Question

A spring–mass system is shown in the figure below. Take the acceleration due to gravity as $g = 9.81 \, m/s^2$. The static deflection due to weight and the time period of oscillations, respectively, are:

• A. 0.392 m and 1.26 s
• B. 0.392 m and 3.52 s
• C. 0.626 m and 1.26 s
• D. 0.626 m and 3.52 s

Hint:

- Static deflection in spring–mass systems is found using $\delta = Mg/k$. - Time period of oscillation depends only on $M$ and $k$, not on gravity: $T = 2\pi \sqrt{M/k}$.

Answer 1

The Correct Option is A

Step 1: Static deflection due to weight.
At equilibrium under the mass weight: \[ k \, \delta = Mg \] \[ \delta = \frac{Mg}{k} \] Substitute values: \[ M = 4 \, kg, g = 9.81 \, m/s^2, k = 100 \, N/m \] \[ \delta = \frac{4 \times 9.81}{100} = \frac{39.24}{100} = 0.392 \, m \]
Step 2: Time period of oscillations.
For a spring–mass system: \[ T = 2\pi \sqrt{\frac{M}{k}} \] Substitute: \[ T = 2\pi \sqrt{\frac{4}{100}} = 2\pi \sqrt{0.04} = 2\pi (0.2) = 1.257 \, s \approx 1.26 \, s \]
Step 3: Interpretation.
Thus, the static deflection = $0.392$ m, and the time period = $1.26$ s. Final Answer: \[ \boxed{0.392 \, m \; \text{and} \; 1.26 \, s} \]