A horizontal stretched string, fixed at two ends, is vibrating in its fifth harmonic…
Physics · JEE Advanced · NTA Exams — Oscillations and Waves
A horizontal stretched string, fixed at two ends, is vibrating in its fifth harmonic according to the equation, y(x, t) = (0.01 m)sin[(62.8 m–1)x] cos[(628 s–1)t]. Assuming π = 3.14, the correct statement(s) is (are).
- The number of nodes is 5 .
- The length of the string is 0.25 m.
- The maximum displacement of the midpoint of the string, from its equilibrium position is 0.01 m.
- The fundamental frequency is 100 Hz.
Answer
(C) The maximum displacement of the midpoint of the string, from its equilibrium position is 0.01 m.
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