Q.9 The time period and the amplitude of an object executing simple harmonic motion, under the restoring force of a spring, are 3.14 seconds and 0.2 m, respectively. If the mass of the object is 2 kg, the maximum force (in Newton) exerted by the spring on the object is
The maximum force exerted by the spring on the object is 1.6 N. This value is calculated using the principles of simple harmonic motion (SHM) for a spring-mass system.
Key Formulas
In SHM, the time period T relates to mass m and spring constant k by T = 2π√(m/k). The angular frequency is ω = 2π/T. The maximum restoring force occurs at maximum displacement (amplitude A) and equals F_max = kA or equivalently F_max = mω²A.
Step-by-Step Solution
Given T = 3.14 s, A = 0.2 m, m = 2 kg:
Compute ω = 2π/T = 2×3.1416/3.14 ≈ 2 rad/s.
Then k = mω² = 2×(2)² = 8 N/m.
Finally, F_max = kA = 8×0.2 = 1.6 N [code execution result].
Option Analysis
| Option | Value (N) | Why Correct/Incorrect |
|---|---|---|
| (A) | 1.6 | Matches F_max = m(2π/T)²A ≈ 1.6. |
| (B) | 3.2 | Double the correct value; possibly from using 2A instead of A. |
| (C) | 4.6 | Incorrect; no direct relation to given parameters. |
| (D) | 5.2 | Incorrect; exceeds calculated force by over 3x. |
Understanding SHM Spring Basics
Simple harmonic motion (SHM) in a spring-mass system follows Hooke’s law, where restoring force F = -kx. The maximum force hits at extreme positions, F_max = kA. Time period T = 2π√(m/k) helps derive k = 4π²m/T².
Detailed Calculation
ω = 2π/3.14 ≈ 2 rad/s.
k = mω² = 2×4 = 8 N/m.
F_max = 8×0.2 = 1.6 N [code execution result].
This matches option (A) precisely.
Common Mistakes in Options
- 3.2 N (B): Might arise from mωA (missing ω) or diameter confusion.
- 4.6 N/5.2 N (C,D): Often from miscalculating π ≈ 3 or wrong ω.
Exam Tips for SHM Problems
- Always use F_max = m(2π/T)²A directly.
- T independent of A in ideal SHM.
- Practice similar SHM spring force questions for CSIR NET physics prep.
