# The oscillation frequency f cycles per second, of a spring is inversely proportional to the square root of the mass m kg of the spring. When m=2.56, f=2, how do you find f when m=4?

Jan 18, 2018

$f = \pm \sqrt{2.56}$

#### Explanation:

$\textcolor{b l u e}{\text{Determining the appropriate values}}$
Inversely proportional $\to a = \frac{k}{b}$ where $k$ is the constant of variation. So we have:

$f = \frac{1}{\sqrt{m}} \times k$

Initial condition:

$f = \frac{1}{\sqrt{m}} \times k \textcolor{w h i t e}{\text{dddd") ->color(white)("dddd}} 2 = \frac{1}{\sqrt{2.56}} \times k$

So we have: $\textcolor{w h i t e}{\text{d}} \to k = 2 \sqrt{2.56}$
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$\textcolor{b l u e}{\text{Answering the question for } m = 4}$

$f = \frac{1}{\sqrt{m}} \times k \textcolor{w h i t e}{\text{ddd")-> color(white)("ddd}} f = \frac{1}{\sqrt{4}} \times 2 \sqrt{2.56}$

$\textcolor{w h i t e}{\text{dddddddddddddd")-> color(white)("ddd}} f = \frac{1}{\pm 2} \times 2 \sqrt{2.56}$

$\textcolor{w h i t e}{\text{dddddddddddddd")-> color(white)("ddd}} f = \pm \sqrt{2.56}$