Question

What would be the extension of a system of a spring and a mass, 'M' where in the spring itself has a mass, 'm'?

Answer 1

It's complex to include the mass of the spring because it expands sectionally.
In most cases, only include the mass of the object. So M*g = k*x
g is accel due to gravity, x is the extension and k is the force constant of the spring.
So, the extension x = M*g/k.

Answer 2

i think we need to go to the centre of mass frame
i'm not sure but i'll try and post it:)

Answer 3

go ahead @rajat97 :)

Answer 4

yup i'm trying but it's midnight (IST) so i'm feeling a bit sleepy
but i'm trying it:)
and what do you think will it be done by the c-frame??

Answer 5

yes

Answer 6

You need to write net force = 0 for an infinitesimal part of the spring of mass dm.

Answer 7

i'll post a drawing of what i thought but i doubt it as the displacement of the centre of mass of the spring may not be equal to the displacement of the c.o.m. of the system

Answer 8

i think we need to go with integration as @Vincent-Lyon.Fr said:)

Answer 9

every infinitesimal part of the spring will experience two forces on is the restoring spring force and other is the weight of the block and that of itself

Answer 10

is it right?

Answer 11

but how will we relate it to the distance??

Answer 12

if i go by the c.o.m. frame , i get the extension=(M+m)g/k
where k is the spring constant

Answer 13

what do you think??

Answer 14

No, you are studying a static problem. There is no different point of view in different frames.

Answer 15

oh!
thanks for that:)
but i don't know what a static problem really is

Answer 16

Use elemental length (at rest) \(dl\) with mass \(dm=\mu \;dl\) and write that net force is zero, callinf \(\vec T(x)\) the tension of the upper part acting on the lower part.

I have no time to derive it now, but I'll do it after dinner.

Answer 17

If you have studied waves already, it is the same kind of demo as when you derive propagation equation on a string.

Answer 18

yeah sure!

Answer 19

i've not studied waves

Answer 20

even i need to go now

Answer 21

My recollection is that Prof. Lewin mentioned the answer to this, found
by a colleague, during one of his mit.edu MIT lectures. One-third of the spring mass was added to the mass at the end of the spring, if I recall correctly.

Answer 22

Let's consider a spring of length at rest L, mass m and constant k.
According to the additivity of spring constants, any part of this spring of length dL will have a constant kL/dL. Let's write e = kL.
e is the equivalent of Young's modulus for this spring.

Let's call x the position of any coil of the spring when it is at rest and u(x), the displacement of this coil when the spring is under tension.

Hooke's law becomes: T(x) = e du/dx

or du/dx = T(x)/e = T(x)/kL (from now, we will not use e anymore)

Then \(u(x)=\int_0^x du=\frac{1}{kL}\int_0^x T(x)dx\)

Let's pull the spring with a given force F.
When the spring has no mass, or when it has a mass but is horizontal, tension T is the same throughout the length of the spring and is equal to F.


Then u(x) is the area under the curve from 0 to x divided by kL.
For x=L, total lengthening of the spring will be: \(\Delta L=u(L)=FL/kL=F/k\)
which is the expected answer.

Now, if the spring has a mass and is hung vertically, then tension will be greatest at the top and decreasing as you go from x=0 to L since it has to balance the weight of the bottom part of the spring.
T(L) = F by choice of applying a force F there;
T(0) = F + mg
The graph looks like:


Now, the total area under the curve will be (F + mg/2)L
and the new elongation will be : F/k + mg/2k

You now see that the mass m of the springs acts as an extra m/2 mass acting on a massless spring.

If F is the weight of a mass M, then extension will be (M + m/2)g/k , which answers cskc97's initial question.