Question

An undamped spring-mass system with mass \( m \) and spring stiffness \( k \) is shown in the figure. The natural frequency and natural period of this system are \( \omega \) rad/s and \( T \) s, respectively. If the stiffness of the spring is doubled and the mass is halved, then the natural frequency and the natural period of the modified system, respectively, are

• A. \( 2\omega \, \text{rad/s} \) and \( T/2 \, \text{s} \)
• B. \( \omega/2 \, \text{rad/s} \) and \( 2T \, \text{s} \)
• C. \( 4\omega \, \text{rad/s} \) and \( T/4 \, \text{s} \)
• D. \( \omega \, \text{rad/s} \) and \( T \, \text{s} \)

Hint:

When modifying a spring-mass system, if the spring stiffness is doubled and the mass is halved, the natural frequency will be multiplied by \( \sqrt{4} = 2 \), and the natural period will be halved.

Answer 1

The Correct Option is A

The natural frequency \( \omega \) for an undamped spring-mass system is given by the formula: \[ \omega = \sqrt{\frac{k}{m}} \] where:
- \( k \) is the spring stiffness, and
- \( m \) is the mass of the system.
The natural period \( T \) is related to the natural frequency by: \[ T = \frac{2\pi}{\omega} \] Now, if the stiffness \( k \) is doubled and the mass \( m \) is halved, the modified values of \( k \) and \( m \) are: - New spring stiffness \( k' = 2k \),
- New mass \( m' = \frac{m}{2} \).
Substitute these new values into the formula for natural frequency: \[ \omega' = \sqrt{\frac{k'}{m'}} = \sqrt{\frac{2k}{m/2}} = \sqrt{\frac{4k}{m}} = 2\omega \] Thus, the new natural frequency is \( 2\omega \, \text{rad/s} \). Next, substitute the new frequency into the formula for the natural period: \[ T' = \frac{2\pi}{\omega'} = \frac{2\pi}{2\omega} = \frac{T}{2} \] Thus, the new period is \( T/2 \, \text{s} \). Therefore, the natural frequency and natural period of the modified system are \( 2\omega \, \text{rad/s} \) and \( T/2 \, \text{s} \), respectively.