Accelerated mass and spring system

In summary, the conversation discusses a problem involving a mass attached to a spring with an instantaneous acceleration of 50m/s². The spring constant and mass of the assembly are given and the system is assumed to be undamped. The goal is to find the force exerted by the spring on the vehicle body, but the relationship between acceleration and deceleration is unclear. Various equations and potential solutions are mentioned, but the problem remains unsolved. The concept of an "instantaneous" acceleration is also discussed, with different interpretations and suggestions for how to approach the problem.
  • #1
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Homework Statement



Imagine a mass attached to a spring, equivalent to a vehicle suspension system, where the mass we are considering is the unsprung mass of the wheel/axle assembly. We are told the unsprung mass is given an instantaneous acceleration of 50m/s². We are given the spring constant of the suspension springs (k = 500kN/mm) and we are given the mass of the wheel/axle assembly (m = 200kg). The system is assumed to be undamped.

I am trying to find the force which is exerted by the spring upward onto the vehicle body.

Homework Equations



This is directly proportional to the compression of the spring but as I see it I just cannot make it work. Conservation of energy seems best suited, however I require the velocity of the wheel assembly (for ke = ½mv²) which I do not know. I have no time step increment with which to work out the velocity either.

The Attempt at a Solution



The acceleration of the wheel/axle assembly is specified in a load case. The problem as I see it is relating this acceleration to the deceleration caused by the spring. I am hopefully just missing something but this problem really has me stuck. If the spring weren’t there then the force on the vehicle body would be a simple F = ma where the acceleration is 50m/s² I think, but this is a much worse than real case.

The mass is accelerated (say +ve direction) at the 'start' of the action and is immediately deccelerated (-ve direction) by the spring. But how this relationship is expressed I can't fathom.

Any advice or views are appreciated, thank you.
 
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  • #2
What is the effect of an 'instantaneous' acceleration ? I don't what it means. Is it a constant acceleration that begins instantaneously ? If so the problem makes sense.
If it is applied for an instant, there won't be any effect.

I thnk you have to go with a constant force, which begins abruptly.
 
  • #3




The situation described in the homework statement is an example of an accelerated mass and spring system, where an external force (50m/s²) is applied to a mass (200kg) connected to a spring (k = 500kN/mm). In order to determine the force exerted by the spring on the vehicle body, we can use the equation F = kx, where F is the force, k is the spring constant, and x is the displacement of the spring from its equilibrium position.

In this case, the displacement of the spring can be calculated using the equation x = (m*a)/k, where m is the mass and a is the acceleration. Substituting the given values, we get x = (200kg * 50m/s²) / (500kN/mm) = 4mm.

Therefore, the force exerted by the spring on the vehicle body is F = (500kN/mm) * (4mm) = 2000N. This force will be directed upwards, opposing the downward force of the mass due to gravity.

In order to calculate the velocity of the wheel assembly, we can use the equation v² = u² + 2as, where u is the initial velocity (which we assume to be zero), a is the acceleration (50m/s²), and s is the displacement of the mass (4mm). Substituting the values, we get v² = 0 + 2*(50m/s²)*(0.004m) = 0.4m²/s². Taking the square root, we get v = 0.63m/s.

Therefore, the kinetic energy of the wheel assembly is KE = (1/2)*(200kg)*(0.63m/s)² = 39.69J. This energy is transferred from the external force to the spring, causing it to compress and exert a force on the vehicle body.

In summary, the force exerted by the spring on the vehicle body is directly proportional to the displacement of the spring, which is in turn proportional to the acceleration of the mass. By using the equations of motion and the conservation of energy, we can determine the force and velocity of the system. I hope this helps to clarify the problem and provide a solution.
 

1. What is an accelerated mass and spring system?

An accelerated mass and spring system is a physical system composed of a mass attached to a spring that is undergoing acceleration. This system is commonly used in physics experiments to study the effects of forces on a mass.

2. How does the mass affect the motion of the spring in an accelerated mass and spring system?

The mass attached to the spring affects the motion of the spring by determining the amplitude and frequency of the oscillations. A heavier mass will result in a slower oscillation with a larger amplitude, while a lighter mass will result in a faster oscillation with a smaller amplitude.

3. What factors affect the acceleration of a mass in an accelerated mass and spring system?

The acceleration of a mass in an accelerated mass and spring system is affected by the force applied to the mass, the mass of the object, and the stiffness of the spring. The greater the force applied or the stiffer the spring, the greater the acceleration will be. The mass of the object also plays a role, as a heavier mass will require more force to accelerate.

4. What is the equation for calculating the acceleration of a mass in an accelerated mass and spring system?

The equation for calculating the acceleration of a mass in an accelerated mass and spring system is a = -kx/m, where a is the acceleration, k is the spring constant, x is the displacement of the mass from its equilibrium point, and m is the mass of the object.

5. How is energy conserved in an accelerated mass and spring system?

In an ideal system, energy is conserved in an accelerated mass and spring system. The potential energy stored in the spring is converted to kinetic energy as the mass oscillates back and forth. As the mass reaches the equilibrium point, all of the potential energy is converted back to kinetic energy, resulting in a constant energy throughout the system.

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