Effective mass (spring–mass system)
In a real spring–mass system, the spring has a non-negligible mass. Since not all of the spring's length moves at the same velocity as the suspended mass, its kinetic energy is not equal to. As such, cannot be simply added to to determine the frequency of oscillation, and the effective mass of the spring,, is defined as the mass that needs to be added to to correctly predict the behavior of the system.
Uniform spring (homogeneous)
The effective mass of the spring in a spring–mass system when using a heavy spring of uniform linear density is of the mass of the spring and is independent of the direction of the spring–mass system. This is because external acceleration does not affect the period of motion around the equilibrium point.The effective mass of the spring can be determined by finding its kinetic energy. For a differential mass element of the spring at a position moving with a speed, its kinetic energy is:
In order to find the spring's total kinetic energy, it requires adding all the mass elements' kinetic energy, and requires the following integral:
If one assumes a homogeneous stretching, the spring's mass distribution is uniform,, where is the length of the spring at the time of measuring the speed. Hence,
The velocity of each mass element of the spring is directly proportional to length from the position where it is attached, i.e., from which it follows:
Comparing to the expected original kinetic energy formula the effective mass of spring in this case is. This result is known as Rayleigh's value, after Lord Rayleigh.
To find the gravitational potential energy of the spring, one follows a similar procedure:
Using this result, the total energy of system can be written in terms of the displacement from the spring's unstretched position :
Note that here is the acceleration of gravity along the spring. By differentiation of the equation with respect to time, the equation of motion is:
The equilibrium point can be found by letting the acceleration be zero:
Defining, the equation of motion becomes:
This is the equation for a simple harmonic oscillator with angular frequency:
Thus, it has a smaller angular frequency than in the ideal spring. Also, its period is given by:
Which is bigger than the ideal spring. Both formulae reduce to the ideal case in the limit.
So 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 to determine the period of oscillation.
Finally, the solution to the initial value problem:
Is given by:
Which is a simple harmonic motion.
General case
As seen above, the effective mass of a spring does not depend upon "external" factors such as the acceleration of gravity along it. In fact, for a non-uniform heavy spring, the effective mass solely depends on its linear density along its length:So the effective mass of a spring is:
This result also shows that, with occurring in the case of an unphysical spring whose mass is located purely at the end farthest from the support.
Three special cases can be considered:
- is the idealised case where the spring has no mass, and.
- is the homogeneous case where Rayleigh's value appears in the equation, i.e.,.
- , where is Dirac delta function, is the extreme case when all the mass is located at, resulting in.
Due to the monotonicity of the integral, it follows that:
With the Lagrangian being:
Real spring
The above calculations assume that the stiffness coefficient of the spring does not depend on its length. However, this is not the case for real springs. For small values of, the displacement is not so large as to cause elastic deformation. In fact for, the effective mass is . Jun-ichi Ueda and Yoshiro Sadamoto have found that as increases beyond, the effective mass of a spring in a vertical spring-mass system becomes smaller than Rayleigh's value and eventually reaches negative values at about. This unexpected behavior of the effective mass can be explained in terms of the elastic after-effect.Comparison with pendulum
Consider the pendulum differential equation:Where is the natural frequency of oscillations. The parameter stands for in an ideal pendulum, and in a compound pendulum, where is the length of the pendulum, is the total mass of the system, is the distance from the pivot point to the pendulum's centre-of-mass, and is the moment of inertia of the system with respect to an axis that goes through the pivot.
Consider a system made of a homogeneous rod swinging from one end, and having attached bob at the other end. Let be the length of the rod, the mass of the rod, and the mass of the bob, thus the linear density is given by, with Dirac's delta function. The total mass of the system is. To find out one must solve by definition of centre-of-mass, whose solution is give by. The moment of inertia of the system is the sum of the two moments of inertia, . Thus the expression can be simplified:
Notice how the final expression is not a function on both the mass of the bob,, and the mass of the rod,, but only on their ratio,. Also notice that initially it has the same structure as the spring–mass system: the product of the ideal case and a correction. Notice that for, the last correction term can be approximated by:
Let's compare both results:
- For the spring–mass system:
- For the pendulum: