A transverse wave is given by y = A sin 2pi l | vedclass

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

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

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(B) The given wave equation is $y = A \sin 2\pi \left( \frac{t}{T} - \frac{x}{\lambda} \right)$.Comparing this with the standard wave equation $y = A \sin (\omega t - kx)$,we have $\omega = \frac{2\pi}{T}$ and $k = \frac{2\pi}{\lambda}$.The particle velocity $v_p$ is given by $\frac{\partial y}{\partial t} = A \omega \cos 2\pi \left( \frac{t}{T} - \frac{x}{\lambda} \right)$.The maximum particle velocity is $v_{p,max} = A \omega = A \left( \frac{2\pi}{T} \right)$.The wave velocity $v$ is given by $v = \frac{\omega}{k} = \frac{2\pi / T}{2\pi / \lambda} = \frac{\lambda}{T}$.According to the problem,$v_{p,max} = 4v$.Substituting the expressions: $A \left( \frac{2\pi}{T} \right) = 4 \left( \frac{\lambda}{T} \right)$.Canceling $T$ from both sides: $2\pi A = 4\lambda$.Therefore,$\lambda = \frac{2\pi A}{4} = \frac{\pi A}{2}$.

$A$ transverse wave is given by $y = A \sin 2\pi \left( \frac{t}{T} - \frac{x}{\lambda} \right)$. The maximum particle velocity is equal to $4$ times the wave velocity when:

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