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Chapter 14: Problem 42

A car horn emits \(380-\mathrm{Hz}\) sound. If the car moves at \(17 \mathrm{m} / \mathrm{s}\) with its horn blasting, what frequency will a person standing in front of the car hear?

Short Answer

Expert verified
The person standing in front of the car will hear the horn at a frequency of approximately \(397 \, Hz\).

Step by step solution

01

Identify the given values

The problem provides the following information: the source frequency \(f = 380 \, Hz\), the speed of the source \(vs = 17 \, m/s\), the speed of sound in air \(v = 343 \, m/s\), and the speed of the observer \(vo = 0 \, m/s\) (because the observer is standing still).
02

Insert the values into the formula

The formula for the observed frequency is \(f' = f \frac{{v + vo}}{{v - vs}}\). We substitute the given values into this formula: \(f' = 380 \, Hz \cdot \frac{{343 \, m/s + 0 \, m/s}}{{343 \, m/s - 17 \, m/s}}\).
03

Calculation

Solving the equation gives: \(f' \approx 397 \, Hz\).
04

Interpret the result

The person standing in front of the car will hear the car horn at a frequency of approximately 397 Hz, which is higher than the actual frequency of the horn. This is due to the Doppler effect.

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