0
$\begingroup$

I'm trying to understand the following question:

The force generated by a spring of spring constant $k$ is $F = -kx$, where $x$ is the displacement from equilibrium. A cube of mass $m$ and length $L$ is attached to the spring, and oscillates around the equilibrium point.

enter image description here

I'm trying to determine which variables effect the frequency. From what I found it is dependent on $k$ and $m$ only. Why is that? I expected it to be effected from $x$. Any insight will be highly appreciated.

$\endgroup$
  • 1
    $\begingroup$ Why do you expect that? Have you solved the equation and actually found the frequency? $\endgroup$ – josepmercadal Nov 8 '17 at 9:06
1
$\begingroup$

If you pull on the cube so that it is a larger distance from the equilibrium point, the spring exerts a larger force on the cube. So, when you let go, the cube will accelerate faster due to the larger force. The cube will be moving faster than a cube with a smaller displacement, so it can cover a longer distance in the same time. For a force like a spring ($F = -kx$), the longer distance and greater acceleration offset each other, so the period of oscillation is the same.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ But the OP wants to know what affects the frequency? Wouldn’t it be changing either K or M? $\endgroup$ – Bill Alsept Nov 8 '17 at 17:06
1
$\begingroup$

Remember this is an idealized system. In practice, the spring constant $k$ will depend on $x$ for large $x$--validating your concerns. In the idealized case, consider dimensional analysis. You have 2 parameters with dimensions (M, mass; L, length; T, time):

$k \rightarrow $ Netwon/meter $\rightarrow \frac{\mathrm{ML/T^2}} {\mathrm L} = \frac{\mathrm M}{\mathrm T^2}$

and

$M \rightarrow {\mathrm M}$,

so that:

$\sqrt{\frac{k}{M}} \rightarrow \frac{1}{\mathrm T}$,

which is a frequency.

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

Adding to Mark's intuitive answer, I want to point out that this is a special property of a restoring force that is linear in the displacement. This is quite special and important in many areas of physics, so this special case gets its own name: (simple) harmonic motion.

In general, one may include higher-order corrections to the restoring force, e.g. cubic: $F=-kx-cx^3$. The additional terms are called anharmonicities and in that case, the resonant frequency does depend on the amplitude.

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

Because the frequency is defined as the number of oscillations(go and back) per second and that is not dependent on diplacement.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.