MHT CET · Physics · Waves and Sound
A string of mass \(0.1 \mathrm{kgm}^{-1}\) has length 0.9 m . It is fixed at both ends and stretched such that it has a tension of 40 N . The string vibrates in three segments with amplitude 0.3 cm . The amplitude (maximum) of the particle velocity is (in m/s)
- A \(\frac{\pi}{2}\)
- B \(\frac{\pi}{3}\)
- C \(\frac{\pi}{5}\)
- D \(\frac{\pi}{6}\)
Answer & Solution
Correct Answer
(C) \(\frac{\pi}{5}\)
Step-by-step Solution
Detailed explanation
\(v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{40}{0.1}} = 20 \mathrm{m/s}\) \(\lambda = \frac{2L}{n} = \frac{2 \times 0.9}{3} = 0.6 \mathrm{m}\)
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