Question

Two vehicles carrying speakers that produce a tone of frequency \(1000.0 \mathrm{~Hz}\) are moving directly toward each other. Vehicle \(\mathrm{A}\) is moving at \(10.00 \mathrm{~m} / \mathrm{s}\) and vehicle \(\mathrm{B}\) is moving at \(20.00 \mathrm{~m} / \mathrm{s}\). Assume the speed of sound in air is \(343.0 \mathrm{~m} / \mathrm{s}\), and find the frequencies that the driver of each vehicle hears.

Short answer

Answer: The driver of vehicle A will hear a frequency of 1065.52 Hz, and the driver of vehicle B will hear a frequency of 1193.41 Hz.

Step-by-step solution

Doppler Effect Formula

The general Doppler effect formula is given by: $$f' = \frac{f (v \pm v_o)}{v \pm v_s}$$ where - \(f'\) is the observed frequency, - \(f\) is the emitted/source frequency, - \(v\) is the speed of sound in the medium, - \(v_o\) is the speed of the observer (positive if moving towards the source and negative if moving away from the source), and - \(v_s\) is the speed of the source (positive if moving away from the observer and negative if moving towards the observer). In our case, \(f = 1000.0 \mathrm{~Hz}\), \(v = 343.0 \mathrm{~m} / \mathrm{s}\), \(v_{oA} = -10.00 \mathrm{~m} / \mathrm{s}\) (since driver A is moving towards the source), \(v_{sA} = 20.00 \mathrm{~m} / \mathrm{s}\) (since vehicle B is moving away from driver A), \(v_{oB} = -20.00 \mathrm{~m} / \mathrm{s}\) (since driver B is moving towards the source), and \(v_{sB} = -10.00 \mathrm{~m} / \mathrm{s}\) (since vehicle A is moving towards driver B).

Calculate frequency heard by driver A

Using the Doppler effect formula, we can find the frequency heard by driver A (\(f_A\)): $$ f_A = \frac{f (v + v_{oA})}{v - v_{sA}}= \frac{1000.0 \mathrm{~Hz} \cdot (343.0 \mathrm{~m} / \mathrm{s} - 10.00 \mathrm{~m} / \mathrm{s})}{343.0 \mathrm{~m} / \mathrm{s} - 20.00 \mathrm{~m} / \mathrm{s}}$$ Solve for \(f_A\): $$ f_A = 1065.52 \mathrm{~Hz} $$

Calculate frequency heard by driver B

Similarly, we can find the frequency heard by driver B (\(f_B\)): $$ f_B = \frac{f (v + v_{oB})}{v + v_{sB}}= \frac{1000.0 \mathrm{~Hz} \cdot (343.0 \mathrm{~m} / \mathrm{s} - 20.00 \mathrm{~m} / \mathrm{s})}{343.0 \mathrm{~m} / \mathrm{s} + 10.00 \mathrm{~m} / \mathrm{s}}$$ Solve for \(f_B\): $$ f_B = 1193.41 \mathrm{~Hz} $$

Final Answer

The driver of vehicle A will hear a frequency of \(1065.52 \mathrm{~Hz}\), and the driver of vehicle B will hear a frequency of \(1193.41 \mathrm{~Hz}\).

Key concepts

Sound Frequency

Sound frequency refers to the number of sound wave cycles that occur per second, measured in hertz (Hz). Imagine waves in the ocean; sound waves travel through the air in much the same way. As these waves reach your ears, they vibrate the eardrum, which is how sound is perceived. In our exercise, the sound frequency emitted by the vehicles' speakers is 1000 Hz. This means each speaker is producing wave cycles 1000 times every second. By understanding sound frequency, we can delve into how motion affects what different observers hear.
In scenarios involving motion, like cars moving, what the observer hears can differ from the original frequency due to what's called the Doppler Effect. This change in frequency occurs because motion affects how fast the sound waves reach the observer. Thus, the frequency you hear might be higher or lower than the source frequency based on relative motion.

Wave Equation

The wave equation forms the backbone of understanding how sound frequency changes with motion. The Doppler Effect relies heavily on this equation for calculating observed frequencies.
To put it simply, the formula used in this context is:

• \( f' = \frac{f (v \pm v_o)}{v \pm v_s} \)
• \( f' \) represents the frequency heard by the observer.
• \( f \) is the original frequency emitted by the source.
• \( v \) stands for the speed of sound in the medium, here it's air with a speed of 343 m/s.
• \( v_o \) is the speed of the observer (positive if moving towards the source and negative if away), and \( v_s \) is the speed of the source.

This equation helps us quantify how sound travels and gets perceived in varying conditions. It divides into components of source, observer and medium, giving a comprehensive measure of sound's journey. Thus, in our problem, it allows us to precisely calculate that driver A hears a frequency of 1065.52 Hz and driver B hears 1193.41 Hz due to their respective movements.

Observational Physics

Observational Physics deals with measurements and perceptions that can be influenced by various factors, like motion.
The Doppler Effect is a fantastic example of how real-world observation can differ from theoretical sound emissions due to motion. In observational physics, attention to these details becomes critical as they affect practical measurements and predictions.
As vehicles A and B move towards each other, their respective speeds alter what each driver hears. This scenario underlines the Doppler Effect's importance in understanding physics associated with moving objects.
By examining how each vehicle's speed affects sound frequency, we recognize observational physics' role in explaining varied perceptions and enhancing comprehension of traveling waves.