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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,${\mathit{f}}_{\mathbf{n}}\mathbf{=}{\mathit{f}}_{\mathbf{1}}\mathbf{+}\mathit{n}\mathbf{鈭�}\mathit{f}\mathbf{,}\mathbf{}\mathit{w}\mathit{h}\mathit{e}\mathit{r}\mathit{e}\mathbf{,}\mathbf{}{\mathit{f}}_{\mathbf{b}\mathbf{e}\mathbf{a}\mathbf{t}}\mathbf{=}\mathit{n}\mathbf{鈭�}\mathit{f}$ ...(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鈥� (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.