## Task 29/63

In a tape recorder, the tape is pulled past the magnetic heads according to the schematic construction shown in the figure.

![](https://cdn.mathpix.com/cropped/2024_06_06_1d2be4837fb839c2b31dg-078.jpg?height=306&width=1010&top_left_y=418&top_left_x=523)

Due to manufacturing inaccuracies, the rotational speed \( n \) of the drive roller fluctuates by a maximum of \(\pm 0.1\%\). Additionally, the radius \( r \) of the drive roller is subject to an error of up to \(\pm 0.01 \, \text{mm}\). With this tape recorder, a 1000-Hz tone is recorded and then played back.

By how many Hertz does the played-back tone fluctuate at most, if no other errors are present besides the specified inaccuracies? Here, \( n = 600 \, \text{min}^{-1} \) and \( r = 3 \, \text{mm} \).

Hint: The frequency \( f \) is proportional to the tape speed \( v \); i.e., \( f = k \cdot v \), where \( k \) is a constant factor.
First, the equation for the belt speed is to be established. The circumference of the drive roller is $U=2 \pi r$. Dividing this by the time $T$ of one revolution yields the circumferential speed, which is equal to the belt speed $v$. Thus, $v=\frac{U}{T}=\frac{2 \pi r}{T}$.

With $T=\frac{1}{n}$, one finally gets $v=2 \pi r n$ (1). The radius $r$ and the rotational speed $n$ are subject to fluctuations of up to $\pm \Delta r$ and $\pm \Delta n$, respectively. As a result, the belt speed $v$ fluctuates by the value $\pm \Delta v$.

The maximum value of the belt speed occurs when $r$ deviates by $+\Delta r$ and $n$ by $+\Delta n$; then

$$
v+\Delta v=2 \pi(r+\Delta r)(n+\Delta n)
$$

The absolute error in the belt speed is then

$$
\Delta v=2 \pi r n+2 \pi(r \Delta n+\Delta r n+\Delta r \Delta n)-2 \pi r n
$$

(due to $v=2 \pi r n$ according to equation 1). Thus,

$$
\Delta v=2 \pi(r \Delta n+\Delta r n+\Delta r \Delta n)
$$

The quantities $\Delta r$ and $\Delta n$ are very small; their product is even smaller, i.e., $\Delta r \Delta n \ll r \Delta n+\Delta r n$. Therefore, $\Delta r \Delta n$ can be neglected compared to $r \Delta n+\Delta r n$ without causing a significant computational error. Thus, the absolute error is

$$
\Delta v=2 \pi(r \Delta n+n \Delta r)
$$

That the neglect of $\Delta r \Delta n$ is reasonable can be confirmed by substituting the given numerical values into equations (2) and (3) and comparing the two results.

During the recording of the frequency $f$, an error is caused due to the fluctuations in the belt speed. It is

$$
f+\Delta f=k(v+\Delta v) \quad ; \quad \Delta f=k(v+\Delta v)-f
$$

and since $f=k v$: $\Delta f=k \Delta v \quad$ (4).

By combining equation (3) with equation (4), one gets

$$
\Delta f=2 k \pi(r \Delta n+n \Delta r)
$$

The constant $k$ can be determined from $f=k v$ to $k=\frac{f}{v}$; with equation (1), this yields $k=\frac{f}{2 \pi r n} \quad$ (6). Substituting equation (6) into equation (5) gives

$$
\Delta f=f\left(\frac{\Delta n}{n}+\frac{\Delta r}{r}\right)
$$

This is the maximum error in the frequency that occurs during recording. In the worst case, the same error is caused again during playback; the frequency thus fluctuates by a maximum value of

$$
\Delta f^{\prime}=2 f\left(\frac{\Delta n}{n}+\frac{\Delta r}{r}\right)
$$

With $\frac{\Delta n}{n}=0.001=0.01\%$ and $\Delta r=0.01 \text{ mm}$ as well as $r=3 \text{ mm}, f=1000 \text{ Hz}$, one finally gets $\Delta f^{\prime}=8.6 \text{ Hz}$

The same deviation occurs with fluctuations downwards. During the playback of the recorded 1000-Hz tone, the tone thus fluctuates by $\pm 8.6 \text{ Hz}$, i.e., the frequency changes within a range from $991.4 \text{ Hz}$ to $1008.6 \text{ Hz}$.