A transverse wave is described by the equatio | vedclass

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(B) The given wave equation is $y = y_0 \sin 2\pi \left( ft - \frac{x}{\lambda} \right)$.The particle velocity $v_p$ is given by the partial derivativ

Vedclass · Accepted Answer

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

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(B) The given wave equation is $y = y_0 \sin 2\pi \left( ft - \frac{x}{\lambda} \right)$.The particle velocity $v_p$ is given by the partial derivative of $y$ with respect to time $t$:$v_p = \frac{\partial y}{\partial t} = y_0 \cdot 2\pi f \cos 2\pi \left( ft - \frac{x}{\lambda} \right)$.The maximum particle velocity is $(v_p)_{\max} = 2\pi f y_0$.The wave velocity $v_w$ is given by $v_w = f \lambda$.According to the problem,the maximum particle velocity is four times the wave velocity:$(v_p)_{\max} = 4 v_w$.Substituting the expressions:$2\pi f y_0 = 4 f \lambda$.Dividing both sides by $4f$:$\lambda = \frac{2\pi f y_0}{4f} = \frac{\pi y_0}{2}$.

$A$ 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:

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