Simple horizontal harmonic oscillator with spring that has a mass.

In summary, the mass of the spring does affect the force applied by the spring, which in turn affects the motion of the block. However, when using the equation F = Ma, the mass of the spring is not explicitly included because it is assumed to have no inertia. To accurately model the motion of the system, the inertia of the spring must be taken into account through a differential equation. This will result in vibrations at multiple frequencies instead of a simple harmonic motion.
  • #1
bgq
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0
Hi,

Consider a block of mass M connected to a spring of mass m and stiffness k horizontally on a frictionless table. We elongate the block some distance, and then release it so that it now oscillates.

According to the theoretical study using energy methods, we see that the mass of the spring affects the motion of the block.

But if we apply Newton's 2nd law to the system formed of the block alone, the mass of the spring do not appear in the equation!

Here it is: F = Ma then -kx = Ma then a=(-k/M)x where the weight and the normal reaction cancel each other.

As we see the mass of the spring does not appear in the equation.

My question is: Does the mass of the spring affect the motion of the block? and if yes, what is wrong in the way I have applied Newton's second law above?

Thanks in advance.
 
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  • #2
The mass of the spring will affect the force applied by the spring, as it requires a force to accelerate this mass. There is nothing wrong with your formula, it is just not sufficient to evaluate the motion of the system.
 
  • #3
The equation "F = -kx" assumes that the spring is stretched uniformly along its length at all times. That is consistent with the assumption that it has no inertia.

It would be more correct to say the force depends on the axial strain at the end of the spring. But if you ignore the inertia of the spring, the strain is uniform along the length, and is measured by x/L where L is the unstretched length of the spring. So the OP's equation F = -kx is correct if you ignore the inertia, and include the effect of the length L as part of the spring constant k.

If you include the inertia of the spring in the model, the spring is not uniformly stretched along its length. To model this you need to set up a differential equation describing the motion of all points along the spring, and the "mass on the end" then becomes part of the boundary conditions for the solution of that equation.

If you include the inertia of the spring, the solution of the equations of motion include vibrations in simple harmonic motion at more than one frequency (in fact, at an infinite number of different frequences), and any linear combination os those solutions - including solutions that will not look like "simple harmonic motion" and may not even be periodic, because the different frequencies for harmonic vibrations might not be in simple integer ratios like 2:1 or 3:2.
 
  • #4
Thank you very much
 
  • #5


I can provide a response to the content you have provided. Firstly, it is important to note that the mass of the spring does indeed affect the motion of the block in a simple horizontal harmonic oscillator. This can be seen through the equation for the period of oscillation, which is dependent on the mass of the spring (T = 2π√(m/k)).

In your application of Newton's second law, you have only considered the forces acting on the block itself (weight and normal reaction). However, in order to fully understand the motion of the block, we must also consider the forces acting on the spring. When the block is elongated and released, the spring exerts a force on the block, causing it to accelerate. This is why the mass of the spring is important in the equation for acceleration (a=(-k/M)x).

In summary, the mass of the spring does affect the motion of the block in a simple horizontal harmonic oscillator. It is important to consider all forces acting on the system, including those acting on the spring, in order to fully understand the motion of the block. I hope this helps clarify any confusion.
 

1. What is a simple horizontal harmonic oscillator?

A simple horizontal harmonic oscillator is a physical system that consists of a mass attached to a spring, with the spring fixed at one end and the mass free to move horizontally. The mass oscillates back and forth around a fixed equilibrium point due to the restoring force of the spring.

2. How does the mass affect the behavior of the oscillator?

The mass of the object attached to the spring affects the frequency of the oscillations. A heavier mass will result in a lower frequency, while a lighter mass will have a higher frequency. However, the mass does not affect the amplitude or the period of the oscillations.

3. What is the role of the spring in a simple horizontal harmonic oscillator?

The spring provides the restoring force that causes the oscillations in a simple horizontal harmonic oscillator. As the mass is displaced from its equilibrium position, the spring exerts a force in the opposite direction, which causes the mass to accelerate back towards the equilibrium point.

4. How is the motion of a simple horizontal harmonic oscillator described?

The motion of a simple horizontal harmonic oscillator is described by the equation x = A*sin(ωt), where x is the displacement of the mass, A is the amplitude of the oscillations, ω is the angular frequency, and t is the time. This equation represents a sinusoidal motion, with the mass oscillating back and forth around the equilibrium point.

5. How does the spring constant affect the behavior of the oscillator?

The spring constant, represented by the symbol k, determines the stiffness of the spring and therefore affects the frequency of the oscillations. A higher spring constant will result in a higher frequency, while a lower spring constant will have a lower frequency. However, the spring constant does not affect the amplitude or the period of the oscillations.

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