You have five tuning forks that oscillate at close but different frequencies. What are the (a) maximum and, (b) minimum number of different beat frequencies you can produce by sounding the forks two at a time, depending on how the frequencies differ?

We can find the maximum possible number of different frequencies produced by sounding the forks at a time by taking the combination. Then using the concept of unique beat frequency, we can find the minimum possible number of different frequencies produced by sounding the forks at a time.

Formula:

The combination formula for selection r values from n values,

${\mathit{n}}_{{\mathbf{C}}_{\mathbf{r}}}\mathbf{=}\frac{\mathbf{n}\mathbf{!}}{\mathbf{r}\mathbf{!}\left(n-r\right)\mathbf{!}}$ ...(ii)

Frequency of nth oscillation,${\mathit{f}}_{\mathbf{n}}\mathbf{=}{\mathit{f}}_{\mathbf{1}}\mathbf{+}\mathit{n}\mathbf{∆}\mathit{f}\mathbf{,}\mathbf{}\mathit{w}\mathit{h}\mathit{e}\mathit{r}\mathit{e}\mathbf{,}\mathbf{}{\mathit{f}}_{\mathbf{b}\mathbf{e}\mathbf{a}\mathbf{t}}\mathbf{=}\mathit{n}\mathbf{∆}\mathit{f}$ ...(ii)

The maximum number of different beat frequencies by sounding two forks at a time can be found by taking the formula of equation (i) and the given data as:

$\begin{array}{rcl}{5}_{C_2}& =& \frac{5!}{3!\left(5-2\right)!}\\ & =& 10\end{array}$

Therefore, the maximum number of different beat frequencies produced by sounding the forks two at a time is 10.

If we have N forks that are evenly spaced with P Hz, the possible beat differences are P, 2P, 3P, 4P… (n-1)P.

In this case, using equation (ii), we can get the frequency as:

$f_n=f_1+nP,wheren=2,3,4,5.$

Therefore, the minimum number of different beat frequencies produced by sounding the forks two at a time is 4.