Effective mass (spring–mass system)

In a real spring–mass system, the spring has a non-negligible mass

, at the other end of the spring, is not moving at all), its kinetic energy is not equal to

to determine the frequency of oscillation, and the effective mass of the spring,

The effective mass of the spring in a spring-mass system when using a heavy spring (non-ideal) of uniform linear density is

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.

The effective mass of the spring can be determined by finding its kinetic energy.

, 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,

is the length of the spring at the time of measuring the speed.

, from which it follows: Comparing to the expected original kinetic energy formula

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 (taking the upwards direction as positive, ignoring constant potential terms and setting the origin of potential energy at

, 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.

Both formulae reduce to the ideal case in the limit

Finally, the solution to the initial value problem: Is given by: Which is a simple harmonic motion.

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

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: To find the corresponding Lagrangian, one must find beforehand the potential gravitational energy of the spring: Due to the monotonicity of the integral, it follows that:

The above calculations assume that the stiffness coefficient of the spring does not depend on its length.

Jun-ichi Ueda and Yoshiro Sadamoto have found[1] that as

, the effective mass of a spring in a vertical spring-mass system becomes smaller than Rayleigh's value

This unexpected behavior of the effective mass can be explained in terms of the elastic after-effect (which is the spring's not returning to its original length after the load is removed).

is the moment of inertia of the system with respect to an axis that goes through the pivot.

by definition of centre-of-mass (this would be an integral equation in the general case,

,but it simplifies to this in the homogeneous case), whose solution is give by

Notice how the final expression is not a function on both the mass of the bob,

Also notice that initially it has the same structure as the spring-mass system: the product of the ideal case and a correction (with Rayleigh's value).

, the last correction term can be approximated by: Let's compare both results: