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BITSAT 2013
Exam
BITSAT
Year
2013
Subject
Physics
Chapter
Waves and Sound

BITSAT 2013 · Physics · Waves and Sound

If equation of transverse wave is y=x_ 0 2 π(nt- x λ ). Maximum...

If equation of transverse wave is y=x0cos⁡2π(nt−xλ).y=x_{0} \cos2 \pi\left(\mathrm{nt}-\frac{\mathrm{x}}{\lambda}\right). Maximum velocity of particle is twice of wave velocity, if λ\lambda is-

Options

  1. Aπ/2x0\pi / 2 \mathrm{x}_{0}
  2. B2πx02 \pi x_{0}
  3. Cπ\pi
  4. Dπx0\pi x_{0}

Correct answer

D. πx0\pi x_{0}

Step-by-step solution

y=x_{0} \cos 2 \pi\left(\right.nt−xλ)nt \left.-\frac{x}{\lambda}\right)\mathrm{y}=\mathrm{x}0cos⁡2πλ(vt−x)[∵_{0} \cos \frac{2 \pi}{\lambda}(\mathrm{vt}-\mathrm{x}) \quad[\because\mathrm{v}=\mathrm{n}λ](dydt)max⁡=x0×2πλv=2v(\lambda]\left(\frac{\mathrm{dy}}{\mathrm{dt}}\right)_{\max }=\mathrm{x}_{0} \times \frac{2 \pi}{\lambda} \mathrm{v}=2 \mathrm{v}( given )∴λ=πx0) \therefore \lambda=\pi \mathrm{x}_{0}
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