Question

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

• A. 1.0 radian
• B. 1.25 radian
• C. 1.57 radian
• D. 0.57 radian

Answer 1

The Correct Option is A

To find the phase difference between the two given wave equations, we start by examining the equations:

The first wave is represented as:

y_1 = a \sin(\omega t + kx + 0.57)

The second wave is represented as:

y_2 = a \cos(\omega t + kx)

To find the phase difference, we need to express both waves in terms of the sine function. We know that:

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

So, for the second wave, which has the cosine, it can be rewritten as:

y_2 = a \cos(\omega t + kx) = a \sin\left(\omega t + kx + \frac{\pi}{2}\right)

Now both wave equations are in terms of the sine function:

• y_1 = a \sin(\omega t + kx + 0.57)
• y_2 = a \sin(\omega t + kx + \frac{\pi}{2})

The phase of y_1 is (\omega t + kx + 0.57) and the phase of y_2 is (\omega t + kx + \frac{\pi}{2}).

The phase difference \Delta \phi is given by:

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

Simplifying the expression gives:

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

Since \frac{\pi}{2} = 1.57 radians, we have:

\Delta \phi = 1.57 - 0.57 = 1.0 radian

Thus, the phase difference between the two waves is 1.0 radian, which matches the correct option given in the question.