A mass $m$ is suspended from a spring of negligible mass and the system oscillates with a frequency $f_1$. The frequency of oscillations if a mass $9 \mathrm{~m}$ is suspended from the same spring is $f_2$. The value of $\frac{f_1}{f_2} \mathrm{i}$ ________.

<p>Let&#39;s start by considering the formula for the frequency of a mass on a spring (a simple harmonic oscillator):</p> <p>$f = \frac{1}{2\pi} \sqrt{\frac{k}{m}}$</p> <p>Where:</p> <ul> <li>$f$ is the frequency of oscillation</li> <li>$k$ is the spring constant</li> <li>$m$ is the mass suspended from the spring</li> </ul> <p>When the mass $m$ is suspended, the frequency $f_1$ is:</p> <p>$f_1 = \frac{1}{2\pi} \sqrt{\frac{k}{m}}$</p> <p>When the mass $9m$ is suspended, the frequency $f_2$ is:</p> <p>$f_2 = \frac{1}{2\pi} \sqrt{\frac{k}{9m}}$</p> <p>We can simplify the square root by taking the 9 inside the root as $3^2$, which gives:</p> <p>$f_2 = \frac{1}{2\pi} \sqrt{\frac{k}{(3^2)m}}$ <br/><br/>$f_2 = \frac{1}{2\pi} \frac{1}{3} \sqrt{\frac{k}{m}}$</p> <p>The ratio of $\frac{f_1}{f_2}$ is therefore:</p> <p>$$ \frac{f_1}{f_2} = \frac{\frac{1}{2\pi} \sqrt{\frac{k}{m}}}{\frac{1}{2\pi} \frac{1}{3} \sqrt{\frac{k}{m}}} $$ <br/><br/>$\frac{f_1}{f_2} = \frac{1}{\frac{1}{3}}$ <br/><br/>$\frac{f_1}{f_2} = 3$</p> <p>So the value of $\frac{f_1}{f_2}$ is $3$.</p>