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JEE Advanced · Physics · 16. Waves & Sound

A horizontal stretched string, fixed at two ends, is vibrating in its fifth harmonic according to the equation, \(y ( x , t )=(0.01 m) \sin \left[\left(62.8 m^{-1}\right) x \right]\) \(\cos \left[\left(628 s^{-1}\right) t \right]\). Assuming \(\pi=3.14\), the correct statement(s) is (are)

  1. A The number of nodes is 5.
  2. B The length of the string is 0.25 m.
  3. C The maximum displacement of the midpoint of the string, from its equilibrium position is 0.01 m.
  4. D The fundamental frequency is 100 Hz.
Verified Solution

Answer & Solution

Correct Answer

(C) The maximum displacement of the midpoint of the string, from its equilibrium position is 0.01 m.

Step-by-step Solution

Detailed explanation


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
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