Question

Two waves are represented by the equation, $y_1 = A \sin(\omega t + kx + 0.57)\ \text{m}$ and $y_2 = A \cos(\omega t + kx)\ \text{m}$, where $x$ is in metre and $t$ is in second. What is the phase difference between them?

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

Hint:

A cosine wave is always exactly $\frac{\pi}{2}$ radians (or $\approx 1.57$ rad) ahead of its corresponding basic sine wave. If the sine wave already has an initial phase angle shift of $0.57$ rad, the remaining phase difference between them is simply the difference from the cosine baseline: $1.57 - 0.57 = 1.0$ rad.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We are given two distinct harmonic wave equations, one written as a sine function and the other as a cosine function. We need to determine the absolute phase angle difference ($\phi$) between them in radians.

Step 2: Key Formula or Approach:
To accurately compare the internal phase arguments of two wave functions, both mathematical expressions must be converted into identical trigonometric forms. We convert the cosine function into a sine function using the standard reduction identity:
$$\cos(\theta) = \sin\left(\theta + \frac{\pi}{2}\right)$$

Step 3: Detailed Explanation:
Let's look at the two wave equations provided:
$$y_1 = A \sin(\omega t + kx + 0.57)$$ $$y_2 = A \cos(\omega t + kx)$$ Using the trigonometric identity, let's rewrite the expression for $y_2$ as a sine function:
$$y_2 = A \sin\left(\omega t + kx + \frac{\pi}{2}\right)$$ Now, isolate the phase angles from both equations to compare them directly:
$$\theta_1 = \omega t + kx + 0.57$$ $$\theta_2 = \omega t + kx + \frac{\pi}{2}$$ The phase difference $\phi$ is found by subtracting the first phase angle from the second:
$$\phi = \theta_2 - \theta_1 = \left(\omega t + kx + \frac{\pi}{2}\right) - (\omega t + kx + 0.57)$$ The time and position variables cancel out completely:
$$\phi = \frac{\pi}{2} - 0.57$$ Substitute the numerical value of $\frac{\pi}{2} \approx 1.57$ radians into the expression:
$$\phi = 1.57 - 0.57 = 1.0\ \text{radian}$$ This subtraction gives a clean phase separation value of exactly 1.0 radian.

Step 4: Final Answer:
The phase difference between the two waves is 1.0 radian, which matches option (B).