If we have a spring with natural length $l$, modulus of elasticity $\lambda$ and it has a weight of $m_1g$ attached to it, then its extension, $x_1$, is $$\frac{m_1gl}{\lambda}$$ If we add on another mass with weight $m_2g$ then we find the change in extension, $x_2$, is $$\frac{m_2gl}{\lambda}$$ after some manipulation.

Hooke's law states that the tension in a spring, $T$, is equal to the spring constant, $k$, multiplied by the extension of the spring, $x$:

$$T=kx=\frac{\lambda x}{l}~~(\text{as the spring constant is equal to $\frac{\lambda }{l}$})$$

Therefore, is it correct to state Hooke's law somewhat differently:

$$\Delta T=\frac{\lambda\Delta x}{l}$$ where $\Delta T$ is the change in tension in the spring.

Mathematically this seems sound; I'd just like to verify this is true.


There seems to be some confusion about the measurements and terms that I have used. To clarify:

The spring constant, $k$, is equal to $\lambda/l$; so my question is equivalently: is the following variation of Hooke's Law correct?

$$\Delta T=k\Delta x$$ where $\Delta T$ is the change in tension in the spring and $\Delta x$ is the change in extension.

  • $\begingroup$ Before you discuss a possible variation of Hooke's law, it will be useful to write the "original" law, with the meaning of the symbols. This will avoid confusion. $\endgroup$
    – nasu
    Commented Dec 24, 2020 at 15:51
  • $\begingroup$ @nasu ok, I'll do that now, thanks for the advice. $\endgroup$ Commented Dec 24, 2020 at 15:52
  • $\begingroup$ What is lambda in your formula? $\endgroup$
    – nasu
    Commented Dec 24, 2020 at 16:30
  • $\begingroup$ @nasu it's the modulus of elasticity; see here: revisionmaths.com/advanced-level-maths-revision/mechanics/… $\endgroup$ Commented Dec 24, 2020 at 16:31
  • $\begingroup$ This is a quite uncommon definition of modulus of ellasticity. At least for me. I'll have to look at what they do there. The definition I know and I think the other people answering or commenting too, is the one given by wiki here: en.wikipedia.org/wiki/Elastic_modulus $\endgroup$
    – nasu
    Commented Dec 24, 2020 at 22:01

1 Answer 1


There is a dimensional mistake, if you write $x_1=\frac{m_1gl}{\lambda}$ you are wrting dimensionally this: $$m=\frac{kg\cdot m\cdot m}{\frac{N}{m}\cdot s^2}$$ you are saying that $m=m^2$ but it's impossbile. Remember that Hooke's law is the following: $$\vec{F}=\lambda(\vec{x}-\vec{x}_0)$$ The force is directly proportional to the displacement.
So if you add a weight you find that $x_1=\frac{m_1g}{\lambda}$ if you add another mass $m_2$ you have the following equation: $$\vec{x}_2=\frac{(m_1+m_2)\vec{g}}{\lambda}$$

  • $\begingroup$ Please see my edit. $\endgroup$ Commented Dec 23, 2020 at 13:12
  • $\begingroup$ Ok, but it's dimensionally wrong. Tension is in newton and at the second member you have newton on meter. It's not the Hooke's law. $\endgroup$ Commented Dec 23, 2020 at 13:15
  • $\begingroup$ I think modulus of elasticity has the Newton as its unit, not Newton/metre. You may be thinking of the spring constant, which is equal to $\lambda/l$. Hooke's law is $T=kx$ where $k$ is the spring consant; $k=\lambda/l$, $T$ is the tension in the string/spring and $x$ is the extension/compression. $\endgroup$ Commented Dec 23, 2020 at 13:51
  • $\begingroup$ I know that I am correct :( Perhaps we have been taught the terminology differently. $\endgroup$ Commented Dec 23, 2020 at 13:56
  • 2
    $\begingroup$ Sorry if I insist, it's not correct. You can find everywhere in internet that Hooke's law doesn't depend on the lenght of the spring. Physics and Math are an universal language. $\endgroup$ Commented Dec 23, 2020 at 14:01

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