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JEE Main 201911 JanuaryMorning Shift
Exam
JEE Main
Year
2019
Date
11 January
Shift
Morning Shift
Subject
Physics
Chapter
Waves and Sound

JEE Main 2019 · 11 January · Morning Shift · Physics · Waves and Sound

Equation of travelling wave on a stretched string of linear density...

Equation of travelling wave on a stretched string of linear density 5 g/m5 \mathrm{~g} / \mathrm{m} is y=0.03sin⁡(450t−9x)\mathrm{y}=0.03 \sin (450 \mathrm{t}-9 \mathrm{x}) where distance and time are measured in SI units. The tension in the string is:

Options

  1. A10 N10 \mathrm{~N}
  2. B7.5 N7.5 \mathrm{~N}
  3. C12.5 N12.5 \mathrm{~N}
  4. D5 N5 \mathrm{~N}

Correct answer

C. 12.5 N12.5 \mathrm{~N}

Step-by-step solution

We have given, y=0.03sin⁡(450t−9x)y=0.03 \sin (450 t-9 x) Comparing it with standard equation of wave, we get ω=\omega= 450k=9450 \mathrm{k}=9 ∴v=ωk=4509=50 m/s\therefore \mathrm{v}=\frac{\omega}{\mathrm{k}}=\frac{450}{9}=50 \mathrm{~m} / \mathrm{s} Velocity of travelling wave on a stretched string is given by v=Tμ⇒Tμ=2500\mathrm{v}=\sqrt{\frac{\mathrm{T}}{\mu}} \Rightarrow \frac{\mathrm{T}}{\mu}=2500 μ=\mu= linear mass density ⇒T=2500×5×10−3\Rightarrow \mathrm{T}=2500 \times 5 \times 10^{-3} ⇒12.5 N\Rightarrow 12.5 \mathrm{~N}
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