Problem 11
Question
\(\bullet\bullet\) (a) At what position is the speed of a mass in a mass-spring system maximum: \((1) x=0,(2) x=-A,\) or (3) \(x=+A ?\) Why? (b) If \(m=0.250 \mathrm{~kg}, k=100 \mathrm{~N} / \mathrm{m}\) and \(A=0.10 \mathrm{~m}\) for such a system, what is the mass's maximum speed?
Step-by-Step Solution
Verified Answer
(a) Speed is maximum at \(x=0\). (b) Maximum speed is \(2 \, \mathrm{m/s}\).
1Step 1: Understanding Maximum Speed in a Mass-Spring System
In a mass-spring system, the speed of the mass is maximum when it passes through the equilibrium position, which is at displacement \(x=0\). This is the point where potential energy is entirely converted into kinetic energy, hence the speed is maximized.
2Step 2: Calculating Maximum Speed Formula
The maximum speed \(v_{max}\) in a mass-spring system is given by \(v_{max} = A \omega\), where \(A\) is the amplitude and \(\omega\) is the angular frequency. The angular frequency \(\omega\) can be calculated using the formula \(\omega = \sqrt{\frac{k}{m}}\).
3Step 3: Substitute Values into the Formula
Given \(m = 0.250 \, \mathrm{kg}\), \(k = 100 \, \mathrm{N/m}\) and \(A = 0.10 \, \mathrm{m}\), first calculate \(\omega\):\[\omega = \sqrt{\frac{100}{0.250}} = \sqrt{400} = 20 \, \mathrm{rad/s}\]Now, substitute the values into the maximum speed formula:\[v_{max} = 0.10 \times 20 = 2 \, \mathrm{m/s}\]
4Step 4: Conclusion
The maximum speed of the mass in the mass-spring system occurs at \(x = 0\) with a value of \(2 \, \mathrm{m/s}\).
Key Concepts
Mass-Spring SystemMaximum SpeedAngular FrequencyEquilibrium Position
Mass-Spring System
A mass-spring system is a classic example of simple harmonic motion. It involves a mass attached to a spring which oscillates back and forth when displaced from its rest position. This system depends on Hooke's Law, which states that the force exerted by the spring is directly proportional to the displacement.
Understanding the dynamics of a mass-spring system is essential as it forms the foundational concept of many physical phenomena, including earthquakes and musical instruments.
- Components: The system consists of a mass, a spring, and a frictionless surface.
- Motion: The mass moves back and forth in a repetitive manner, which is called oscillation, due to the restoring force of the spring.
Understanding the dynamics of a mass-spring system is essential as it forms the foundational concept of many physical phenomena, including earthquakes and musical instruments.
Maximum Speed
In any oscillating system like a mass-spring, the speed of the mass is not constant. It changes as the spring force varies with position. The critical point of maximum speed occurs when the mass is at the equilibrium position. At this position, the displacement (\(x = 0\) ) is zero.
This can also be understood without complex details. Simply put, it's like a swing: it's fastest at the bottom and slowest at the top of its arc. The formula for this speed, derived from conservation laws, is: \[v_{max} = A \omega\], where \(A\) represents amplitude and \(\omega\) is the angular frequency.
- Kinetic Energy: At the equilibrium, all potential energy is converted to kinetic energy, maximizing speed.
This can also be understood without complex details. Simply put, it's like a swing: it's fastest at the bottom and slowest at the top of its arc. The formula for this speed, derived from conservation laws, is: \[v_{max} = A \omega\], where \(A\) represents amplitude and \(\omega\) is the angular frequency.
Angular Frequency
Angular frequency, \(\omega\), is a measure of how quickly the mass oscillates in the system. It's crucial in understanding how fast the oscillations are and is defined as the rate of change of the phase of the system. For a mass-spring system, angular frequency is calculated as: \[\omega = \sqrt{\frac{k}{m}}\]
The higher the angular frequency, the faster the oscillations occur, making it a central parameter in designing and analyzing systems like watches and sensors.
- Components: \(k\) is the spring constant, reflecting spring stiffness.
- Mass Influence: \(m\) is the mass of the object, helping determine the system's responsiveness.
The higher the angular frequency, the faster the oscillations occur, making it a central parameter in designing and analyzing systems like watches and sensors.
Equilibrium Position
The equilibrium position in a mass-spring system is where the spring is perfectly balanced, neither compressed nor stretched. It's located at displacement \(x = 0\). At this point, the forces acting on the mass are equal and opposite, resulting in no net force.
Understanding this concept is key in realizing how energy transforms in the system. The process is dynamic, shifting between potential and kinetic energy as the mass moves through the equilibrium, emphasizing the beautiful harmony of physical principles at play.
- Significance: It is the reference point from which displacement is measured.
- Energy Intersection: Kinetic energy peaks, while potential energy is at its lowest.
Understanding this concept is key in realizing how energy transforms in the system. The process is dynamic, shifting between potential and kinetic energy as the mass moves through the equilibrium, emphasizing the beautiful harmony of physical principles at play.
Other exercises in this chapter
Problem 9
\(\bullet\bullet\) A particle of mass \(0.10 \mathrm{~kg}\) is attached to a spring of spring constant \(10 \mathrm{~N} / \mathrm{m}\). If the maximum accelerat
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\(\bullet\bullet\) (a) At what position is the magnitude of the force on a mass in a mass-spring system minimum: \((1) x=0\), (2) \(x=-A,\) or \((3) x=+A ?\) Wh
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\(\bullet\bullet\) A mass-spring system is in SHM in the horizontal direction. If the mass is \(0.25 \mathrm{~kg}\), the spring constant is \(12 \mathrm{~N} / \
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\(\bullet\bullet\) A horizontal spring on a frictionless level air track has a \(0.150-\mathrm{kg}\) object attached to it and it is stretched \(6.50 \mathrm{~c
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