The equation of a wave travelling on a string is $$y = \sin[20\pi x + 10\pi t]$$, where x and t are in SI units. The minimum distance between two points having the same oscillating speed is :

The given transverse wave on the string is
$$y(x,t)=\sin\!\left(20\pi x+10\pi t\right)$$ where $$x$$ is in metres and $$t$$ is in seconds.

For any particle of the string the instantaneous transverse (oscillatory) speed is obtained by differentiating $$y$$ with respect to time:

Formula: $$v_y=\frac{\partial y}{\partial t}= \omega \cos\!\left(kx+\omega t\right)$$ where $$k$$ is the wave-number and $$\omega$$ is the angular frequency.

Here $$k=20\pi$$ and $$\omega=10\pi$$, so
$$v_y=10\pi \cos\!\left(20\pi x+10\pi t\right)$$ $$-(1)$$

At a fixed instant $$t$$, two points $$x_1$$ and $$x_2$$ will have the same oscillatory speed magnitude when

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

Let the corresponding phase angles be $$\phi_1$$ and $$\phi_2$$: $$\phi_1 = 20\pi x_1+10\pi t,\qquad \phi_2 = 20\pi x_2+10\pi t$$

Condition for equal absolute value of a cosine:
$$\left|\cos\phi_1\right|=\left|\cos\phi_2\right| \; \Longrightarrow \; \phi_2=\phi_1\pm n\pi,\; n\in\mathbb{Z}$$ (The plus/minus $$\pi$$ shift changes the sign of the cosine but keeps its magnitude the same.)

Therefore the phase difference between the two points is
$$\Delta\phi = \phi_2-\phi_1 = n\pi$$

Since $$\Delta\phi = k\,\Delta x$$, we get
$$k\,\Delta x = n\pi \quad \Longrightarrow \quad \Delta x = \frac{n\pi}{k}$$ $$-(2)$$

With $$k = 20\pi$$, equation $$(2)$$ becomes
$$\Delta x = \frac{n\pi}{20\pi} = \frac{n}{20}\; \text{metre}$$

The minimum non-zero distance corresponds to $$n = 1$$:

$$\Delta x_{\min} = \frac{1}{20}\; \text{m} = 0.05\; \text{m} = 5.0\; \text{cm}$$

Hence, the minimum separation between two points on the string having the same oscillating speed is 5.0 cm.

Option A is correct.