A longitudinal wave is represented by $$y = 10 \sin 2\pi\left(nt - \frac{x}{\lambda}\right)$$ cm. The maximum particle velocity will be four times the wave velocity if the determined value of wavelength is equal to

A longitudinal wave is given by $$y = 10\sin 2\pi\left(nt - \frac{x}{\lambda}\right)$$ cm. We need to find the wavelength when the maximum particle velocity equals four times the wave velocity.

First, rewrite the wave equation in standard form by expanding the argument. This gives $$y = 10\sin\left(2\pi nt - \frac{2\pi x}{\lambda}\right) \text{ cm}$$. Comparing this with the standard form $$y = A\sin(\omega t - kx)$$ shows that the amplitude is $$A = 10$$ cm, the angular frequency is $$\omega = 2\pi n$$, and the wave number is $$k = \frac{2\pi}{\lambda}$$.

Next, the particle velocity is obtained by differentiating $$y$$ with respect to time, yielding $$v_p = \frac{\partial y}{\partial t} = 10 \cdot 2\pi n \cdot \cos\left(2\pi nt - \frac{2\pi x}{\lambda}\right) \text{ cm/s}$$. Therefore, the maximum particle velocity is $$v_{p,\max} = A\omega = 10 \times 2\pi n = 20\pi n \text{ cm/s}$$.

Similarly, the wave velocity (phase velocity) is given by $$v_w = \frac{\omega}{k} = \frac{2\pi n}{2\pi / \lambda} = n\lambda \text{ cm/s}$$.

Then, applying the condition that $$v_{p,\max} = 4v_w$$ leads to $$20\pi n = 4 \times n\lambda$$. Dividing both sides by $$n$$ (assuming $$n \neq 0$$) gives $$20\pi = 4\lambda$$, so $$\lambda = \frac{20\pi}{4} = 5\pi \text{ cm}$$.

Hence, the correct answer is Option B: $$5\pi$$.