A transverse wave is described by the equation $Y = Y_0 \sin 2\pi \left(ft - \frac{X}{\lambda}\right)$. The…
A transverse wave is described by the equation $Y = Y_0 \sin 2\pi \left(ft - \frac{X}{\lambda}\right)$. The maximum particle velocity is equal to four times the wave velocity, if
$\lambda = \frac{\pi Y_0}{4}$
$\lambda = \frac{\pi Y_0}{2}$
$\lambda = \pi Y_0$
$\lambda = 2\pi Y_0$
Solution
We have, $(v_P)_{\text{max}} = 4v$ or $Y_0\omega = 4(f\lambda)$ or $Y_0(2\pi f) = 4f\lambda$
$\therefore$ Wavelength, $\lambda = \frac{\pi Y_0}{2}$