Question

The natural frequency of a simple spring-mass system with static deflection \(9.81\ \text{cm}\) is

• A. \(10\ \text{rad/s}\)
• B. \(100\ \text{rad/s}\)
• C. \(9.81\ \text{rad/s}\)
• D. \(1\ \text{rad/s}\)

Hint:

Using static deflection, \[ \boxed{ \omega_n=\sqrt{\frac{g}{\delta}} } \] where \[ \delta \] must always be expressed in metres.

Answer 1

The Correct Option is A

Step 1: Recall the relation between natural frequency and static deflection. For a vertical spring-mass system, \[ \boxed{ \omega_n=\sqrt{\frac{g}{\delta}}, } \] where
• \(\omega_n\) = natural frequency (rad/s),
• \(g=9.81\ \text{m/s}^2\),
• \(\delta\) = static deflection.

Step 2: Substitute the given values. Given, \[ \delta=9.81\ \text{cm}=0.0981\ \text{m}. \] Hence, \[ \omega_n = \sqrt{\frac{9.81}{0.0981}} = \sqrt{100} = 10\ \text{rad/s}. \] Therefore, \[ \boxed{10\ \text{rad/s}} \] is the correct answer. Thus, \[ \boxed{(A)} \] is the correct answer.