Given wave equation

$y = \sin(20\pi x + 10\pi t)$

This is of the form

$y = \sin(kx + \omega t)$

so,

$k = 20\pi \;\text{rad/m}, \qquad \omega = 10\pi \;\text{rad/s}$

Find the wavelength

Using

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

we get

$\lambda = \frac{2\pi}{20\pi} = \frac{1}{10}\text{ m} = 0.1\text{ m} = 10\text{ cm}$

Oscillating speed of a particle of the string

The transverse velocity of a particle is

$v_y = \frac{\partial y}{\partial t}$

Therefore,

$v_y = 10\pi \cos(20\pi x + 10\pi t)$

So the oscillating speed is the magnitude:

$|v_y| = 10\pi \left|\cos(20\pi x + 10\pi t)\right|$

Condition for same oscillating speed at two points

For two points on the string at the same time, they have the same oscillating speed if

$\left|\cos(20\pi x_1 + 10\pi t)\right| = \left|\cos(20\pi x_2 + 10\pi t)\right|$

The minimum non-zero phase difference for which $|\cos \theta|$ repeats is

$\Delta \phi = \pi$

because

$|\cos(\theta + \pi)| = |-\cos\theta| = |\cos\theta|$

Convert phase difference to distance

Since phase difference due to separation $\Delta x$ is

$\Delta \phi = k\Delta x$

we get

$20\pi \Delta x = \pi$

$\Delta x = \frac{\pi}{20\pi} = \frac{1}{20}\text{ m} = 0.05\text{ m} = 5\text{ cm}$

Answer

The minimum distance between two points having the same oscillating speed is

$\boxed{5\text{ cm}}$

Hence, the correct option is D.