Suggested languages for you:

Americas

Europe

Q103P

Expert-verified
Found in: Page 513

### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718

# A police car is chasing a speeding Porsche 911. Assume that the Porsche’s maximum speed is ${\mathbf{80}}{\mathbf{.0}}{\mathbf{}}\frac{\mathbf{\text{m}}}{\mathbf{\text{s}}}$ and the police car’s is ${\mathbf{54}}{\mathbf{.0}}{\mathbf{}}\frac{\mathbf{\text{m}}}{\mathbf{\text{s}}}$. At the moment both cars reach their maximum speed, what frequency will the Porsche driver hear if the frequency of the police car’s siren is ${\mathbf{\text{440 Hz}}}$? Take the speed of sound in air to be .${\mathbf{340}}{\mathbf{}}\frac{\mathbf{\text{m}}}{\mathbf{\text{s}}}$

The frequency heard by the Porsche driver is $400\text{Hz}$

See the step by step solution

## Step 1: The given data

1. The maximum speed of Porsche is ${v}_{D}=80.0\frac{\text{m}}{\text{s}}$
2. The speed of the police car is ${v}_{s}=54.0\frac{\text{m}}{\text{s}}$
3. The frequency of the police car’s siren is $f=440\text{Hz}$
4. The speed of sound in air is $v=340\frac{\text{m}}{\text{s}}$

## Step 2: Determine the concept of the Doppler Effect

Use the concept of the Doppler effect of sound. We can set the sign according to the rule. The detector is the Porsche’s driver and the source is the siren of the police car. If the detector moves away, then use the sign in the numerator to get a shift down which means the frequency would decrease. If the source moves towards the detector, use the minus sign in the denominator to get a shift up which means the frequency would increase.

Formula:

The frequency received by the observer or the source according to Doppler’s Effect, (observer moving away from source and source moving towards observer) $f\text{'}=f\left(\frac{v-{v}_{D}}{v-{v}_{s}}\right)$ (i)

## Step 3: Calculate the frequency heard by the Porsche driver

As, the driver is moving away from the police car and the police car is approaching the driver, the frequency of the wave heard by the driver is given using the equation (i) as:

$\begin{array}{c}{f}^{\text{'}}=440\text{\hspace{0.17em}Hz}\left(\frac{340\frac{\text{m}}{\text{s}}-80.0\frac{\text{m}}{\text{s}}}{340\frac{\text{m}}{\text{s}}-54.0\frac{\text{m}}{\text{s}}}\right)\\ =40\text{0\hspace{0.17em}Hz}\end{array}$

Hence, the frequency of the sound wave heard by the driver is $400\text{\hspace{0.17em}Hz}$