Q114 / 200 · NEET 2002 · Physics

A wave travelling in the positive $x$-direction with amplitude $\mathrm{A}=0.2 \mathrm{~m}$, velocity $v=360 \mathrm{~ms}^{-1}$ and wavelength $\lambda$ $=60 \mathrm{~m}$, then the correct expression for the wave is:

  • A
    $y=0.2 \sin \left[2 \pi\left(6 t+\frac{x}{60}\right)\right]$
  • B
    $y=0.2 \sin \left[\pi\left(6 t+\frac{x}{60}\right)\right]$
  • Correct answer
    C
    $y=0.2 \sin \left[2 \pi\left(6 t-\frac{x}{60}\right)\right]$
  • D
    $y=0.2 \sin \left[\pi\left(6 t-\frac{x}{60}\right)\right]$

Solution

The general equation of wave travelling in $x$-direction is given as $$ y=A \sin \left[\frac{2 \pi}{\lambda}(V t-x)\right] $$ Where $A=0.2 \mathrm{~m}, v=360 \mathrm{~ms}^{-1}, \lambda=$ $60 \mathrm{~m}$ $\therefore$ The equation becomes $$ \begin{aligned} & y=0.2 \sin \left[\frac{2 \pi}{60}(360 t-x)\right] \\ & y=0.2 \sin \left[2 \pi\left(6 t-\frac{x}{60}\right)\right] \end{aligned} $$

Hence, the correct answer is option: 'C'.

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