Class 11 Physics Chapter 13: Oscillations case study question with answers
One case-based question of the kind CBSE-pattern school exams ask: a passage and a table, then 3 parts worth 4 marks in all, with the answer and the marking scheme for each. Written from the NCERT chapter.
| t (s) | 0 | 0.1 | 0.2 | 0.3 | 0.4 |
|---|---|---|---|---|---|
| x (cm) | 6 | 0 | −6 | 0 | 6 |
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Answer: (c) 0.4 s
The glider starts at +6 cm and first returns to +6 cm, moving the same way, at t = 0.4 s; that is one period. 0.2 s is only the time from one extreme to the other.
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Answer: (b) 94 cm s⁻¹
A = 6 cm and ω = 2π/0.4 = 5π rad s⁻¹, so vm = ωA = 30π ≈ 94 cm s⁻¹, reached at t = 0.1 s and 0.3 s when x = 0. 30 cm s⁻¹ forgets the π; 47 cm s⁻¹ uses half of ω.
Answer and marking scheme
A = 6 cm, ω = 2π/T = 5π ≈ 15.7 rad s⁻¹ and x = +A at t = 0, so φ = 0 and x = 6 cos(5πt) cm. At t = 0.2 s, x = −6 cm, and a = −ω²x = −(5π)² × (−6) ≈ +1480 cm s⁻² ≈ 14.8 m s⁻², directed along +x, towards the equilibrium position.
- x = 6 cos(5πt) cm, with ω = 5π rad s⁻¹ (1)
- a ≈ 1480 cm s⁻² (14.8 m s⁻²) along +x, towards equilibrium (1)
Answer and marking scheme
x = 6 cos(5πt) cm with ω = 5π ≈ 15.7 rad s⁻¹ (A = 6 cm, φ = 0 since x = +A at t = 0). Setting x = 3: cos(5πt) = 0.5, so the earliest 5πt = π/3 and t = 1/15 s ≈ 0.0667 s.
- x = 6 cos(5πt) cm, with ω = 5π rad s⁻¹ (1)
- cos(5πt) = 0.5, t = 1/15 s ≈ 0.067 s (1)