## 225. Math Puzzle $2 / 84$

In the quartz clock, an oscillator generates a vibration with a frequency of 32768 Hz (Hertz), which is converted by a 15-stage frequency divider so that one pulse is generated every second to drive the display mechanism.

By how many seconds does the display error increase per day if the oscillator is incorrectly set and generates $32769 \mathrm{~Hz}$?
At a frequency of $32769 \mathrm{~Hz}$, the clock runs fast because the period $T$ decreases by

$$
\Delta T=\frac{1}{32768}-\frac{1}{32769} \approx \frac{1}{32768^{2}} \quad \text { or } \quad \Delta \underline{T}=\frac{1}{32768}
$$

This quotient applies to both a single oscillation and many consecutive oscillations. For $T=24 \mathrm{~h}=86,4 \cdot 10^{3} \mathrm{~s}$, we get

$$
T=\frac{86,4 \cdot 10^{3}}{32768}=2.64 \mathrm{~s}
$$

The clock runs fast by $2.64 \mathrm{~s}$ per day if it is not adjusted.