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The Spring-Mass System

Are you willing to learn about what is the spring-mass system and what are its various applications? What is the horizontal spring-mass system equation? Then this guide will help you to learn all about it.

Spring mass system

In a real spring-mass system, the mass of the spring is not negligible m. The total length of the spring is not moving at the same velocity v as the suspended mass M, so its kinetic energy is not equal to ½ mv2. Therefore, it is not possible to simply add m to M to determine the vibration frequency. The effective mass of the spring is defined as the mass added to it for correctly predicting system behaviour.

The effective mass of a spring in a spring-mass system with an ideal uniform linear density spring is 1/3 of the mass of the spring and is independent of the orientation of the spring-mass system. H. Horizontal, vertical and diagonal systems all have the same effective mass. This is because the external acceleration does not affect the travel time around the equilibrium point.

K = ∫m ½ u² dm

Since spring is uniform then,

dm = (dy / L) m

here, L is the length of the spring when measuring the speed

K = L∫0 ½ u² (dy /L) m

   = ½ m / L L∫0 u² dy

The velocity of each mass element of the spring is directly proportional to the length from where the spring is attached (faster when near the block, slower when near the ceiling).

u = vy / L

so,

K = ½ m /L L∫0  (vy / L)² dy

= ½ m / L³ v² L∫0  y² dy

= ½ m / L³ v² [y³/3]L0 

= ½ m/3 v²

In this case, the effective mass of the spring is m / 3. Using this result, the total energy of the system can be expressed as the displacement x from the unstretched position of the spring (ignoring certain potential terms and assuming upwards are positive).

T (total energy of the system) = ½ (m/3) v² + ½ Mv²  + ½ kx² – ½ mgx – Mgx

Here, g is the acceleration of gravity along the spring

(-m/3 – M) a = kx – ½ mg – Mg

Xeq = 1/k ( ½ mg + Mg)

(m/3 + M) a = -kx’

τ = 2Π (M + m/3 /k)½ 

Adding the effective mass of the spring to the mass of the load gives the “total effective mass” of the system. It should be used in standard expressions 2Π (m/k) ½. This is a period of oscillation.

Applications of the spring-mass system

General case

As mentioned above, the effective mass of a spring does not depend on “external” factors such as gravitational acceleration along the spring. In fact, the effective mass of a non-uniform spring depends only on the linear density p(x) along its length.

K = ∫m ½ u² dm

= L∫0 ½ u² p(x) dx

= L∫0 ½ (vx/L)² p(x) dx

= ½ [L∫0 x²/L² p(x) dx] v²

Then the effective mass of spring is

meff = L∫0 x²/L² p(x) dx

Real spring

The above calculation assumes that the stiffness coefficient of the spring does not depend on the length of the spring. However, this is not the case with real feathers. If the value of M / m is small, the displacement is not large enough to cause elastic deformation. Junichi Ueda and Yoshiyuki Sadamoto discovered that when M / m exceeds 7, the effective mass of vertical spring-mass springs falls below the Rayleigh value m / 3, eventually reaching a negative value. This unexpected behaviour of effective mass can be explained by the elastic sequelae (that is, the spring does not return to its original length after unloading).

Horizontal spring-mass equation

Imagine a mass connected to a frictionless horizontal spring. Understand the vibrational motion of the system

ΣF = -ma 

or, -kx = ma

Now, the acceleration in second time derivative of position is

d²x / dt² = -k/m x

in the generic form 

x(t) = A cos Ñ  t

or

x(t) = B sin Ñ  t

here, A is the amplitude of oscillation and Ñ  is the angular frequency

two-time derivatives of x(t) give

v(t) = dx/dt = -AÑ  sin Ñ  t

a(t) = d²x/dt² = dv/dt = -A Ѡ² cos Ѡ t

= a = – Ѡ² x

By comparing this equation, we get

Angular frequency Ѡ is related to the period T of the oscillation 

T = 2Π / Ѡ

Conclusion: 

The movement of the connected mass is described by two second-order differential equations. In the simplest case, friction and drag can be ignored and only elastic forces that follow Hooke’s law can be considered. It turns out that even such a simplified system has non-trivial dynamic characteristics. In general, the characteristics of motion are determined by two natural frequencies that depend on the system parameters “that is, the mass of the object and the spring constant”. In addition, the movement of the mass is strongly dependent on the initial conditions.

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