The frequency of the digital watch is:
[tex]f=32.8 kHz = 3280 Hz[/tex]
The period of oscillation is equal to the reciprocal of the frequency; therefore, in this problem, the period of oscillation of the digital watch is:
[tex]T=\frac{1}{f}=\frac{1}{3280 Hz}=3.05 \cdot 10^{-4}s [/tex]
