Question

Two waves are represented by the equations, \(y_1 = asin(ωt+kx+0.57)\) m and \(y_2 = acos(ωt+kx)\) m, where \(x\) is in metre and \(t\) is in second. What is the phase difference between them? (\(π = 3.14\))

• A. \(0.57\) radian
• B. \(1.57\) radian
• C. \(1.25\) radian
• D. \(1.0\) radian

Hint:

Convert the cosine to a sine with an extra \(\pi/2\) phase.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
The phase difference is found when both waves are written in the same form, such as \(\sin(\omega t + kx + \phi)\).

Step 2: Convert:
\(y_2 = a\cos(\omega t + kx) = a\sin\left(\omega t + kx + \frac{\pi}{2}\right)\).
The phase of \(y_1\) is \(0.57\), and the phase of \(y_2\) is \(\frac{\pi}{2} = 1.57\) (with \(\pi = 3.14\)).

Step 3: Difference:
\[ \Delta\phi = 1.57 - 0.57 = 1.0\ \text{rad} \]
Option A forgets the shift between sine and cosine. Option B uses \(\frac{\pi}{2}\) alone.

Final Answer:
The phase difference is \(1.0\) radian, option (D). \[ \boxed{1.0\ \text{rad}} \]