tuned}}}{F_{\mathrm{initial}}}=\left(\frac{440}{434}\right)^2=1.027841.$$Increase tension by about $2.78\%$. A beat count alone supplies $|f-f_{\mathrm{ref}}|$; the controlled change determines which side of the reference is occupied.

Physical check

Increasing tension must raise the pitch of this initially flat string, consistent with the observed reduction in beat rate.