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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Product $(A)$ and $(B)$ is differentiated by
