# Where does this period formula come from?

I extend a warm greeting to all of you. I just carried out a simple harmonic motion experience at lab. We're told to find out the relationship between Period and Mass. That thing is I've always found the period formula as:

$$T=2π\sqrt{\frac{m }{k} }$$

However, I found this formula at my lab's guide.

$$T= 2π\sqrt{\frac{m + \frac{M}{3}}{k} }$$

Were $M$ is the mas of the spring. I have not idea where this can come from. I only want to know some hints in case you know something about it.

I appreciate you helping me out.

• he formula that you first listed, $T=2\pi \sqrt{\frac{m}{k}}$ is only valid if the spring used is massless. This is an ideal case. If you consider that the springs mass as $M$, the time period will definitely vary. Sep 16, 2015 at 3:10
• Thank you! Well, I only know the SHM equations came from a differential equation, thing which I can't understand yet. But I also wonder why the mass of the spring was never used. Anyway, is there any book where I can find how to reach this period equation? Thank you again.
– Omar
Sep 16, 2015 at 3:22

This phenomena is called Effective mass on a spring–mass system.
You can see an easy explanation on wikipedia.

It basically says that:

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.

This is where the $M \over 3$ term comes in.

Now, the complete term of $m + {M \over 3}$ is when you find the effective mass of the spring by finding its kinetic energy. For a complete satisfying explanation, you must be familiar with lagrangian mechanics. But I hope this is ok for now.

• thank you! Just one question regarding the first equation. When is it useful or when should I use the second?
– Omar
Sep 16, 2015 at 3:10
• The easy answer is: Any time that the mass of the spring is not negligible. This is, it's not small enough compared with the other mass (how much is enough, it's up to you). Sep 16, 2015 at 3:17

With a mass-spring system, the mass attached to the spring, $m$, is not the only mass which is accelerating. The spring also has some accelerating mass, but each part of the spring stretches and accelerates differently. If one does a careful analysis of each loop of a spring, how it contributes to the total spring constant and how much mass each loop has, the ideal effect is that $M/3$ should be added to the mass $m$.

If the spring is not perfect (a tapered spring or some of the loops have been stretched), the spring mass contribution will be different.

My students did this experiment today with tapered springs and the springs contributed between $0.25M$ and $0.30M$.