The phase difference between the instantaneous velocity and acceleration of a particle executing simple harmonic motion is (a) π (b) 0.707π (c) zero (d) 0.5π

21. The phase difference between the instantaneous velocity and acceleration of a particle executing simple harmonic motion is:

(a) \( \pi \)

(b) \( 0.707\pi \)

(c) zero

(d) \( 0.5\pi \)

Detailed Answer:

The phase difference between the velocity and acceleration of a particle undergoing Simple Harmonic Motion (SHM) is always a constant value. In SHM, the velocity vector is always ahead of the acceleration vector by a quarter cycle, or \( 90^\circ \). This corresponds to a phase difference of \( 0.5\pi \).

Mathematically, the velocity \( v \) and acceleration \( a \) in SHM are related to the displacement \( x \) by the following equations:

• Velocity: \( v = A\omega \cos(\omega t + \phi) \)
• Acceleration: \( a = -A\omega^2 \sin(\omega t + \phi) \)

Where:

• \( A \) is the amplitude
• \( \omega \) is the angular frequency
• \( t \) is time
• \( \phi \) is the phase constant

As you can see from the equations, the velocity and acceleration are not in phase but have a phase difference of \( 90^\circ \), which is equivalent to \( 0.5\pi \) radians.

The correct answer is (d) \( 0.5\pi \).

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