# Why does the spring constant not depend on the mass of the object attached?

It is said that:

$$F = -m\omega^2 x = -kx,$$

so $$k=m\omega^2$$. Since $$k$$ is the spring constant it doesn't depend on the mass of the object attached to it, but here $$m$$ signifies the mass of the object. Then how is $$k$$ independent of the mass attached?

• math.meta.stackexchange.com/q/5020 Hi. Use Latex to render formulas.
– Gert
Aug 20 '20 at 15:12
• Because $\omega$ isn't a constant, and it depends on mass itself. Aug 20 '20 at 15:48

Then how $$k$$ is independent of mass attached?

The clue is in :

$$F=kx$$

It states simply that the spring, when extended by $$x$$, will provide a restoring force $$F=kx$$.

The force needed to affect the extension (displacement) $$x$$ can be provided by almost anything. A mass (its weight) can do it but is just one way, one way of many.

$$\omega$$ isn't a constant of the spring, but it actually depends on the mass you attach to the spring. $$\omega$$ refers to the frequency of oscillation of the attached mass. The formula for $$\omega$$ for an attached mass $$m$$ is $$\sqrt{\frac{k}{m}}$$, where $$k$$ is the spring constant. If you use $$\omega=\sqrt{\frac{k}{m}}$$ in the formula, $$m$$ cancels out leaving only $$k$$

The unit of $$k$$ is $$\frac N m=\frac{kgm}{{sec}^2}/m=\frac{kg}{{sec}^2}$$, so $$kx$$ has the unit of a force, which is explicitly stated in Hooke's law, $$F=ma=kx$$. This means you can divide the mass out on both sides of the equation. So Hooke's law doesn't depend on the mass attached to a string.