Problem 51
Question
\(\bullet\)Dolphins and bats determine the location of their prey using echolocation (see Conceptual Question 15\()\). If it takes \(15 \mathrm{~ms}\) for a bat to receive the ultrasonic sound wave reflected off a mosquito, how far is the mosquito from the bat? Take the speed of sound as \(345 \mathrm{~m} / \mathrm{s}\)
Step-by-Step Solution
Verified Answer
The mosquito is approximately 2.59 meters away from the bat.
1Step 1: Understanding the Problem
We need to find the distance to the mosquito based on the time it takes for the sound wave to return. The bat emits a sound, which hits the mosquito and reflects back to the bat.
2Step 2: Calculate Time for One-Way Trip
Since the total time of 15 ms is for a round trip (to the mosquito and back), we must divide this time by 2 to find the time for a one-way trip. \[ t_{one-way} = \frac{15 \text{ ms}}{2} = 7.5 \text{ ms} \]
3Step 3: Convert Milliseconds to Seconds
Convert 7.5 ms to seconds since the speed of sound is given in meters per second.\[ t_{one-way} = 7.5 \text{ ms} = 0.0075 \text{ s} \]
4Step 4: Use the Speed and Time to Find Distance
Use the formula for distance: \( \text{Distance} = \text{Speed} \times \text{Time} \). With speed = 345 m/s and time = 0.0075 s, calculate the distance.\[ d = 345 \text{ m/s} \times 0.0075 \text{ s} = 2.5875 \text{ m} \]
5Step 5: Round and State the Result
The distance to the mosquito should be stated with appropriate significant figures. The answer is approximately 2.59 m.
Key Concepts
Speed of SoundDistance CalculationUltrasonic Waves
Speed of Sound
The speed of sound plays a crucial role in echolocation, as it determines how quickly sound waves can travel through a medium like air. In our exercise, the speed of sound is considered to be 345 meters per second (m/s). This means that sound waves can cover 345 meters in one second when traveling through air at room temperature.
Understanding the speed of sound is essential for accurately calculating distances, especially in applications such as echolocation used by bats and dolphins. The speed of sound can vary based on different factors, such as the medium it is traveling through (e.g., air, water, or a solid) and the environmental conditions (like temperature and pressure). Generally, sound will travel faster in water and solids than in air.
Understanding the speed of sound is essential for accurately calculating distances, especially in applications such as echolocation used by bats and dolphins. The speed of sound can vary based on different factors, such as the medium it is traveling through (e.g., air, water, or a solid) and the environmental conditions (like temperature and pressure). Generally, sound will travel faster in water and solids than in air.
- In gases, sound speeds up with increasing temperature.
- In liquids and solids, sound generally travels faster due to compact molecules.
Distance Calculation
Calculating distance using the speed of sound involves understanding how sound waves travel to and from a target object. This process includes determining the time taken for a sound to make a round trip and using the speed of sound in various mediums to find the distance.
In the exercise, the bat emits sound waves that travel to the mosquito and reflect back. Since the total measured time, 15 milliseconds (ms), includes the trip to the mosquito and back, we divide it by two to find the one-way time, resulting in 7.5 ms.
It is crucial to convert time into consistent units for calculation, so 7.5 ms is converted to 0.0075 seconds. The formula used for distance calculation is: \[ \text{Distance} = \text{Speed} \times \text{Time} \]
For the given values, \( \text{Distance} = 345 \text{ m/s} \times 0.0075 \text{ s} = 2.5875 \text{ m} \). It is important to use proper units and significant figures. Thus, the final distance is rounded appropriately to 2.59 meters, indicating how far the mosquito is from the bat.
In the exercise, the bat emits sound waves that travel to the mosquito and reflect back. Since the total measured time, 15 milliseconds (ms), includes the trip to the mosquito and back, we divide it by two to find the one-way time, resulting in 7.5 ms.
It is crucial to convert time into consistent units for calculation, so 7.5 ms is converted to 0.0075 seconds. The formula used for distance calculation is: \[ \text{Distance} = \text{Speed} \times \text{Time} \]
For the given values, \( \text{Distance} = 345 \text{ m/s} \times 0.0075 \text{ s} = 2.5875 \text{ m} \). It is important to use proper units and significant figures. Thus, the final distance is rounded appropriately to 2.59 meters, indicating how far the mosquito is from the bat.
Ultrasonic Waves
Ultrasonic waves are sound waves with a frequency higher than the upper audible limit of human hearing. Typically above 20,000 hertz (Hz), these waves are employed in various applications, including biological echolocation and medical imaging.
Bats and dolphins use ultrasonic waves to hunt and navigate. The high frequency allows for better resolution and detailing of the objects being detected. Ultrasonic waves travel like any other sound wave but can provide more information due to their frequency.
The advantage of using ultrasonics in echolocation is that they allow animals to detect small objects, such as insects, with precision. For instance, they enable bats to find a mosquito even in complete darkness.
Bats and dolphins use ultrasonic waves to hunt and navigate. The high frequency allows for better resolution and detailing of the objects being detected. Ultrasonic waves travel like any other sound wave but can provide more information due to their frequency.
The advantage of using ultrasonics in echolocation is that they allow animals to detect small objects, such as insects, with precision. For instance, they enable bats to find a mosquito even in complete darkness.
- High frequency means shorter wavelength, suitable for detecting small objects.
- Ultrasonic waves can penetrate objects and provide internal details in medical diagnostics.
Other exercises in this chapter
Problem 49
\(\bullet\) A wave on a rope that measures \(10 \mathrm{~m}\) long takes \(2.0 \mathrm{~s}\) to travel the whole rope. If the wavelength of the wave is \(2.5 \m
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\(\bullet\) A student reading his physics book on a lake dock notices that the distance between two incoming wave crests is about \(0.75 \mathrm{~m},\) and he t
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\(\bullet\) Light waves travel in a vacuum at a speed of \(3.00 \times 10^{8} \mathrm{~m} / \mathrm{s} .\) The frequency of blue light is about \(6 \times 10^{1
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\(\bullet\bullet\) A sonar generator on a submarine produces periodic ultrasonic waves at a frequency of \(2.50 \mathrm{MHz}\). The wavelength of the waves in s
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