Two waves are represented by the equations $y_1=a\sin(\omega t+kx+0.57) ext{ m}$ and $y_2=a\cos(\omega t+kx)\

Question

Two waves are represented by the equations $y_1=a\sin(\omega t+kx+0.57)	ext{ m}$ and $y_2=a\cos(\omega t+kx)	ext{ m}$, where $x$ is in metre and $t$ in second. The phase difference between them is:

Accepted Answer

1. **Identify the Given Wave Equations:** The given equations are $y_1 = a \sin(\omega t + kx + 0.57)$ and $y_2 = a \cos(\omega t + kx)$.
2. **Convert to the Same Trigonometric Function:** To easily compare phases, both equations should be in the same trigonometric form. We can convert $y_2$ from cosine to sine using the identity $\cos(	heta) = \sin(	heta + \frac{\pi}{2})$.
   $$y_2 = a \sin\left(\omega t + kx + \frac{\pi}{2}ight)$$
3. **Substitute the Value of $\pi$:** Since $\pi \approx 3.14$, $\frac{\pi}{2} \approx 1.57 	ext{ rad}$.
   $$y_2 = a \sin(\omega t + kx + 1.57)$$
4. **Calculate the Phase Difference:** The phase of the first wave is $\phi_1 = \omega t + kx + 0.57$ and the phase of the second wave is $\phi_2 = \omega t + kx + 1.57$.
   The phase difference $\Delta \phi = \phi_2 - \phi_1 = (\omega t + kx + 1.57) - (\omega t + kx + 0.57) = 1.0 	ext{ rad}$.

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