Question

A person in a parked car sounds the horn. The frequency of the horn's sound is \(489 \mathrm{~Hz}\). A driver in an approaching car measures the frequency of the horn's sound as \(509.4 \mathrm{~Hz}\). What is the speed of the approaching car? (Use \(343 \mathrm{~m} / \mathrm{s}\) for the speed of sound.)

Short answer

Answer: The approximate speed of the approaching car is \(11.32\,\text{m/s}\).

Step-by-step solution

First, let's write down the values that we know: - Source frequency \((f_{source}) = 489\,\text{Hz}\) - Observed frequency \((f_{obs}) = 509.4\,\text{Hz}\) - Speed of sound \((v) = 343\,\text{m/s}\) - Source speed \((v_s) = 0\,\text{m/s}\) (since the car is parked) We need to find the speed of the observer (approaching car), \(v_{o}\). #Step 2: Use the Doppler effect formula#

Recall the Doppler effect formula: \(f_{obs} = f_{source} \frac{v + v_{o}}{v + v_{s}}\) As the source is stationary, we have \(v_{s}=0\). So, the formula becomes: \(f_{obs} = f_{source} \frac{v + v_{o}}{v}\) Now, plug in the known values and solve for \(v_{o}\): \(509.4 = 489 \frac{343 + v_{o}}{343}\) #Step 3: Solve for \(v_{o}\)#

To solve for \(v_{o}\), first, divide both sides of the equation by \(489\): \(\frac{509.4}{489} = \frac{343 + v_{o}}{343}\) Now, cross-multiply: \(343(\frac{509.4}{489}) = 343 + v_{o}\) Subtract \(343\) from both sides: \(343(\frac{509.4}{489}) - 343 = v_{o}\) Finally, calculate \(v_{o}\): \(v_{o} \approx 11.32\,\text{m/s}\) #Step 4: Write down the answer#

We found that the speed of the approaching car, \(v_{o}\), is approximately \(11.32\,\text{m/s}\).