Spring: does it have mass or is it massless?

[TjGrinnell]

Homework Statement

A spring has a spring constant of 10.0 N/m. I suspend a mass of 0.250 kg from the spring and set it in motion with an amplitude of 3.50 cm. I measure the period of the motion to be 1.15 s.
Is the spring massless? How do you know? If it is not massless, what is its effective mass?
If I suspend a mass of 0.500 kg from the spring and set it in motion with an amplitude of 2.75 cm, what will the period be?

Relevant Equations

F=-kx
T=2pi*sqrt(m/k)

I can easily do the second problem if only I knew the answer to the first. I am just not sure how I would go about figuring out if the spring has mass or not. And if it does, how would I calculate that mass?

 

[etotheipi]

If the spring in the question were massless, what would the period of the motion be?

 

[TjGrinnell]

If the spring in the question were massless, what would the period of the motion be?

Ok... I was definitely trying to over complicate this. So to get the mass of the spring I would just have to solve for m and subtract the .250 kg mass, is this right? 1.15=2pi*sqrt(m/10)

 

[etotheipi]

It's the right idea, unfortunately life is a little bit more complicated. The effective mass of a spring-mass system if $M$ is the mass of the hanging object and $m$ is the mass of the spring is $M + \frac{m}{3}$.

That is to say that the time period of a spring-mass system where the spring has mass $m$ is $$T = 2\pi \sqrt{\frac{M+\frac{m}{3}}{k}}$$

 

[TjGrinnell]

It's the right idea, unfortunately life is a little bit more complicated. The effective mass of a spring-mass system if $M$ is the mass of the hanging object and $m$ is the mass of the spring is $M + \frac{m}{3}$.

That is to say that the time period of a spring-mass system where the spring has mass $m$ is $$T = 2\pi \sqrt{\frac{M+\frac{m}{3}}{k}}$$

I see, I really appreciate the help.

 

[etotheipi]

If you're interested, you can find a proof of it here.

 

[Chestermiller]

The analysis in Wiki makes the tacit assumption that the deformation of the spring is homogeneous. Otherwise, the displacement distribution along the spring is much more complicated, and described by the wave equation.

 

[Merlin3189]

So when they ask for the "effective mass", do they want m or m/3 ?

 

[etotheipi]

So when they ask for the "effective mass", do they want m or m/3 ?

I believe the effective mass of the spring is termed $\frac{m}{3}$, as opposed to the real mass of the spring $m$. I suppose you're right in that we could've answered the question without knowing how this term related to the actual mass!

My mistake for not reading the question carefully enough

 

[Merlin3189]

No criticism intended -1
I didn't remember the m/3 and had just thought the rest of question might be answered with an additional mass added to the actual mass. I wondered if that's what they meant by effective mass.