A transverse wave is described by the equatio | vedclass

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

Vedclass · Accepted Answer

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

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

$A$ 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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