# SHM with mass of spring included

• Alettix
In summary: This is the same as the frequency for a point mass on a massless spring, from Hooke's law. But the amplitude is not the same. For a point mass on a massless spring, the amplitude would be ##Z=A \sin\left(\omega\sqrt{\frac{\rho}{k}}x\right)##. The amplitude depends on the mass of the spring. And the mass of the spring is only a fraction of the total mass when the spring contains a large attached mass.In summary, there are two formulas for calculating the period of simple harmonic motion (SMH) for an object on a spring. The first formula does not take into account the mass of the spring, while the
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! :)

Alettix said:
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.

Last edited:
PeroK said:
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?

Alettix said:
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.

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}##.

## 1. How does the mass of a spring affect its SHM?

The mass of a spring affects its SHM by changing its natural frequency. The natural frequency is the frequency at which the spring oscillates without any external forces acting on it. A heavier mass on the spring will result in a lower natural frequency and a slower oscillation.

## 2. What is the relationship between the mass of the spring and its period?

The mass of the spring and its period have an inverse relationship. This means that as the mass of the spring increases, the period of its oscillations will decrease. This relationship is expressed by the equation T = 2π√(m/k) where T is the period, m is the mass of the spring, and k is the spring constant.

## 3. Can the mass of the spring affect its amplitude?

The mass of the spring does not directly affect its amplitude, which is the maximum displacement from equilibrium. However, a heavier mass on the spring can cause the spring to stretch more, resulting in a larger amplitude. The amplitude of a spring's SHM is primarily affected by the initial conditions and the amount of energy put into the system.

## 4. How does the mass of the spring affect its energy?

The mass of the spring affects its energy by changing the amount of potential and kinetic energy it has. As the mass increases, the potential energy of the spring also increases, while the kinetic energy decreases. This is because a heavier mass will require more energy to move, resulting in a slower oscillation and lower kinetic energy.

## 5. Is there a limit to how much mass a spring can hold during SHM?

Yes, there is a limit to how much mass a spring can hold during SHM. This is due to the maximum weight that a spring can support without permanently deforming or breaking. This limit is known as the elastic limit and varies for different types of springs. Once this limit is exceeded, the spring will no longer exhibit SHM and may become damaged.

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