Frequency Increases when Stiffness Increases

From ProofWiki
Jump to navigation Jump to search

Theorem

Problem Definition

CartOnSpring.png

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 $Q = \dfrac 1 {2 \pi \sqrt m}$

From Square Root is Strictly Increasing, the square root is strictly increasing on positive real numbers.

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

$\blacksquare$


Sources