Question

Doppler Effect The Doppler effect (named after Christian Doppler) is the change in the pitch (frequency) of the sound from a source \((s)\) as heard by an observer \((o)\) when one or both are in motion. If we assume both the source and the observer are moving in the same direction, the relationship is $$ f^{\prime}=f_{a}\left(\frac{v-v_{o}}{v-v_{s}}\right) $$ where \(f^{\prime}=\) perceived pitch by the observer $$ \begin{aligned} f_{a} &=\text { actual pitch of the source } \\ v &=\text { speed of sound in air (assume } 772.4 \mathrm{mph}) \\ v_{o} &=\text { speed of the observer } \\ v_{s} &=\text { speed of the source } \end{aligned} $$ Suppose that you are traveling down a road at \(45 \mathrm{mph}\) and you hear an ambulance (with siren) coming toward you from the rear. The actual pitch of the siren is 600 hertz \((\mathrm{Hz})\) (a) Write a function \(f^{\prime}\left(v_{s}\right)\) that describes this scenario. (b) If \(f^{\prime}=620 \mathrm{~Hz}\), find the speed of the ambulance. (c) Use a graphing utility to graph the function. (d) Verify your answer from part (b).

Short answer

The speed of the ambulance is approximately 68.63 mph.

Step-by-step solution

- Identify Variables

First, identify and note down all variables and constants given in the problem. The given constants and variables are: - Actual pitch of the siren, \( f_a = 600 \text{ Hz} \)- Speed of sound in air, \( v = 772.4 \text{ mph} \)- Speed of the observer, \( v_o = 45 \text{ mph} \)- We need to write the function in terms of the speed of the source, \( v_s \)

- Write the Doppler Effect Equation

The Doppler Effect formula is given as: \[ f' = f_a \left( \frac{v - v_o}{v - v_s} \right) \]Since we need to write the function in terms of \( v_s \), we have \[ f'(v_s) = f_a \left( \frac{v - v_o}{v - v_s} \right) \]

- Substitute the Given Values

Plug the given values into the equation: - \( f_a = 600 \text{ Hz} \)- \( v = 772.4 \text{ mph} \)- \( v_o = 45 \text{ mph} \)The function becomes: \[ f'(v_s) = 600 \left( \frac{772.4 - 45}{772.4 - v_s} \right) \]

- Solve for the Speed of the Ambulance

Given \( f' = 620 \text{ Hz} \) and using the function we just derived:\[ 620 = 600 \left( \frac{772.4 - 45}{772.4 - v_s} \right) \]To solve for \( v_s \), rearrange the equation:\[ 620 = 600 \left( \frac{727.4}{772.4 - v_s} \right) \]Divide both sides by 600:\[ \frac{620}{600} = \frac{727.4}{772.4 - v_s} \]Simplify to get:\[ 1.0333 = \frac{727.4}{772.4 - v_s} \]Cross-multiply to solve for \( v_s \):\[ 1.0333(772.4 - v_s) = 727.4 \]Solve this equation step-by-step:\[ 798.27772 - 1.0333v_s = 727.4 \]Subtract 727.4 from both sides:\[ 70.87772 = 1.0333v_s \]Finally, divide both sides by 1.0333:\[ v_s = \frac{70.87772}{1.0333} \approx 68.63 \text{ mph} \]

- Graph the Function

Use a graphing utility to plot the function \[ f'(v_s) = 600 \left( \frac{727.4}{772.4 - v_s} \right) \]The x-axis represents the speed of the ambulance, \( v_s \), and the y-axis represents the perceived frequency, \( f' \). Note the intersection point where \( f' = 620 \text{ Hz} \).

- Verify the Answer

Check that the derived speed \( v_s = 68.63 \text{ mph} \) matches the criteria in the problem \( f' = 620 \text{ Hz} \). Substitute \( v_s \) back into the function:\[ f(v_s) = 600 \left( \frac{727.4}{772.4 - 68.63} \right) = 620 \text{ Hz} \]. This verifies the calculation is correct.

Key concepts

Doppler Effect

The Doppler Effect refers to the change in the frequency or pitch of a sound as perceived by an observer moving relative to the source of the sound. Imagine you're standing by a road and an ambulance with its siren on passes by. As it approaches you, the sound is higher-pitched. Once it passes and starts moving away, the pitch lowers. This effect is due to the compression and elongation of sound waves.
In our exercise's equation \( f^{\prime} = f_a \left( \frac{v - v_o}{v - v_s} \right) \):

• \f^{\prime}\text{ is the perceived pitch}
• \f_a\text{ is the actual pitch of the source}
• \text{v}\text{ is the speed of sound in air}
• \text{v_o}\text{ is the speed of the observer}
• \text{v_s}\text{ is the speed of the source}

Frequency

Frequency refers to how often a wave repeats in a given time period and is measured in Hertz (Hz). It's crucial in understanding the Doppler Effect because the perceived pitch of a sound changes with frequency. For instance, in our example, the ambulance siren's actual frequency is \( 600 \text{ Hz} \).
When the sound source (siren) and observer (you) move relative to each other, the frequency you hear shifts. Higher frequencies correspond to higher pitches, and lower frequencies correspond to lower pitches. This change in perception due to motion is what the Doppler Effect quantifies.

Function Graphing

Graphing functions visually represents the relationship between variables. In our problem, graphing the function \( f^{\prime}(v_s) = 600 \left( \frac{727.4}{772.4 - v_s} \right) \) shows how the perceived frequency changes with the ambulance's speed.
Using graphing utilities or graphing calculators, you can plot this function to observe key points. For example, we can see where the perceived frequency \( f^{\prime} \) equals \( 620 \text{ Hz} \). This graph helps confirm our calculations since the intersection will match our calculated ambulance speed of approximately 68.63 mph. Graphing thus aids in visual verification and analysis of mathematical relationships.

Sound Wave Equations

Sound waves, which carry our concocted sirens from the ambulance, are an essential part of understanding the Doppler Effect. The equation used in our problem represents a classic scenario where sound waves change due to relative motion.
Key elements of the sound wave equation for Doppler Effect:

• Numerator: \( v - v_o \), which emphasizes the speed difference affecting the perceived frequency
• Denominator: \( v - v_s \), showing how the speed of the source impacts the wave’s perceived frequency

By substituting actual values, the equation quantifies pitch change, assisting in problem-solving and comprehension. Understanding this allows visualization of why an approaching sound has a higher pitch while receding sound lowers in pitch.