Question

The minimum phase difference between two simple harmonic motions is
\(x_1 = \frac{1}{\sqrt{2}}sinωt+\frac{1}{\sqrt{2}}cosωt\)
\(x_2 = sinωt+cosωt\)     \([sin\frac{π}{4} = cos\frac{π}{4} = \frac{1}{\sqrt{2}}]\)

• A. zero
• B. \(\frac{π^C}{3}\)
• C. \(\frac{π^C}{4}\)
• D. \(\frac{π^C}{5}\)

Hint:

Write each expression as a single sine with a phase constant.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
Combine sine and cosine using \(\sin\omega t + \cos\omega t = \sqrt2\sin\left(\omega t + \frac\pi4\right)\).

Step 2: Rewrite both
\[ x_1 = \frac{1}{\sqrt2}(\sin\omega t + \cos\omega t) = \sin\left(\omega t + \frac\pi4\right) \]
\[ x_2 = \sin\omega t + \cos\omega t = \sqrt2\sin\left(\omega t + \frac\pi4\right) \]

Step 3: Compare phases
Both have the same phase constant \(\pi/4\). They differ only in amplitude (1 and \(\sqrt2\)). So the phase difference is zero.

Final Answer:
The phase difference is zero, option (A). \[ \boxed{0} \]