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JEE Main 2019 (11 Jan, Shift 1) · Progressive Wave Equation

Progressive Wave Equation question from JEE Main 2019 (11 Jan, Shift 1)

Q84JEE Main 2019 · 11 JAN · SHIFT 1Medium

Progressive Wave Equation

Equation of travelling wave on a stretched string of linear density 5 g/m is y = 0.03 sin(450 t – 9x) where distance and time are measured in SI units. The tension in the string is:

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Answer: (d)

  1. v=ωk=4509=50 m/sv = \frac{\omega}{k} = \frac{450}{9} = 50\text{ m/s}
  2. μ=5 g/m=5×10−3 kg/m\mu = 5\text{ g/m} = 5\times10^{-3}\text{ kg/m}
  3. T=μv2=5×10−3×2500=12.5 NT = \mu v^2 = 5\times10^{-3} \times 2500 = 12.5\text{ N}
  4. So the answer is (d) 12.5 N.
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