In a transverse wave the distance between a crest and neighbouring trough at the same instant is 4.0 cm and the distance between a crest and trough at the same place is 1.0 cm. The next crest appears at the same place after a time interval of 0.4 s. The maximum speed of the vibrating particles in the medium is:

In this transverse wave problem, we need to find the maximum speed of the vibrating particles. Let's break down the given information step by step.

The distance between a crest and a neighboring trough at the same instant is given as 4.0 cm. In a wave, the distance from a crest to the next trough is half the wavelength. Therefore, we can write:

$$\frac{\lambda}{2} = 4.0 \text{ cm}$$

Solving for the wavelength $$\lambda$$:

$$\lambda = 4.0 \times 2 = 8.0 \text{ cm}$$

Next, the distance between a crest and a trough at the same place is 1.0 cm. This refers to the vertical distance at a fixed point in space. At a given location, the particle oscillates between maximum displacement (crest) and minimum displacement (trough). The amplitude $$A$$ is the maximum displacement from the mean position. The distance between crest and trough positions is the difference between $$+A$$ and $$-A$$, which is $$2A$$. So:

$$2A = 1.0 \text{ cm}$$

Solving for the amplitude $$A$$:

$$A = \frac{1.0}{2} = 0.5 \text{ cm}$$

The next crest appears at the same place after 0.4 s. This is the time taken for one complete cycle, so it is the time period $$T$$:

$$T = 0.4 \text{ s}$$

The particles in the medium undergo simple harmonic motion. The maximum speed $$v_{\text{max}}$$ occurs when the particle passes through the mean position and is given by:

$$v_{\text{max}} = \omega A$$

where $$\omega$$ is the angular frequency. The angular frequency is related to the time period by:

$$\omega = \frac{2\pi}{T}$$

Substituting the values:

$$\omega = \frac{2\pi}{0.4} = \frac{2\pi}{\frac{4}{10}} = \frac{2\pi \times 10}{4} = \frac{20\pi}{4} = 5\pi \text{ rad/s}$$

Now, using $$v_{\text{max}} = \omega A$$:

$$v_{\text{max}} = 5\pi \times 0.5 = 5\pi \times \frac{1}{2} = \frac{5\pi}{2} \text{ cm/s}$$

Comparing with the options:

A. $$\frac{3\pi}{2}$$ cm/s

B. $$\frac{5\pi}{2}$$ cm/s

C. $$\frac{\pi}{2}$$ cm/s

D. $$2\pi$$ cm/s

The value $$\frac{5\pi}{2}$$ cm/s matches option B. Hence, the correct answer is Option B.