Simple Harmonic Motion Question

In summary, the conversation discusses the behavior of a mass attached to a horizontal spring and a loose mass placed in front of it. It is assumed that there is no friction and the spring is ideal. The masses will separate when the spring starts to slow down and the 4.5 KG mass will stop at a height of 0.4 m above its initial position if it shoots vertically.
  • #1
descartes75
3
0
I am going brain dead on a rather simple question. Appreciate some help.


A mass of 4.5 KG attatched to a horizontal spring of constan 75 KN/m is compressed 0.4 m from Equilibirium and then a loose mass of 2.5 KG is placed in front of it . IF the mass is now released:

1. at what point do the masses separate.
2. How far apart are they when the 4.5 KG mass stops?


Assume no friction.
 
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  • #2
part 1:
the mass moves at the same speed with the spring while they are in contact, so, they will separate at the point when the spring started to slow down (deccelarate) [assuming the spring is ideal, massless, and ignore the effect of gravity]
part 2:
the mass will not stop if it shoots horizontally... coz you assume no friction...
if it shoots vertically, you can use conservation of energy... the energy stored in spring is 1/2 kx^2
 
  • #3


1. The masses will separate when the spring force acting on the 4.5 KG mass is equal to the weight of the 2.5 KG mass. This can be calculated using the equation F = kx, where F is the force, k is the spring constant, and x is the displacement from equilibrium. Plugging in the given values, we get F = (75 KN/m)(0.4 m) = 30 KN. Since the weight of the 2.5 KG mass is 24.5 N, the masses will separate when the displacement of the 4.5 KG mass is 30 KN/24.5 N = 1.22 m from equilibrium.

2. To find the distance between the masses when the 4.5 KG mass stops, we need to use the conservation of energy principle. At the point where the 4.5 KG mass stops, all the potential energy stored in the spring will be converted into kinetic energy of the masses. We can set up the equation as follows:

1/2kx^2 = 1/2mv^2

Where k is the spring constant, x is the displacement of the 4.5 KG mass, m is the total mass (4.5 KG + 2.5 KG = 7 KG), and v is the velocity of the masses. Solving for x, we get x = √(mv^2/k). Plugging in the values, we get x = √(7 KG)(30 KN/7 KG)/75 KN/m = 0.86 m. Therefore, the distance between the masses when the 4.5 KG mass stops is 0.86 m.

Note: It is always important to check the units and make sure they are consistent throughout the calculations. In this case, we converted the units of the spring constant from KN/m to N/m to match the units of the other quantities.
 

What is Simple Harmonic Motion?

Simple Harmonic Motion (SHM) is a type of periodic motion in which an object oscillates back and forth around an equilibrium point, with a constant amplitude and a constant period. It occurs when the restoring force on the object is directly proportional to the object's displacement from the equilibrium point.

What is an example of Simple Harmonic Motion?

A pendulum is a common example of Simple Harmonic Motion. As it swings back and forth, its motion follows a sinusoidal curve and its period remains constant as long as the amplitude of oscillation is small.

What is the equation for Simple Harmonic Motion?

The equation for Simple Harmonic Motion is x(t) = A cos(ωt + φ), where x(t) is the displacement of the object at time t, A is the amplitude of oscillation, ω is the angular frequency, and φ is the phase constant.

What is the difference between Simple Harmonic Motion and Uniform Circular Motion?

The main difference between Simple Harmonic Motion and Uniform Circular Motion is that SHM involves a back and forth motion along a straight line, while UCM involves a constant circular motion around a fixed point. Additionally, the acceleration in SHM is directly proportional to the displacement, while the acceleration in UCM is constant and always perpendicular to the velocity.

How is Simple Harmonic Motion related to energy?

In Simple Harmonic Motion, the total energy (kinetic + potential) remains constant. As the object oscillates between its extreme positions, the energy is constantly being converted between kinetic and potential energy. At the equilibrium point, the energy is entirely potential, while at the maximum displacement, the energy is entirely kinetic.

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