Problem Analysis:

We are given the equation of a transverse wave: $y = y_0 \sin 2\pi(ft - \frac{x}{\lambda})$. We are also given the condition that the maximum particle velocity is four times the wave velocity. We need to find the relationship between the wavelength ($\lambda$) and amplitude ($y_0$).

Deriving Particle Velocity

The particle velocity ($v_p$) is the rate of change of displacement ($y$) with respect to time ($t$). We find it by differentiating the wave equation with respect to $t$:

$ v_p = \frac{\partial y}{\partial t} = \frac{\partial}{\partial t} \left[ y_0 \sin 2\pi(ft - \frac{x}{\lambda}) \right] $

Using the chain rule:

$ v_p = y_0 \cos 2\pi(ft - \frac{x}{\lambda}) \cdot \frac{\partial}{\partial t} \left[ 2\pi(ft - \frac{x}{\lambda}) \right] $

$ v_p = y_0 \cos 2\pi(ft - \frac{x}{\lambda}) \cdot (2\pi f) $

$ v_p = 2\pi f y_0 \cos 2\pi(ft - \frac{x}{\lambda}) $

Finding Maximum Particle Velocity

The maximum value of the cosine function ($\cos \theta$) is 1. Therefore, the maximum particle velocity ($v_{p,max}$) is:

$ v_{p,max} = 2\pi f y_0 $

Determining Wave Velocity

The wave equation can be written as $y = y_0 \sin(2\pi f t - \frac{2\pi x}{\lambda})$.

Comparing this with the standard form $y = A \sin(\omega t - kx)$, we identify:

• Angular frequency $\omega = 2\pi f$
• Wave number $k = \frac{2\pi}{\lambda}$

The wave velocity ($v_w$) is given by $v_w = \frac{\omega}{k}$:

$ v_w = \frac{2\pi f}{\frac{2\pi}{\lambda}} = f \lambda $

Applying the Condition and Solving for Wavelength

The problem states that $v_{p,max} = 4 \cdot v_w$. Substituting the expressions we found:

$ 2\pi f y_0 = 4 \cdot (f \lambda) $

Assuming $f \neq 0$, we can cancel $f$ from both sides:

$ 2\pi y_0 = 4 \lambda $

Now, we solve for $\lambda$:

$ \lambda = \frac{2\pi y_0}{4} $

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

Conclusion

The relationship derived matches Option B.