MCQ
A transverse wave is described by the equation $y = A \,\,sin\,[2\pi (f t - x/\lambda ) ]$.The maximum particle velocity is equal to four times the wave velocity if:
  • A
    $\lambda = \pi A/4$
  • $\lambda = \pi A/2$
  • C
    $\lambda = \pi A$
  • D
    $\lambda = 2\pi A$

Answer

Correct option: B.
$\lambda = \pi A/2$
b
In a plane progressive transverse wave particles of the medium oscillate simple harmonically about their mean positions. Hence, the concepts of $SHM$ are applicable to the particles.

Hence, the maximum particle velocity $=\pm A \omega$ which is at the mean position.

From the given equation of wave$:$

$y=A \sin \left(2 \pi f t-2 \pi \frac{x}{\lambda}\right)$

From the wave equation, $\omega=2 \pi f$

and wave velocity, $v=\frac{\omega}{k}=\frac{2 \pi f}{\frac{2 \pi}{\lambda}}$

$v=\lambda f$

From the given condition$:$

$A \times 2 \pi f=4 \lambda f$

$\lambda=\frac{\pi A}{2}$

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