Why do different parts of a massive spring expand proportional to their distance while the spring has some mass hung on the bottom which is comparably very less than the mass of the spring? If we take a massless (or practically negligible mass) spring then the elongation of the spring is uniform. But why does it make an exception when the spring has mass? A theoretical analytical answer will be appreciated. Now you might say it's an approximation but the approximation should be nearly true.

My book, The Physics of Waves and Oscillations by N. K. Bajaj, says:

Let us assume that the various parts of the spring undergo displacements proportional to their distance from the fixed end as indicated in the diagram. We can now compute the total kinetic energy of the spring at an instant when the mass M has a displacement $\phi$. Now the displacement per unit length of the spring is $\phi/l$.

(from page 40)

  • $\begingroup$ I think this is a good question.. wondered the same thing when I was an undergrad.. I remember there being an additional assumption that the mass per unit length or something was held constant $\endgroup$ Commented Sep 16, 2018 at 2:24
  • $\begingroup$ @InertialObserver in addition they have said that the mass of the string is more than the hanged mass . And yep the mass of the spring is distributed uniformly. But after hanging the mass in the bottom this should not be the same . As already the book stated that the different parts of the spring will expand at different amount so the densities shouldn't be uniform. $\endgroup$ Commented Sep 16, 2018 at 3:23
  • $\begingroup$ Is your question why different parts of the spring have different stretch? Or is your question why the variation in the stretch is linear ("proportional") rather than some other function? The first is pretty easy to argue, since the top of a massive dangling spring is supporting more weight than the bottom. The second is more complicated. $\endgroup$
    – rob
    Commented Sep 17, 2018 at 18:41

3 Answers 3


Although the easiest method of experimenting with the oscillations of a mass on a spring is to hang the spring vertically, the simplest scenario for analysing these oscillations is to have the spring and mass resting on a frictionless horizontal surface. The weight of the spring then has no effect on the extension of the spring, which can be assumed to be uniform over its length.

Although the author's diagram Fig. 1.24 displays the spring vertically in the book, the extension is portrayed as being uniform over the length of the spring, with coils spaced at a constant distance apart, as in the horizontal case. The result which he obtains is only approximate for a vertical spring, as in his diagram; however, it is exact for a horizontal spring, which he does not appear to realise. (The exercise which he is solving does not state that the spring is suspended vertically, nor does it mention the weight of the load or spring, only their mass.)

At the end of the calculation on p 42 the author states :

The above calculation is not exact because we have assumed that the extension of an element of the spring is proportional to the distance from the fixed end and that the velocity $d\phi/dt$ is the same for all elements of the spring. In fact, different elements undergo different accelerations. The expression for the time period will hold if $m \ll M$, in which case the stretching force does not vary appreciably with distance along the spring and can be treated as roughly constant.

Source : Google Books preview

For a spring of finite mass $m$ which is hanging vertically and carrying load $M$, the weight supported by each section of the spring (hence the tension in each section) increases linearly from $Mg$ at the bottom to $(M+m)g$ at the top. The extension of each section increases from the bottom to the top, in proportion to the tension. The extension is not uniform along the length of the spring, as it is in the horizontal case : it increases linearly from bottom to top. The average tension is $(M+\frac12 m)g$ so the total extension is $x$ where $(M+\frac12m)g=kx$. The effective mass of the spring is different for static extension ($\frac12 m$) than it is for dynamic extension, ie oscillation ($\frac13 m$).

In fact even for the horizontal case the author's calculation of the period applies only for one possible mode in which the spring oscillates - ie all sections being in phase with each other. Other modes are possible, with phase differences between sections, and different periods for each mode.


The stretched length is a smooth function of the mass placed at the end of the spring.

Then by using Taylor's theorem, to a first approximation, this smooth function will be linear. This means it will have a constant term and a term proportional to the mass.

On physical grounds we can discount the constant term - since including this would mean that the spring will stretch itself when there is no mass attached.

This then just leaves a term that is proportional to the mass. The constant of proportionality here is Hooke's constant.

This means that the real physics of Hooke's law really relies on understanding why the assumption of smoothness can be made, the validity of the approximation and why we can remove the constant term - which is essentially Newton's first law.

  • $\begingroup$ Please elaborate with some equations. $\endgroup$ Commented Sep 16, 2018 at 11:05
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    $\begingroup$ @Nobody recognizeable: but would you recognise them if I did? The point of my answer is to show where the physics of the situation lies and not carry out the calculation. If you can follow the reasoning above - and it's not complicated - a bright sixteen year old should be able to follow it - then you should be able to fill out the rest yourself. $\endgroup$ Commented Sep 16, 2018 at 11:09
  • $\begingroup$ I'll more happy if you explain the full phonemena with theoretically and mathematically. But please try to define every term. I'll try my best to understand. But if I can't I'll have to strive and that's it . @moziburullah $\endgroup$ Commented Sep 16, 2018 at 11:12

I think it might be because the helical springs when compressed are not actually compressed, they are twisted.The spring as a whole might be acted upon by an axial compressive load but the wire is actually twisted.With twisting, the deformation at the far end(away from fixed) will be more than the fixed end.But Iam not sure about this.


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