A mass of 0.5 kg stretches a spring 0.09 m. The mass is in a medium that exerts a viscous resistance of 34 m N when the mass has a velocity of 4 - The viscous resistance is proportional to the speed of the object. S Suppose the object is displaced an additional 0.04 m and released. m Find an function to express the object's displacement from the spring's natural position, in m after t seconds. Let positive displacements indicate a stretched spring, and use 9.8 as the acceleration due to gravity. 8² u(t) =
A mass of 0.5 kg stretches a spring 0.09 m. The mass is in a medium that exerts a viscous resistance of 34 m N when the mass has a velocity of 4 - The viscous resistance is proportional to the speed of the object. S Suppose the object is displaced an additional 0.04 m and released. m Find an function to express the object's displacement from the spring's natural position, in m after t seconds. Let positive displacements indicate a stretched spring, and use 9.8 as the acceleration due to gravity. 8² u(t) =
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![### Understanding the Spring's Displacement for Educational Purposes
Consider a mass of 0.5 kg that stretches a spring by 0.09 m. This mass is situated in a medium that provides a viscous resistance of 34 N when the mass has a velocity of \(4 \, \frac{m}{s}\). The viscous resistance in this context is proportional to the speed of the object.
Suppose the object is further displaced by an additional 0.04 m and then released.
Our goal is to determine a function that expresses the displacement \( u(t) \) of the object from the spring's natural position over time \( t \) in seconds. We assume positive displacements indicate a stretched spring and use the acceleration due to gravity as \( 9.8 \, \frac{m}{s^2} \).
To find the function, we will use the following format:
\[ u(t) = \]
**Given:**
- Mass (\( m \)) = 0.5 kg
- Initial stretch of the spring (\( x \)) = 0.09 m
- Viscous resistance (\( F \)) = 34 N when the velocity (\( v \)) is \( 4 \, \frac{m}{s} \)
- Additional displacement (\( \Delta x \)) = 0.04 m
- Acceleration due to gravity (\( g \)) = \( 9.8 \, \frac{m}{s^2} \)
This kind of problem explores the principles of harmonic motion and damping in a physical system and can be used to illustrate the dynamics of mechanical vibrations.
**Note:** The box in the diagram is meant for the final function representing displacement \( u(t) \), which will be derived through calculations using principles from differential equations and Newton's laws of motion.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa86b2d75-dc89-44ee-816d-e64b811f53cb%2Fd88b60ae-3233-4ca9-a97f-89f3264196f8%2Fw0aqxde_processed.png&w=3840&q=75)
Transcribed Image Text:### Understanding the Spring's Displacement for Educational Purposes
Consider a mass of 0.5 kg that stretches a spring by 0.09 m. This mass is situated in a medium that provides a viscous resistance of 34 N when the mass has a velocity of \(4 \, \frac{m}{s}\). The viscous resistance in this context is proportional to the speed of the object.
Suppose the object is further displaced by an additional 0.04 m and then released.
Our goal is to determine a function that expresses the displacement \( u(t) \) of the object from the spring's natural position over time \( t \) in seconds. We assume positive displacements indicate a stretched spring and use the acceleration due to gravity as \( 9.8 \, \frac{m}{s^2} \).
To find the function, we will use the following format:
\[ u(t) = \]
**Given:**
- Mass (\( m \)) = 0.5 kg
- Initial stretch of the spring (\( x \)) = 0.09 m
- Viscous resistance (\( F \)) = 34 N when the velocity (\( v \)) is \( 4 \, \frac{m}{s} \)
- Additional displacement (\( \Delta x \)) = 0.04 m
- Acceleration due to gravity (\( g \)) = \( 9.8 \, \frac{m}{s^2} \)
This kind of problem explores the principles of harmonic motion and damping in a physical system and can be used to illustrate the dynamics of mechanical vibrations.
**Note:** The box in the diagram is meant for the final function representing displacement \( u(t) \), which will be derived through calculations using principles from differential equations and Newton's laws of motion.
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