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Effective mass (spring–mass system)
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Effective mass (spring–mass system) : ウィキペディア英語版
Effective mass (spring–mass system)
In a real spring–mass system, the spring has a non-negligible mass m. Since not all of the spring's length moves at the same velocity u as the suspended mass M, its kinetic energy is not equal to m u^2 /2. As such, m cannot be simply added to M to determine the frequency of oscillation, and the effective mass of the spring is defined as the mass that needs to be added to M to correctly predict the behavior of the system.
==Ideal uniform spring==

The effective mass of the spring in a spring-mass system when using an ideal spring of uniform linear density is 1/3 of the mass of the spring and is independent of the direction of the spring-mass system (i.e., horizontal, vertical, and oblique systems all have the same effective mass). This is because external acceleration does not affect the period of motion around the equilibrium point.
We can find the effective mass of the spring by finding its kinetic energy. This requires adding all the length elements' kinetic energy, and requires the following integral:
:T =\int_m\fracu^2\,dm
Since the spring is uniform, dm=\left(\frac\right)m, where L is the length of the spring. Hence,
:T = \int_0^L\fracu^2\left(\frac\right)m\!
::=\frac\frac\int_0^L u^2\,dy
The velocity of each mass element of the spring is directly proportional to its length, i.e. u=\frac, from which it follows:

:T =\frac\frac\int_0^L\left(\frac\right)^2\,dy
:=\frac\fracv^2\int_0^L y^2\,dy
:=\frac\fracv^2\left()_0^L
:=\frac\fracv^2
Comparing to the expected original kinetic energy formula \fracmv^2, we can conclude that effective mass of spring in this case is ''m''/3. Using this result, we can write down the equation of motion of the spring by writing the Lagrangian in terms of the displacement x from the spring's unstretched position (ignoring constant potential terms and taking the upwards direction as positive):
: L = T - V
:: = \frac\frac\dot^2 + \fracM \dot^2 - \frac k x^2 - \frac - M g x
Note that g here is the acceleration of gravity along the spring. Using the Euler–Lagrange equation, we arrive at the equation of motion:
:\left( \frac+M \right) \ddot = -kx - \frac - Mg
We can find the equilibrium point x_} = -\frac\left( \frac + Mg \right)
Defining \bar = x - x_+M \right) \ddot
This is the equation for a simple harmonic oscillator with period:
:\tau = 2 \pi \left( \frac \right)^
So we can see that the effective mass of the spring added to the mass of the load gives us the "effective total mass" of the system that must be used in the standard formula 2 \pi \left( \frac \right)^ to determine the period of oscillation.

抄文引用元・出典: フリー百科事典『 ウィキペディア(Wikipedia)
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