Frequency Decreases when Mass Increases

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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 mass of $C$ increases, the frequency of the motion of $C$ decreases.


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:

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

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

From Reciprocal Function is Strictly Decreasing, the reciprocal function is strictly decreasing on positive real numbers.

Hence, if the force constant of $S$ is constant (as is implicit in the statement of the problem), increasing $m$ causes a decrease in $\nu$.

$\blacksquare$


Sources