Physics · Oscillations and waves · chapter 14 of 14
Class 11 Physics Chapter 14: Waves 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.
Case study · 4 marks
Meera, a student in Pune, studies a sitar-like string on a board. A movable bridge lets her change the vibrating length L while the tension stays the same. For each length she finds, with a set of tuning forks, the frequency at which the string vibrates in its fundamental mode. Her readings are in the table. The linear mass density of the string is 2.5 × 10⁻³ kg m⁻¹.
| L (m) | Fundamental frequency (Hz) |
|---|---|
| 0.40 | 300 |
| 0.50 | 240 |
| 0.60 | 200 |
| 0.80 | 150 |
Part 1 · 1 mark
The speed of transverse waves on the string is
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Answer: (b) 240 m s⁻¹
Fundamental ν = v/2L, so v = 2Lν = 2 × 0.40 × 300 = 240 m s⁻¹; every row gives the same value.
Part 2 · 1 mark
The readings show that, at fixed tension, the fundamental frequency
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Answer: (d) is inversely proportional to L
The product Lν is 120 m s⁻¹ in every row, so ν ∝ 1/L, as ν = v/2L requires.
Part 3 · 2 marks
Meera wants the 0.50 m length to sound 300 Hz as its fundamental. By how much must she raise the tension?
Answer and marking scheme
The required speed is v' = 2Lν = 2 × 0.50 × 300 = 300 m s⁻¹, needing T' = μv'² = 2.5 × 10⁻³ × 300² = 225 N. The present tension is μv² = 2.5 × 10⁻³ × 240² = 144 N, so the tension must be raised by 225 − 144 = 81 N.
- v' = 300 m s⁻¹ and T' = μv'² = 225 N (1)
- present T = 144 N; raise by 81 N (1)
Or
Find the length at which the string's fundamental frequency would be 360 Hz, and the frequency of the second harmonic at that length.
Answer and marking scheme
L = v/2ν = 240/(2 × 360) = 1/3 m ≈ 0.33 m. The second harmonic is 2 × 360 = 720 Hz.
- L = v/2ν ≈ 0.33 m (1)
- second harmonic 720 Hz (1)