Correct answer is 3

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 $3$.

Correct answer is 12

${\mathrm{k}}_{\mathrm{eq}}=\frac{2\mathrm{k}\cdot \mathrm{k}}{3\mathrm{k}}+\mathrm{k}=\frac{5\mathrm{k}}{3}$

Angular frequency of oscillation $(\omega )=\sqrt{\frac{{\mathrm{k}}_{\mathrm{eq}}}{\mathrm{m}}}$

$(\omega )=\sqrt{\frac{5\mathrm{k}}{3\mathrm{m}}}$

Period of oscillation $(\tau )=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{3\mathrm{m}}{5\mathrm{k}}}$

$=\pi \sqrt{\frac{12\mathrm{m}}{5\mathrm{k}}}$