$$\mathrm{y}_2 = \mathrm{A} \cos (\omega \mathrm{t} + \mathrm{kx}) \mathrm{m} $$

we need to express both waves in a similar trigonometric form to compare their phases directly. The cosine function can be rewritten in terms of the sine function using the identity:

$$\cos(\theta) = \sin\left(\theta + \frac{\pi}{2}\right)$$

Hence, we can rewrite $$\mathrm{y}_2$$ as follows:

$$\mathrm{y}_2 = \mathrm{A} \cos (\omega \mathrm{t} + \mathrm{kx}) = \mathrm{A} \sin \left( (\omega \mathrm{t} + \mathrm{kx}) + \frac{\pi}{2} \right) $$

Now, compare $$\mathrm{y}_1$$ and the rewritten form of $$\mathrm{y}_2$$:

$$\mathrm{y}_1 = \mathrm{A} \sin (\omega \mathrm{t} + \mathrm{kx} + 0.57) \mathrm{m}$$

$$\mathrm{y}_2 = \mathrm{A} \sin \left( (\omega \mathrm{t} + \mathrm{kx}) + \frac{\pi}{2} \right) \mathrm{m}$$

The wave $$\mathrm{y}_1$$ has a phase term of $$(\omega \mathrm{t} + \mathrm{kx} + 0.57)$$, and the rewritten wave $$\mathrm{y}_2$$ has a phase term of $$(\omega \mathrm{t} + \mathrm{kx} + \frac{\pi}{2})$$.

Thus, the phase difference between the two waves, $$\phi$$, is the difference between these two phase terms:

$$\phi = \left( (\omega \mathrm{t} + \mathrm{kx} + \frac{\pi}{2}) \right) - \left( \omega \mathrm{t} + \mathrm{kx} + 0.57 \right) $$

$$\phi = \frac{\pi}{2} - 0.57 $$

Now, we need to evaluate $$\frac{\pi}{2}$$ in radians. Recall that $$\pi \approx 3.14159265$$. Therefore,

$$\frac{\pi}{2} \approx \frac{3.14159265}{2} \approx 1.57 \, \text{radians}$$

Substituting this value, we get:

$$\phi = 1.57 - 0.57 = 1.0 \, \text{radians}$$

Therefore, the phase difference between the two waves is:

Option B

1.0 radian

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