# Question #d9e02

May 13, 2017

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:
$\omega = \sqrt{\frac{k}{m}}$
and
$\omega = 2 \pi f$

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

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

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

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

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

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

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

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