Question #d9e02

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Question #d9e02

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Answer 1 · 2017-05-13T03:13:56

See below.

Explanation:

In the answer below, I have assumed the question refers to simple harmonic motion:

For simple harmonic motion, we have the formulas:
$ω = \sqrt{\frac{k}{m}}$ $omega=sqrt(k/m)$
and
$ω = 2 π f$ $omega=2pif$

where $ω$ $omega$ is the angular velocity of the object, $k$ $k$ is the spring constant, $m$ $m$ is the mass of the object, and $f$ $f$ is frequency

By combining the two equations and solving for $f$ $f$ , we get:
$2 π f = \sqrt{\frac{k}{m}}$ $2pif=sqrt(k/m)$
$f = \frac{1}{2 π} \sqrt{\frac{k}{m}}$ $f=1/(2pi)sqrt(k/m)$

Since the only values we care about in this problem are $m$ $m$ and $f$ $f$ , we can disregard the constant $\frac{1}{2 π}$ $1/(2pi)$ and let $k$ $k$ be some arbitrary constant, say $1$ $1$ , just to make this easier:
$f = \sqrt{\frac{1}{m}}$ $f=sqrt(1/m)$

Now we can substitute $m$ $m$ for ${m, \frac{1}{4} m, 4 m}$ ${m, 1/4m, 4m}$

If $m = m$ $m=m$ :
$f = \sqrt{\frac{1}{m}}$ $f=sqrt(1/m)$
this is our value to which we will compare quartering and quadrupling the mass to

If $m = \frac{1}{4} m$ $m=1/4m$
$f = \sqrt{\frac{1}{\frac{1}{4} m}}$ $f=sqrt(1/(1/4m))$
$f = \sqrt{\frac{4}{m}}$ $f=sqrt(4/m)$
$f = 2 \sqrt{\frac{1}{m}}$ $f=2sqrt(1/m)$
which is a frequency $2$ $2$ times the original frequency.

if $m = 4 m$ $m=4m$
$f = \sqrt{\frac{1}{4 m}}$ $f=sqrt(1/(4m))$
$f = \frac{1}{2} \sqrt{\frac{1}{m}}$ $f=1/2sqrt(1/m)$
which is a frequency $\frac{1}{2}$ $1/2$ times the original frequency.

Therefore, when we take one-fourth of the mass, we have a frequency $2$ $2$ times the original frequency and when we take four times the mass, we have a frequency $\frac{1}{2}$ $1/2$ times the original frequency.

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Question #d9e02

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