Question

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 \(9m\) is suspended from the same spring is \(f_2\). The value of \(\frac{f_1}{f_2}\) is ___.

Answer 1

Correct Answer: 3

To solve the problem, we begin by understanding the relationship between the frequency of oscillations and the mass attached to the spring, given by the formula for the frequency of a mass-spring system: \( f = \frac{1}{2\pi}\sqrt{\frac{k}{m}} \), where \( k \) is the spring constant and \( m \) is the mass. 

For mass \( m \), the frequency is \( f_1 = \frac{1}{2\pi} \sqrt{\frac{k}{m}} \).

For mass \( 9m \), the frequency is \( f_2 = \frac{1}{2\pi} \sqrt{\frac{k}{9m}} = \frac{1}{2\pi} \frac{1}{3} \sqrt{\frac{k}{m}} = \frac{1}{3} \cdot f_1 \).

Thus, \(\frac{f_1}{f_2} = \frac{f_1}{\frac{1}{3}f_1} = 3\).

Hence, the value of \(\frac{f_1}{f_2}\) is clearly \( 3 \), which falls within the given range 3,3.

Answer 2

Given: - Mass of first system: \( m \) - Mass of second system: \( 9m \) - Frequencies: \( f_1 \) and \( f_2 \)

Step 1: Frequency of a Mass-Spring System

The frequency of oscillation of a mass-spring system is given by:

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

where \( k \) is the spring constant and \( m \) is the mass.

Step 2: Calculating \( f_1 \) and \( f_2 \)

For the first system with mass \( m \):

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

For the second system with mass \( 9m \):

\[ f_2 = \frac{1}{2\pi} \sqrt{\frac{k}{9m}} = \frac{1}{2\pi} \cdot \frac{1}{3} \sqrt{\frac{k}{m}} = \frac{f_1}{3} \]

Step 3: Finding the Ratio \( \frac{f_1}{f_2} \)

\[ \frac{f_1}{f_2} = \frac{f_1}{\frac{f_1}{3}} = 3 \]

Conclusion: The value of \( \frac{f_1}{f_2} \) is \( 3 \).