For a wave equation Y = Y_0 sin 2pi left( ft | vedclass

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(B) The given wave equation is $Y = Y_0 \sin(2\pi ft - \frac{2\pi x}{\lambda})$.Comparing this with the standard form $Y = A \sin(\omega t - kx)$,we g

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$\lambda = \frac{\pi Y_0}{2}$

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(B) The given wave equation is $Y = Y_0 \sin(2\pi ft - \frac{2\pi x}{\lambda})$.Comparing this with the standard form $Y = A \sin(\omega t - kx)$,we get amplitude $A = Y_0$,angular frequency $\omega = 2\pi f$,and wave number $k = \frac{2\pi}{\lambda}$.The maximum particle velocity $v_{p,max} = A\omega = Y_0(2\pi f)$.The wave velocity $v_w = \frac{\omega}{k} = \frac{2\pi f}{2\pi / \lambda} = f\lambda$.According to the problem,$v_{p,max} = 4v_w$.Substituting the values: $Y_0(2\pi f) = 4(f\lambda)$.Canceling $f$ from both sides: $2\pi Y_0 = 4\lambda$.Therefore,$\lambda = \frac{2\pi Y_0}{4} = \frac{\pi Y_0}{2}$.

For a wave equation $Y = Y_0 \sin 2\pi \left( ft - \frac{x}{\lambda} \right) \text{ cm}$,the condition for the maximum particle velocity to be four times the wave velocity is:

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