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Sagot :
Answer:
Approximately [tex]1061\; {\rm Hz}[/tex] (rounded to the nearest whole number,) assuming that the speed of sound in the air is [tex]330\; {\rm m \cdot s^{-1}}[/tex].
Explanation:
Let [tex]v[/tex] denote the speed of this sound wave: [tex]v = 330\; {\rm m\cdot s^{-1}}[/tex] by assumption.
Let [tex]\Delta v[/tex] denote the speed at which the source (the bus) moves away from the observer (the cyclist.) In this question, [tex]\Delta v = 25\; {\rm m\cdot s^{-1}} - 5\; {\rm m\cdot s^{-1}} = 20\; {\rm m\cdot s^{-1}}[/tex].
Let [tex]d[/tex] denote the distance between the source (the bus) and the observer (the cyclist) when the first crest leaves the source. At a speed of [tex]v[/tex], it would take [tex]d / v[/tex] before the first crest arrive at the observer.
Let [tex]T[/tex] denote the period of this wave. If a crest of this wave leaves the source at [tex]t = 0[/tex], the very next one would leave at [tex]t = T[/tex]. Since the source is moving away from the observer at [tex]\Delta v[/tex], the distance between the two would have since increased from [tex]d[/tex] to [tex](d + \Delta v\times T)[/tex]. At a speed of [tex]v[/tex], this crest would now need [tex](d + \Delta v\times T) / v[/tex] to reach the observer.
In summary:
- The first crest leaves the source at [tex]t = 0[/tex] and arrives at the observer at [tex]t = d / v[/tex].
- The second crest leaves the source at [tex]t = T[/tex] and arrives at the observer at [tex]t = T + (d + \Delta v\times T) / v[/tex].
The time difference between the two crests is [tex]T[/tex] at the source. However, at the observer, this difference would have become:
[tex]\begin{aligned} &T + \frac{d + \Delta v\times T}{v} - \frac{d}{v} \\ =\; & T + \frac{\Delta v\times T}{v} \\ =\; & T\times \frac{v + \Delta v}{v}\end{aligned}[/tex].
Therefore, the ratio between the period at the source and at the observer would be:
[tex]\begin{aligned}\frac{T(\text{observer})}{T(\text{source})} &= \frac{\displaystyle T \times \frac{v + \Delta v}{v}}{T} \\ &= \frac{v + \Delta v}{v}\end{aligned}[/tex].
The frequency [tex]f[/tex] of a wave is equal to the reciprocal of its period [tex]T[/tex]. In other words, [tex]f = (1 / T)[/tex]. Therefore:
[tex]\begin{aligned}\frac{f(\text{observer})}{f(\text{source})} &= \frac{T(\text{source})}{T(\text{observer})} \\ &= \frac{v}{v + \Delta v}\end{aligned}[/tex].
Since [tex]f(\text{observer}) = 1000\; {\rm Hz}[/tex], [tex]\Delta v = 20\; {\rm m\cdot s^{-1}}[/tex], and [tex]v = 330\; {\rm m\cdot s^{-1}[/tex] by assumption:
[tex]\begin{aligned} & f(\text{source})\\ =\; & f(\text{observer}) \times \frac{v + \Delta v}{v} \\ =\; & 1000\; {\rm m\cdot s^{-1}} \times \frac{330\; {\rm m\cdot s^{-1}} + 30\; {\rm m\cdot s^{-1}}}{330\; {\rm m\cdot s^{-1}}} \\ \approx\; & 1061\; {\rm m\cdot s^{-1}}\end{aligned}[/tex].
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