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JEE Main 2008
Exam
JEE Main
Year
2008
Subject
Physics
Chapter
Waves and Sound

JEE Main 2008 · Physics · Waves and Sound

A wave travelling along the x -axis is described by the equation...

A wave travelling along the xx-axis is described by the equation y(x,t)=0.005cos⁡(αx−βt)y(x, t)=0.005 \cos (\alpha x-\beta t). If the wavelength and the time period of the wave are 0.08 m0.08 \mathrm{~m} and 2.0 s2.0 \mathrm{~s}, respectively, then α\alpha and β\beta in appropriate units are

Options

  1. Aα=25.00π,β=π\alpha=25.00 \pi, \beta=\pi
  2. Bα=0.08π,2.0π\alpha=\frac{0.08}{\pi}, \frac{2.0}{\pi}
  3. Cα=0.04π,β=1.0π\alpha=\frac{0.04}{\pi}, \beta=\frac{1.0}{\pi}
  4. Dα=12.50π,β=π2.0\alpha=12.50 \pi, \beta=\frac{\pi}{2.0}

Correct answer

A. α=25.00π,β=π\alpha=25.00 \pi, \beta=\pi

Step-by-step solution

y=0.005 \cos (α\alpha x-βt)\beta \mathrm{t}) comparing the equation with the standard form, y=Acos⁡[(xλ−tT)2π]2π/λ=α and 2π/T=βα=2π/0.08=25.00πβ=π\begin{aligned}& y=A \cos \left[ \left(\frac{\mathrm{x}}{\lambda}-\frac{\mathrm{t}}{\mathrm{T}}\right)2 \pi\right]\\ & 2 \pi/ \lambda=\alpha \text{ and }2 \pi/ T=\beta\\ & \alpha=2 \pi/ 0.08=25.00 \pi\\ & \beta=\pi \end{aligned}
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