For a sinusoidal mechanical wave $y=A\sin(\omega t\pm kx)$, the maximum particle speed is

Particle velocity is $v_p=\dfrac{dy}{dt}=A\omega\cos(\omega t\pm kx)$. Since the maximum value of cosine is 1, the maximum particle speed is $(v_p)_{\max}=A\omega$. Therefore option C is correct.