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

The problem is addressed by first analyzing the relationship between oscillation frequency and attached mass, defined by the mass-spring system frequency formula: \( f = \frac{1}{2\pi}\sqrt{\frac{k}{m}} \). Here, \( k \) represents the spring constant, and \( m \) denotes the mass.

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

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

Consequently, the ratio \(\frac{f_1}{f_2}\) is calculated as \(\frac{f_1}{\frac{1}{3}f_1} = 3\).

Therefore, the value of \(\frac{f_1}{f_2}\) is \( 3 \), which is within the specified range of 3 to 3.