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

Question

The root mean square velocity ($v_{\text{rms}}$) of a gas is given by:

Answer 1

Step 1: Amplitude form
Each is a sum of equal-weight sine and cosine, so both are sines shifted by $45^{\circ}$.

Step 2: Phase
The shift is $\pi/4$ for both; only the amplitudes differ.

Step 3: Result
Phase difference $= 0$, option (A).

Final Answer:
Option (A). \[ \boxed{0} \]

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

Show Hint

The Correct Option is A

Solution and Explanation

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