SHM with mass of spring included

[Alettix]

Normally in high school physics-textbooks, the following formula for the period of simple harmonic motion (SMH) for a object on a spring is derived:

T²= 1/(4π²k)*m

where T is the period, k the springconstant and m the mass of the object on the spring. This is usually acquired by setting up a force-equation for the object and solving the obtained differentialequation.
However, the mass of the spring itself isn't included in this formula. So today, I stumbled across the following formula for the period of SHM:

T²= 1/(4π²k)*(m+1/3m₀)

where m₀ is the mass of the spring.

My question is, do any of you know how this formula is derived? I understand that the "whole mass" of the spring won't affect the period of the motion, because every part of the spring doesn't have the same acceleration (not the same force equation). What I wonder about is why it's exactly 1/3.

Thank you! :)

 

[PeroK]

Normally in high school physics-textbooks, the following formula for the period of simple harmonic motion (SMH) for a object on a spring is derived:

T²= 1/(4π²k)*m

where T is the period, k the springconstant and m the mass of the object on the spring. This is usually acquired by setting up a force-equation for the object and solving the obtained differentialequation.
However, the mass of the spring itself isn't included in this formula. So today, I stumbled across the following formula for the period of SHM:

T²= 1/(4π²k)*(m+1/3m₀)

where m₀ is the mass of the spring.

My question is, do any of you know how this formula is derived? I understand that the "whole mass" of the spring won't affect the period of the motion, because every part of the spring doesn't have the same acceleration (not the same force equation). What I wonder about is why it's exactly 1/3.

Thank you! :)

If you take the mass of the spring into account, then you have a mass with a variable velocity along its length. If you assume uniform linear density of the spring, then the formula is not too hard to derive.

Regarding the factor of 1/3, it's the same factor as the moment of inertia of a rod about one end.

 

[Alettix]

If you take the mass of the spring into account, then you have a mass with a variable velocity along its length. If you assume uniform linear density of the spring, then the formula is not too hard to derive.

Can uniform linear density be assumed, even if the downmost parts of the spring will stretch more than the upper ones? Or will they?

Regarding the factor of 1/3, it's the same factor as the moment of inertia of a rod about one end.

I shouldn't try to make somekind of analogy between SMH and rotation of a rod, should I?

 

[PeroK]

Can uniform linear density be assumed, even if the downmost parts of the spring will stretch more than the upper ones? Or will they?
I shouldn't try to make somekind of analogy between SMH and rotation of a rod, should I?

If you imagine stretching a spring it should still be uniform.

There's a mathematical analogy there rather than a physical one.

 

[haruspex]

I tried this from first principles, but in the special case of no attached mass. I get a different result.
I see the result quoted in the OP derived at https://en.wikipedia.org/wiki/Effective_mass_(spring–mass_system), but this line bothers me:

The velocity of each mass element of the spring is directly proportional to its length, i.e.

,​

This is treating the spring as if uniformly stretched at all times. This appears to be an approximation only valid when the spring's mass is a small fraction of the attached mass. A static hanging massive spring with no attached mass will not be stretched at all at its lowest element.

In this situation it is better to work in terms of elastic modulus, $\lambda$, than spring constant. If the spring is mass m, length L, density $\rho$ then $m=L\rho$, $\lambda=kL$.
If y = y(x,t) is the position (measuring downwards) of the element that would be x from the top in the relaxed spring, then
$\left(\frac {\partial^2y}{\partial t^2}-g\right)\rho = \lambda\frac{\partial^2y}{\partial x^2}$
In general, there may be harmonic oscillations within the spring. The static solution is $y = Y(x) = x + \frac{g\rho x}{\lambda}(L-\frac x2)$
I.e. the displacement of the 'x' element is $\frac{g\rho x}{\lambda}(L-\frac x2)$.
The simplest oscillation corresponds to $y = Y(x)+Z(x)\sin(\omega t)$.
From that I get $Z=A \sin\left(\omega\sqrt{\frac{\rho}{\lambda}}x\right)$. Applying boundary condition $Z'(L)=0$ (and Z' nonzero for any x between 0 and L), $\omega=\frac{\pi}2\sqrt{\frac km}$.