TS EAMCET · Physics · Waves and Sound
An air column in a tube of length 50 cm, closed at one end is vibrating in its fifth harmonic. The phase difference between a particle at the open end and a particle at 42 cm from the open end is
- A \(90^{\circ}\)
- B \(180^{\circ}\)
- C \(0^{\circ}\)
- D \(270^{\circ}\)
Answer & Solution
Correct Answer
(C) \(0^{\circ}\)
Step-by-step Solution
Detailed explanation
\(L = 50 \text{ cm}\), fifth harmonic for closed pipe \(\implies L = 5 \frac{\lambda}{4}\) \(\lambda = \frac{4L}{5} = \frac{4 \times 50}{5} = 40 \text{ cm}\) \(k = \frac{2\pi}{\lambda} = \frac{2\pi}{40} = \frac{\pi}{20} \text{ cm}^{-1}\) Displacement for open end at \(x=0\):…
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