JEE Advanced2013PhysicsWaves and SoundActual
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)
Options
- AThe number of nodes is 5.
- BThe length of the string is 0.25 m.
- CThe maximum displacement of the midpoint of the string, from its equilibrium position is 0.01 m.
- DThe fundamental frequency is 100 Hz.
Correct answer
B. The length of the string is 0.25 m.
Step-by-step solution
Nodes = 6 λ = 0.1 ∴ 5 th harmonic , 5 λ 2 = L L = 0.25 At L 2 y max = 0.01 sin in 62.8 × 0.25 2 = 0.01 Interfering wave velocity = w k = 6 2 8 62.8 = 1 0 ∴ fundamental frequency = ν 2 L = 2 0