A transverse wave is represented by the equation y = A sin(kx - omega t). For what value of the wavelength is the maximum particle velocity equal to…

The maximum particle velocity is given by $v_{\text{max}} = A\omega$. The wave velocity is $v = \frac{\omega}{k}$. Given that $v_{\text{max}} = 3v$, we have $A\omega = 3\left(\frac{\omega}{k}\right)$, which simplifies to $k = \frac{3}{A}$. Since wave number $k = \frac{2\pi}{\lambda}$, we get $\frac{2\pi}{\lambda} = \frac{3}{A}$, leading to $\lambda = \frac{2\pi A}{3}$.