NEET Physics Time Period & Frequency Match the Following Question
Match Column-I with Column-II.
| Column-I | Column-II |
|---|---|
| (a) Simple Pendulum Period | (i) T = 2π√(L/g) |
| (b) Spring Constant Relation | (ii) T = 2π√(m/k) |
| (c) Higher Frequency | (iii) Shorter Time Period |
| (d) Damping Effect | (iv) Decreased Amplitude |
a-i, b-ii, c-iii, d-iv
a-ii, b-i, c-iv, d-iii
a-iii, b-iv, c-i, d-ii
a-iv, b-iii, c-ii, d-i
Correct Answer
Detailed Explanation
The period of a simple pendulum is given by T = 2π√(L/g), which matches with (a-i). The spring constant relation for a mass-spring system is T = 2π√(m/k), correctly matching (b-ii). Higher frequency correlates with a shorter time period, hence (c-iii) is accurate. The damping effect leads to decreased amplitude, matching (d-iv).
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