# Variation of Hooke's Law Confirmation

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.

EDIT:

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.

• 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.
– nasu
Dec 24, 2020 at 15:51
• @nasu ok, I'll do that now, thanks for the advice. Dec 24, 2020 at 15:52
• What is lambda in your formula?
– nasu
Dec 24, 2020 at 16:30
• @nasu it's the modulus of elasticity; see here: revisionmaths.com/advanced-level-maths-revision/mechanics/… Dec 24, 2020 at 16:31
• 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
– nasu
Dec 24, 2020 at 22:01

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: $$\vec{x}_2=\frac{(m_1+m_2)\vec{g}}{\lambda}$$
• 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. Dec 23, 2020 at 13:51