Given: - Mass of first system: \( m \) - Mass of second system: \( 9m \) - Frequencies: \( f_1 \) and \( f_2 \)
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.
For the first system with mass \( m \):
\[ f_1 = \frac{1}{2\pi} \sqrt{\frac{k}{m}} \]
For the second system with mass \( 9m \):
\[ f_2 = \frac{1}{2\pi} \sqrt{\frac{k}{9m}} = \frac{1}{2\pi} \cdot \frac{1}{3} \sqrt{\frac{k}{m}} = \frac{f_1}{3} \]
\[ \frac{f_1}{f_2} = \frac{f_1}{\frac{f_1}{3}} = 3 \]
Conclusion: The value of \( \frac{f_1}{f_2} \) is \( 3 \).
A body of mass 1000 kg is moving horizontally with a velocity of 6 m/s. If 200 kg extra mass is added, the final velocity (in m/s) is: