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(B) The given wave equation is $y = y_0 \sin 2 \pi \left( 
u t - \frac{x}{\lambda} ight)$.Comparing this with the standard wave equation $y = y_0 \

Vedclass · Accepted Answer

$\lambda = \frac{\pi y_0}{2}$

Vedclass · Answer

(B) The given wave equation is $y = y_0 \sin 2 \pi \left( 
u t - \frac{x}{\lambda} ight)$.Comparing this with the standard wave equation $y = y_0 \sin (\omega t - kx)$,we have $\omega = 2 \pi 
u$ and $k = \frac{2 \pi}{\lambda}$.The wave velocity $v_w$ is given by $v_w = \frac{\omega}{k} = \frac{2 \pi 
u}{2 \pi / \lambda} = 
u \lambda$.The particle velocity $v_p$ is the derivative of displacement with respect to time: $v_p = \frac{\partial y}{\partial t} = y_0 (2 \pi 
u) \cos 2 \pi \left( 
u t - \frac{x}{\lambda} ight)$.The maximum particle velocity is $v_{p, 	ext{max}} = 2 \pi 
u y_0$.According to the problem,$v_{p, 	ext{max}} = 4 v_w$.Substituting the expressions: $2 \pi 
u y_0 = 4 (
u \lambda)$.Solving for $\lambda$: $2 \pi y_0 = 4 \lambda$,which gives $\lambda = \frac{2 \pi y_0}{4} = \frac{\pi y_0}{2}$.

$A$ transverse wave is described by the equation $y = y_0 \sin 2 \pi \left( \nu t - \frac{x}{\lambda} \right)$. The maximum particle velocity is equal to four times the wave velocity if $\lambda =$

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