Question

Two simple harmonic motions are represented by equations \(y_1 = 0.5 \sin [200\pi t + \frac{\pi}{3}]\) and \(y_2 = 0.5 \cos \pi t\). The phase difference of the velocity of particle 1 with respect to the velocity of particle 2 is:

• A. \( \frac{\pi}{2} \)
• B. \( \frac{\pi}{6} \)
• C. \( -\frac{\pi}{6} \)
• D. \( \frac{\pi}{3} \)

Hint:

Velocity in SHM always leads displacement by \(\frac{\pi}{2}\). Use this directly for phase comparisons.

Answer 1

The Correct Option is C

Step 1: Write displacement equations.
\[ y_1 = 0.5 \sin(200\pi t + \frac{\pi}{3}), \quad y_2 = 0.5 \cos(\pi t) \]

Step 2: Convert cosine to sine form.
\[ \cos(\pi t) = \sin\left(\pi t + \frac{\pi}{2}\right) \]
So,
\[ y_2 = 0.5 \sin\left(\pi t + \frac{\pi}{2}\right) \]

Step 3: Velocity expressions.
Velocity is derivative of displacement:
\[ v = \frac{dy}{dt} \]
For SHM:
\[ v \propto \cos(\omega t + \phi) \]

Step 4: Phase of velocity.
Velocity leads displacement by \(\frac{\pi}{2}\):
\[ \phi_{v1} = \frac{\pi}{3} + \frac{\pi}{2} \] \[ \phi_{v2} = \frac{\pi}{2} + \frac{\pi}{2} = \pi \]

Step 5: Phase difference.
\[ \Delta \phi = \phi_{v1} - \phi_{v2} \]
\[ \Delta \phi = \left(\frac{\pi}{3} + \frac{\pi}{2}\right) - \pi \]

Step 6: Simplify.
\[ \Delta \phi = \frac{2\pi + 3\pi}{6} - \frac{6\pi}{6} = -\frac{\pi}{6} \]

Step 7: Final conclusion.
\[ \boxed{-\frac{\pi}{6}} \] Hence, correct answer is option (C).