Finding Spring Stiffness and Effective Mass for Suspended Beam

In summary, the first conversation involves finding the maximum dynamic tension and frequency of vibration for a 400 kg safe being lowered at 6 m/s using Newton's 2nd law and the equation ω=k/m. In the second conversation, the task is to find the spring stiffness of two springs and the effective mass of a uniform beam using the parallel spring equation and the period of vertical vibration with and without a load.
  • #1
rock.freak667
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Homework Statement


A cable is used to suspend a 400 kg safe. It is being lowered at 6 m/s, when it stops suddenly. Find the maximum dynamic tension and the frequency of vibration. k=2(105 N/m


Homework Equations



Newton's 2nd law

The Attempt at a Solution



Well I easily found the frequency using ω=k/m and got the correct answer.

If I find the resultant force in the direction of mg, I get mg-T=ma. I am not given a, I am given, v.


Homework Statement


A uniform beam is suspended by 2 springs A & B. Unloaded, the period of vertical vibration is 0.83s. When 50kg is loaded onto its centre, the period is 1.52s. Find the spring stiffness of each spring and the effective mass of the beam


Homework Equations





The Attempt at a Solution



The springs would be in parallel so

keq=2k

ω=√keq/m ⇒(2π/0.83)^2 (50) = 2k, but this does not give me the correct answer. k=608 N/m. I get 1432.66 N/m
 
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  • #2
1) Right, the velocity they give you will be an initial condition that you can use with the general equation for a mass on a spring. Max tension should be at max deceleration.

2) It gives you the period for the unloaded beam. Once the beam is loaded with more mass it will have a different period. :)
 
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FAQ: Finding Spring Stiffness and Effective Mass for Suspended Beam

1. What is Newton's second law?

Newton's second law states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. This can be expressed as the equation F=ma, where F is the net force, m is the mass of the object, and a is the acceleration.

2. How does Newton's second law apply to simple harmonic motion (SHM)?

In SHM, the force acting on an object is directly proportional to its displacement from its equilibrium position. This force is known as the restoring force and is responsible for the periodic motion of the object. According to Newton's second law, this restoring force is directly proportional to the acceleration of the object, which in turn is directly proportional to its displacement. This relationship is expressed as F = -kx, where k is the spring constant and x is the displacement from equilibrium.

3. What is the relationship between mass and period in SHM?

According to Newton's second law, the acceleration of an object in SHM is directly proportional to its mass. This means that a heavier object will have a smaller acceleration and therefore a longer period compared to a lighter object. The relationship between mass and period can be expressed as T = 2π√(m/k), where T is the period, m is the mass, and k is the spring constant.

4. How does amplitude affect the motion in SHM?

Amplitude, which is the maximum displacement from equilibrium, affects the motion in SHM by determining the maximum velocity and acceleration of the object. The larger the amplitude, the larger the velocity and acceleration will be. However, the period of the motion will remain the same regardless of the amplitude. This is because the restoring force is directly proportional to the displacement, not the amplitude.

5. Can Newton's second law be applied to non-mechanical systems in SHM?

Yes, Newton's second law can be applied to non-mechanical systems in SHM as long as there is a restoring force proportional to the displacement. This can be seen in systems such as a mass attached to a spring, a pendulum, or an LC circuit. In these cases, the net force acting on the object is still proportional to its acceleration, even though the nature of the force may differ from traditional mechanical systems.

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