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A mass moving in one dimension is attached to a horizontal spring. It slides on the surface below it, with equal coefficients of static and kinetic friction, \(\mu_{k}=\mu_{s} .\) The equilibrium position is \(x=0 .\) If the mass is pulled to some initial position and released from rest, it will complete some number of oscillations before friction brings it to a stop. When released from \(x=a(a>0)\), it completes exactly \(1 / 4\) of an oscillation, i.e., it stops precisely at \(x=0 .\) Similarly, define \(b>0\) as the greatest \(x\) from which it could be released and comlete \(1 / 2\) of an oscillation, stopping on the far side and not coming back toward equilibrium. Find \(b / a\). Hint: To keep the algebra simple, set every fixed parameter of the system equal to \(1 .\)

Short Answer

Expert verified
The ratio \( b/a \) is 1.

Step by step solution

01

Understanding the Problem

A mass is attached to a spring on a surface with friction, released from rest at position \( x = a \), and stops at \( x = 0 \) after completing 1/4 of an oscillation. We need to find the ratio \( b/a \), where \( b \) is the maximum position from which the mass completes 1/2 an oscillation and stops on the opposite side.
02

Analyze Forces and Motion

The force exerted by the spring is \( F_{spring} = -kx \) and the frictional force is \( F_{friction} = \, \mu mg \), where both constants \( k \), \( \mu \), and \( m \) are set to 1 for simplicity. The equation of motion is \( m\ddot{x} = -kx - \mu mg \). Here, \( x \) is the displacement.
03

Energy Considerations

Initially, at position \( x = a \), the total energy is the potential energy stored in the spring: \( U = \frac{1}{2}kx^2 = \frac{1}{2}a^2 \). For the system to stop at \( x = 0 \), all this energy must be dissipated by friction, meaning \( \mu mg \cdot a = \frac{1}{2}a^2 \). Solving, we get \( a = 2 \mu mg \).
04

Solving for Stopping After 1/4 of an Oscillation

Since \( a = 2 \mu mg \) and \( \mu = 1 \), this becomes \( a = 2 \). This confirms the mass slides from \( x = 2 \) to \( x = 0 \) to stop after 1/4 of an oscillation.
05

Solving for 1/2 Oscillation

When pulled from \( x = b \) to complete 1/2 an oscillation stopping on the opposite side, potential energy \( \frac{1}{2}b^2 = \mu mg \cdot (2b) \) must dissipate over twice the displacement. Substituting \( \mu \), \( m \), and \( g \) with 1, \( b^2 = 2b \), resulting in \( b = 2 \).
06

Find b/a

Using the values found, \( a = 2 \) and \( b = 2 \), then \( \frac{b}{a} = \frac{2}{2} = 1 \).

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Key Concepts

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

Equation of Motion
In simple harmonic motion, the equation of motion is fundamental in understanding how objects move. For an object attached to a spring, this involves both restoring and opposing forces.
When a mass is connected to a spring, the spring force is given by Hooke's law as \( F_{spring} = -kx \) where \( k \) is the spring constant and \( x \) is the displacement from the equilibrium position.
Because friction is present, it's necessary to introduce a frictional force into the equation of motion. Suitable for one-dimensional motion over a surface, the frictional force is represented as \( F_{friction} = \, \mu mg \), where \( \mu \) is the coefficient of friction, \( m \) is the mass, and \( g \) is the acceleration due to gravity. Given that constants \( k \), \( \mu \), and \( m \) are all set to 1 for simplicity, the equation of motion is simplified to:\[ m\ddot{x} = -kx - \mu mg \]This equation dictates the motion, combining the spring's restoring force and the opposing frictional force into one understandable format, guiding us through solving the exercise.
Potential Energy
Potential energy in a spring-mass system is key as it helps determine how much energy is stored and possibly dissipated as heat due to friction. When the mass is displaced by a distance \( x \) from its equilibrium, potential energy \( U \) associated with the spring is calculated by the formula:
\[ U = \frac{1}{2}kx^2 \]
In our specific problem, because the spring constant \( k \) and the initial displacement \( a \) are simplified to 1 for ease of calculation, the expression for potential energy at the starting position becomes \( U = \frac{1}{2}a^2 \). This quantifies the total stored energy when the spring is compressed or stretched.
As the mass moves back towards the equilibrium position, from \( x = a \) to \( x = 0 \), all this potential energy must be converted to overcome friction and eventually stop the mass. In this exercise, it’s explicit that this energy conversion is responsible for the 1/4 and 1/2 oscillations as described.
Frictional Force
Frictional force significantly affects motion through its opposing nature, working against the direction of movement. This force is constant if the coefficient of frictions, static and kinetic, equal \( \mu \). As stated, the force of friction is determined by \( F_{friction} = \, \mu mg \).
In scenarios of motion on a surface, friction dissipates the energy stored in the system, usually as heat, and is the main reason objects in motion eventually come to a stop.
In this particular case, the mass undergoes oscillations until all potential energy is used to counteract the friction. Initially, the setup presupposes the parameters like mass \( m \), gravity \( g \), and \( \mu = 1 \). Thus \( F_{friction} \) simplifies to \( F_{friction} = 1 \).
  • When released from \( x = a \), friction ensures the mass stops precisely at \( x = 0 \) after 1/4 of an oscillation, maintaining the energy balance.
  • For 1/2 oscillation, it acts to halt motion at the opposite side when reaching maximum allowed displacement \( x = b \).
Friction moderates the motion, conditioning the extent of oscillations based on initial energy.
Oscillation
Oscillation is the repeated back and forth motion about an equilibrium position. In physics, particularly when dealing with springs and masses, it describes how objects move in simple harmonic motion.
In our context, oscillation is characterized by moving 1/4 or 1/2 of its complete path before stopping. Released from position \( x = a \), the system completes 1/4 of an oscillation before friction brings it to complete stop at \( x = 0 \).
With the exercise simplifying parameters such as \( k \), \( m \), and \( g \), a crucial point to grasp is:
  • Moving from \( 1/4 \) to \( 1/2 \) an oscillation suggests a doubling of the effective distance due to dissipating potential energy over a broader range.
  • The inherent friction ensures that energy conversion aligns with these set distances to achieve specified stopping points.
Oscillation in this problem reflects how energy input in the system by initial displacement gets carefully controlled, providing an understanding of energy dynamics and motion affected by friction.

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Most popular questions from this chapter

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