QJEEM24Ph8916: JEE Main 2024 Physics - Class 11 JEEM24Ph8916 Solution: A mass is suspended from a spring of negligible mass and the…

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 $3$ .