# I Acceleration of mass depends upon K and L of a spring?

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1. Dec 8, 2018

### John McAndrew

I have a question that appears elementary, but bizarre in its conclusion:

A mass $M$ is accelerated by a spring of length $L$, wave-speed $v_p$, spring-constant $K$ and a constant force $F$ at the other end. As $K$ increases, the extension of the spring $dx$ decreases as does the energy stored $\frac{1}{2} k{dx}^2$. The max rate at which work can be done by the spring on the mass is of the order:

$P$ = (energy stored in spring)/(propagation time over spring's $L$)
= $\frac{1}{2} k{dx}^2v_p/L$

Hence as $K\rightarrow \infty, P \rightarrow 0$: the mass can't accelerate!

Is this correct?
The physics seems to say yes, but I don't quite believe it and may have gone wrong somewhere.

Last edited: Dec 8, 2018
2. Dec 9, 2018

### jfizzix

For a constant initial pull-back force $F$
The initial displacement, and displacement amplitude of the oscillation decreases as $1/k$
(coming from $F=-k (x-x_{0})$).
The frequency of the oscillation increases with $\sqrt{k}$
(where $\omega=\sqrt{k/m}$).
The oscillation of the velocity of the mass has an amplitude of frequency times the displacement amplitude.
(if $x(t)=x_{0} + A sin(\omega t)$ then $dx/dt=A\omega cos(\omega t)$).

Therefore, the velocity amplitude decreases as $1/\sqrt{k}$, and goes to zero as $k\rightarrow\infty$

However, if we take another derivative in time, we see that the amplitude for acceleration is actually constant for changing values of k. So, the mass does always accelerate (in this model), and with a higher frequency for higher $k$. Effectively, the velocity is changing back and forth so fast (with constant amplitude) that the over time the average change in velocity is insignificant.

3. Dec 9, 2018

### John McAndrew

Your spring model assumes a wave velocity $v_p = \infty$ which means energy can be transported through the spring infinitely fast; whereas I'm thinking of something like a slinky spring with a finite $v_p$ transporting energy at a finite rate. Let's say the oscillations have died down and the spring has a constant extension accelerating the mass with a constant force; and the force suddenly goes to zero. The mass at the other end will continue to accelerate until the change in force reaches it a time $L/v_p$, with a maximum available energy during this period equal to that stored in the spring just before the change in force. If this stored energy is close to zero for a large enough $k$, then the rate at which energy can be transported to accelerate the mass, and therefore the acceleration, appears to be limited IMO.

4. Dec 9, 2018

### A.T.

Is that consistent with spring-constant → ∞ ?

5. Dec 10, 2018

### John McAndrew

I'm not sure, it was just a suggestion to get my picture of a finite $v_p$ across. Another suggestion would be steel wire giving $v_p = \sqrt{\frac{T}{\rho}}$ where $\rho$ is the linear mass density.

6. Dec 10, 2018

### A.T.

How is $v_p$ related to the stiffness of the spring or wire material?

7. Dec 11, 2018

### John McAndrew

The stiffness $k = AE/L$ where $A$ is the cross sectional area, $E$ is Young's modulus and $L$ is the length. Young's modulus $E = \sigma/\epsilon$ where $\sigma$ is the uniaxial force per unit area, and $\epsilon$ is the change in length divided by the original length.

$v_p$ depends upon the linear mass density which appears to me to be independent of either $\epsilon$ or $\sigma$, and therefore independent of the stiffness $k$.

8. Dec 12, 2018 at 12:04 AM

### A.T.

Is it consistent with your observation that the propagation speed in a spring doesn't depend on its stiffness?

9. Dec 12, 2018 at 6:29 PM

### John McAndrew

I think I understand what you're getting at now; before I thought you were criticizing my use of a slinky spring. This is where things got a little confusing for me because looking further, there appeared to be at least two possible expressions for the wave velocity in a steel wire:

$v_p = \sqrt{{\frac{T}{\mu}}}$ http://www.animations.physics.unsw.edu.au/jw/wave_equation_speed.htm
$v_p = \sqrt{{\frac{E}{\rho}}}$ https://en.wikipedia.org/wiki/Wave_equation#Derivation_of_the_wave_equation

The first was derived for 1-D transverse waves which I carelessly assumed would work for longitudinal also. So to answer your original question, $v_p \propto \sqrt{k}$ from the second expression. This now makes things more complicated since the wave velocity can go to the speed of light $c$, requiring a relativistic analysis.