# Frequency Increases when Stiffness Increases

## Theorem

### Problem Definition

Consider a cart $C$ of mass $m$ attached to a wall by means of a spring $S$.

Let $C$ be free to move along a straight line with zero friction.

Let the force constant of $S$ be $k$.

Let the displacement of $C$ at time $t$ from the equilibrium position be $\mathbf x$.

Then as the force constant (or "stiffness") of $k$ increases, the frequency of the motion of $C$ likewise increases.

## Proof

From Position of Cart attached to Wall by Spring, the horizontal position of $C$ is given as:

$(1): \quad x = C_1 \cos \alpha t + C_2 \sin \alpha t$

where:

$C_1$ and $C_2$ depend upon the conditions of $C$ at time $t = 0$
$\alpha = \sqrt {\dfrac k m}$

From the definition of frequency:

 $\displaystyle \nu$ $=$ $\displaystyle \dfrac \alpha {2 \pi}$ $\displaystyle$ $=$ $\displaystyle \dfrac 1 {2 \pi} \sqrt {\dfrac k m}$ $\displaystyle$ $=$ $\displaystyle Q \sqrt k$ where $Q = \dfrac 1 {2 \pi \sqrt m}$

Hence, if the mass is constant (as is implicit in the statement of the problem), increasing $k$ causes an increase in $\nu$.

$\blacksquare$