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JEE Advanced 2013Paper 1
Exam
JEE Advanced
Year
2013
Paper
Paper 1
Subject
Physics
Chapter
Waves and Sound

JEE Advanced 2013 · Paper 1 · Physics · Waves and Sound

A horizontal stretched string, fixed at two ends, is vibrating in...

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

Options

  1. AThe number of nodes is 5.
  2. BThe length of the string is 0.25 m.
  3. CThe maximum displacement of the midpoint of the string, from its equilibrium position is 0.01 m.
  4. DThe fundamental frequency is 100 Hz.

Correct answer

B. The length of the string is 0.25 m.

Step-by-step solution

A horizontal stretched string, fixed at two ends, is vibrating in its fifth harmonic according to the equation, y (x ,
Nodes=6\text{Nodes} = 6
λ=0.1\lambda = \text{0.1}
∴5thharmonic,5λ2=L∴ {5}^{\text{th}} harmonic , 5 \dfrac{\lambda}{2} = \text{L}
L=0.25\text{L} = \text{0.25}
AtL2\text{At} \dfrac{\text{L}}{2}
ymax=0.01 sin in(62.8×0.252){\text{y}}_{\text{max}} = \text{0.01 sin in} \left(\text{62.8} \times \dfrac{\text{0.25}}{2}\right)
=0.01= \text{0.01}
Interfering wave velocity=wk=62862.8=10\text{Interfering wave velocity} = \dfrac{\text{w}}{\text{k}} = \dfrac{6 2 8}{\text{62.8}} = 1 0
∴fundamental frequency=ν2L=20∴ \text{fundamental frequency} = \dfrac{ν}{2 \text{L}} = 2 0
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