Submarine A is moving at 9.20 m/s toward submarine B which is at rest. Submarine A emits a 3.50 MHz ultrasound. a. What frequency would the Submarine B detect? The speed of sound in water is 1482 m/s. W
Properties of sound
A sound wave is a mechanical wave (or mechanical vibration) that transit through media such as gas (air), liquid (water), and solid (wood).
Quality Of Sound
A sound or a sound wave is defined as the energy produced due to the vibrations of particles in a medium. When any medium produces a disturbance or vibrations, it causes a movement in the air particles which produces sound waves. Molecules in the air vibrate about a certain average position and create compressions and rarefactions. This is called pitch which is defined as the frequency of sound. The frequency is defined as the number of oscillations in pressure per second.
Categories of Sound Wave
People perceive sound in different ways, like a medico student takes sound as vibration produced by objects reaching the human eardrum. A physicist perceives sound as vibration produced by an object, which produces disturbances in nearby air molecules that travel further. Both of them describe it as vibration generated by an object, the difference is one talks about how it is received and other deals with how it travels and propagates across various mediums.
![**Physics Problem: Ultrasound Detection by Submarines**
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**Problem Statement:**
Submarine A is moving at 9.20 m/s toward submarine B, which is at rest. Submarine A emits a 3.50 MHz ultrasound.
**Question:**
a. What frequency would Submarine B detect? The speed of sound in water is 1482 m/s.
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**Detailed Explanation:**
This problem involves the Doppler effect, which occurs when there is a relative motion between a source of sound and an observer. In this scenario, Submarine A is the source of the sound and is moving toward Submarine B, which is the observer at rest. The following formula is used to calculate the frequency detected by the observer:
\[ f' = f \left( \frac{v + v_0}{v - v_s} \right) \]
Where:
- \( f' \) is the frequency observed by Submarine B
- \( f \) is the emitted frequency by Submarine A (3.50 MHz)
- \( v \) is the speed of sound in water (1482 m/s)
- \( v_0 \) is the speed of the observer (Submarine B, which is 0 m/s)
- \( v_s \) is the speed of the source (Submarine A, which is 9.20 m/s)
**Solution:**
1. Substitute the given values into the formula:
\[ f' = 3.50 \text{ MHz} \left( \frac{1482 \text{ m/s} + 0 \text{ m/s}}{1482 \text{ m/s} - 9.20 \text{ m/s}} \right) \]
2. Perform the calculations within the equation:
\[ f' = 3.50 \text{ MHz} \left( \frac{1482}{1472.80} \right) \]
3. Simplify the fraction:
\[ f' = 3.50 \text{ MHz} \left( 1.0062 \right) \]
4. Multiply to find the detected frequency:
\[ f' \approx 3.5217 \text{ MHz} \]
Therefore, the frequency detected by Submarine B is approximately 3.52 MHz.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8b2b1966-cbdc-46cc-aafa-333a90f3e9b8%2F04cb4d8c-64b9-483d-ad04-d3dbfb871e15%2F7vkhrud_processed.jpeg&w=3840&q=75)
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