/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 130 A student holds an inexpensive s... [FREE SOLUTION] | 91影视

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Chapter 16: Problem 130

A student holds an inexpensive sonic range finder and uses the range finder to find the distance to the wall. The sonic range finder emits a sound wave. The sound wave reflects off the wall and returns to the range finder. The round trip takes 0.012 s. The range finder was calibrated for use at room temperature \(T=20^{\circ} \mathrm{C}\), but the temperature in the room is actually \(T=23^{\circ} \mathrm{C}\). Assuming that the timing mechanism is perfect, what percentage of error can the student expect due to the calibration?

Short Answer

Expert verified
The student can expect a percentage error of approximately \(0.52\%\) due to calibration.

Step by step solution

01

Calculate the speed of sound at given temperatures.

First, we need to calculate the speed of sound at 20掳C and 23掳C. We can use the following formula to find the speed of sound in air: \(v = 331.4 + 0.6T\) Where v is the speed of sound, and T is the temperature in Celsius. Calculate the speed of sound for both temperatures using this formula. At 20掳C: \(v_{20} = 331.4 + 0.6(20) = 331.4 + 12 = 343.4 \,\text{m/s}\) At 23掳C: \(v_{23} = 331.4 + 0.6(23) = 331.4 + 13.8 = 345.2 \,\text{m/s}\)
02

Calculate the actual distance to the wall.

Next, we need to find the actual distance to the wall using the round trip time and the speed of sound at 23掳C. Since the round trip took 0.012 seconds, we can divide that time by 2 to find the time it took for the sound wave to travel to the wall. Then, multiply the time by the speed of sound to find the distance. Time to wall: \(\frac{0.012 \,\text{s}}{2} = 0.006 \,\text{s}\) Distance to wall at 23掳C: \(d_{23} = v_{23} \times \text{time} = 345.2 \,\text{m/s} \times 0.006 \,\text{s} = 2.0712 \,\text{m}\)
03

Calculate the expected distance using calibration.

Now, we need to calculate the distance to the wall using the calibrated speed of sound at 20掳C. We can use the same method as in Step 2. Distance to wall at 20掳C: \(d_{20} = v_{20} \times \text{time} = 343.4 \,\text{m/s} \times 0.006 \,\text{s} = 2.0604 \,\text{m}\)
04

Calculate the percentage of error.

Finally, we need to calculate the percentage of error due to the calibration difference. We can use the following formula: Percentage Error \(=\frac{\text{Actual Distance - Expected Distance}}{\text{Actual Distance}} \times 100\) Percentage Error \(= \frac{2.0712 \,\text{m} - 2.0604 \,\text{m}}{2.0712 \,\text{m}} \times 100 \approx 0.52 \%\) So the student can expect a percentage error of approximately 0.52% due to calibration.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Speed of Sound
The speed of sound in air is not a constant number; it changes with temperature.
This is because sound waves move faster in warmer air. Generally, the speed of sound increases by about 0.6 meters per second (m/s) for each degree Celsius (掳C) rise in temperature. In physics, you can calculate the speed of sound using the formula: \(v = 331.4 + 0.6T\)
where \(v\) is the speed of sound and \(T\) is the temperature in 掳C.
This formula helps us understand that the basic speed of sound at 0掳C is 331.4 m/s, and it increases as the air warms.
  • For temperatures of 20掳C, the speed becomes 343.4 m/s.
  • At 23掳C, it becomes 345.2 m/s.
Knowing how to calculate the speed of sound is important for understanding how sound waves behave in different environments and it is often used in experiments and technology related to sound and acoustics.
Temperature Adjustment
Temperature adjustments are crucial when using devices like sonic range finders because these devices often assume a standard temperature for accurate measurements.
In this case, the standard temperature used for calibration is 20掳C. However, if the room temperature is different, such as 23掳C, calculations of distance using sound waves might need adjustments.
This is because the speed of sound also differs with temperature.
As shown earlier, the speed of sound at 23掳C is higher than at 20掳C.
  • This affects the time it takes for the sound wave to travel back.
  • The change in speed impacts calculated distances, leading to potential measurement errors.
By recalculating based on the actual temperature and speed of sound, one can ensure more accuracy in measurement.
Therefore, accounting for temperature is an essential step in experiments involving sound.
Percentage Error Calculation
Percentage error is a way to express how much a measured or expected value deviates from the actual value in a percentage form.
In practical use, it gives an idea of the accuracy of a measurement or calculation.To find the percentage error, we use this formula:\[\text{Percentage Error} = \frac{\text{Actual Distance - Expected Distance}}{\text{Actual Distance}} \times 100\]In the specific problem:
  • The actual distance was calculated based on the true speed of sound at 23掳C, resulting in 2.0712 meters.
  • The expected distance, based on a standard calibration temperature of 20掳C, was 2.0604 meters.
Calculating the percentage error gives us:\[\text{Percentage Error} = \frac{2.0712 \, \text{m} - 2.0604 \, \text{m}}{2.0712 \, \text{m}} \times 100 \approx 0.52\%\]This tells us that the error introduced by not accounting for the change in temperature amounts to about 0.52%.
Remembering to calculate percentage error is important in experiments to judge the reliability of results and calibrations.

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Most popular questions from this chapter

Your ear is capable of differentiating sounds that arrive at each ear just 0.34 ms apart, which is useful in determining where low frequency sound is originating from. (a) Suppose a low-frequency sound source is placed to the right of a person, whose ears are approximately 18 \(\mathrm{cm}\) apart, and the speed of sound generated is \(340 \mathrm{m} / \mathrm{s}\). How long is the interval between when the sound arrives at the right ear and the sound arrives at the left ear? (b) Assume the same person was scuba diving and a low- frequency sound source was to the right of the scuba diver. How long is the interval between when the sound arrives at the right ear and the sound arrives at the left ear, if the speed of sound in water is \(1500 \mathrm{m} / \mathrm{s}\) ? (c) What is significant about the time interval of the two situations?

The speed of light in air is approximately \(v=3.00 \times 10^{8} \mathrm{m} / \mathrm{s}\) and the speed of light in glass is \(v=2.00 \times 10^{8} \mathrm{m} / \mathrm{s} .\) A red laser with a wavelength of \(\lambda=633.00 \mathrm{nm}\) shines light incident of the glass, and some of the red light is transmitted to the glass. The frequency of the light is the same for the air and the glass.(a) What is the frequency of the light? (b) What is the wavelength of the light in the glass?

A quality assurance engineer at a frying pan company is asked to qualify a new line of nonstick-coated frying pans. The coating needs to be \(1.00 \mathrm{mm}\) thick. One method to test the thickness is for the engineer to pick a percentage of the pans manufactured, strip off the coating, and measure the thickness using a micrometer. This method is a destructive testing method. Instead, the engineer decides that every frying pan will be tested using a nondestructive method. An ultrasonic transducer is used that produces sound waves with a frequency of \(f=25 \mathrm{kHz}\). The sound waves are sent through the coating and are reflected by the interface between the coating and the metal pan, and the time is recorded. The wavelength of the ultrasonic waves in the coating is \(0.076 \mathrm{m}\). What should be the time recorded if the coating is the correct thickness \((1.00 \mathrm{mm}) ?\)

A string with a mass of \(0.30 \mathrm{kg}\) has a length of \(4.00 \mathrm{m}\). If the tension in the string is \(50.00 \mathrm{N},\) and a sinusoidal wave with an amplitude of \(2.00 \mathrm{cm}\) is induced on the string, what must the frequency be for an average power of \(100.00 \mathrm{W}\) ?

Use the linear wave equation to show that the wave speed of a wave modeled with the wave function \(y(x, t)=0.20 \mathrm{m} \sin \left(3.00 \mathrm{m}^{-1} x+6.00 \mathrm{s}^{-1} t\right)\) .is \(v=2.00 \mathrm{m} / \mathrm{s} .\) What are the wavelength and the speed of the wave?
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