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JEE Main 2025 (23 Jan, Shift 2) · Progressive Wave Equation

Progressive Wave Equation question from JEE Main 2025 (23 Jan, Shift 2)

Q27JEE Main 2025 · 23 JAN · SHIFT 2Easy

Progressive Wave Equation

The equation of a transverse wave travelling along a string is

y(x,t)=4.0sin⁡[20×10−3x+600t]mmy(x, t) = 4.0\sin\left[20 \times 10^{-3}x + 600t\right]\text{mm}, where xx is in mm and tt is in second. The velocity of the wave is:

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Answer: (d)

  1. k=20×10−3 mm−1k = 20 \times 10^{-3}\text{ mm}^{-1}, ω=600 s−1\omega = 600\text{ s}^{-1}
  2. v=ωk=6000.02=3×104 mm/s=30 m/sv = \dfrac{\omega}{k} = \dfrac{600}{0.02} = 3 \times 10^{4}\text{ mm/s} = 30\text{ m/s}
  3. kxkx and ωt\omega t have the same sign, so the wave moves along −x-x.
  4. So the answer is (d) −30 m/s-30\text{ m/s}.
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