Question

If a mass of 10 kg suspended from a spring causes a static deflection of 1 cm, then the natural frequency of the system is

• A. 2.98 Hz
• B. 3.98 Hz
• C. 4.98 Hz
• D. 5.98 Hz

Hint:

Use \( f = \frac{1}{2\pi} \sqrt{g/\delta} \) for vertical spring-mass systems.

Answer 1

The Correct Option is C

To determine the natural frequency of the system, we need to use the relationship between the static deflection caused by a mass on a spring and the natural frequency. When a mass \( m \) is suspended from a spring, it causes a deflection \( \delta \). The stiffness of the spring \( k \) is given by Hooke's Law:

\( k = \frac{F}{\delta} = \frac{mg}{\delta} \)

Where:

• \( m = 10 \) kg (mass)
• \( g = 9.81 \) m/s\(^2\) (acceleration due to gravity)
• \( \delta = 0.01 \) m (deflection)

Substituting the values, we find:

\( k = \frac{10 \times 9.81}{0.01} = 9810 \) N/m

The natural frequency \( f_n \) is calculated using the formula for the natural frequency of a spring-mass system:

\( f_n = \frac{1}{2\pi} \sqrt{\frac{k}{m}} \)

Substituting the known values:

\( f_n = \frac{1}{2\pi} \sqrt{\frac{9810}{10}} \)

\( f_n = \frac{1}{2\pi} \sqrt{981} \)

\( f_n \approx \frac{1}{2\pi} \times 31.32 \)

\( f_n \approx 4.98 \) Hz

Therefore, the natural frequency of the system is 4.98 Hz.