Let's start by considering the formula for the frequency of a mass on a spring (a simple harmonic oscillator):

$$ f = \frac{1}{2\pi} \sqrt{\frac{k}{m}} $$

Where:

• $$ f $$ is the frequency of oscillation
• $$ k $$ is the spring constant
• $$ m $$ is the mass suspended from the spring

When the mass $$ m $$ is suspended, the frequency $$ f_1 $$ is:

$$ f_1 = \frac{1}{2\pi} \sqrt{\frac{k}{m}} $$

When the mass $$ 9m $$ is suspended, the frequency $$ f_2 $$ is:

$$ f_2 = \frac{1}{2\pi} \sqrt{\frac{k}{9m}} $$

We can simplify the square root by taking the 9 inside the root as $$3^2$$, which gives:

$$ f_2 = \frac{1}{2\pi} \sqrt{\frac{k}{(3^2)m}} $$

$$ f_2 = \frac{1}{2\pi} \frac{1}{3} \sqrt{\frac{k}{m}} $$

The ratio of $$ \frac{f_1}{f_2} $$ is therefore:

$$ \frac{f_1}{f_2} = \frac{\frac{1}{2\pi} \sqrt{\frac{k}{m}}}{\frac{1}{2\pi} \frac{1}{3} \sqrt{\frac{k}{m}}} $$

$$ \frac{f_1}{f_2} = \frac{1}{\frac{1}{3}} $$

$$ \frac{f_1}{f_2} = 3 $$

So the value of $$ \frac{f_1}{f_2} $$ is $$3$$.