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?

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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?

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Answer 1 · 2018-01-18T10:36:01

$f = \pm \sqrt{2.56}$ $f =+-sqrt(2.56)$

Explanation:

$Determining the appropriate values$ $color(blue)("Determining the appropriate values")$
Inversely proportional $\to a = \frac{k}{b}$ $->a=k/b$ where $k$ $k$ is the constant of variation. So we have:

$f = \frac{1}{\sqrt{m}} \times k$ $f=1/(sqrt(m))xxk$

Initial condition:

$f = \frac{1}{\sqrt{m}} \times k dddd \to dddd 2 = \frac{1}{\sqrt{2.56}} \times k$ $f=1/(sqrt(m))xxk color(white)("dddd") ->color(white)("dddd")2=1/sqrt(2.56)xxk$

So we have: $d \to k = 2 \sqrt{2.56}$ $color(white)("d")->k=2sqrt(2.56)$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$Answering the question for m = 4$ $color(blue)("Answering the question for "m=4)$

$f = \frac{1}{\sqrt{m}} \times k ddd \to ddd f = \frac{1}{\sqrt{4}} \times 2 \sqrt{2.56}$ $f=1/(sqrt(m))xxk color(white)("ddd")-> color(white)("ddd") f=1/sqrt(4)xx2sqrt(2.56)$

$dddddddddddddd \to ddd f = \frac{1}{\pm 2} \times 2 \sqrt{2.56}$ $color(white)("dddddddddddddd")-> color(white)("ddd")f=1/(+-2)xx2sqrt(2.56)$

$dddddddddddddd \to ddd f = \pm \sqrt{2.56}$ $color(white)("dddddddddddddd")-> color(white)("ddd")f =+-sqrt(2.56)$

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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?

1 Answer

Explanation:

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