Question

Average velocity of a particle executing SHM in one complete vibration is :

• A. $\frac{A\,\omega^2}{2}$
• B. zero
• C. $\frac{A_\omega}{2}$
• D. $A_\omega$

Answer 1

The Correct Option is B

The average velocity of a particle executing Simple Harmonic Motion (SHM) over one complete vibration is asked in the question. To solve this, let's explore the concept of SHM and how average velocity is determined in this context:

• In SHM, a particle oscillates back and forth over a mean position following a sinusoidal pattern. One complete vibration is the motion from one extreme to the other and back again.
• Velocity in SHM is given by v = \pm A \omega \sin(\omega t + \phi), where \( A \) is the amplitude, \( \omega \) is the angular frequency, \( t \) is time, and \( \phi \) is the phase constant.
• Over one complete vibration (or cycle), the particle returns to its original position with all net displacements canceling out.

The average velocity is calculated as the total displacement divided by the total time. Over a complete cycle in SHM:

• Total Displacement: The particle returns to its starting position, making the net displacement zero.
• Total Time: A complete cycle corresponds to a period \( T \).

Thus, the average velocity \( V_{\text{avg}} \) is:

V_{\text{avg}} = \frac{\text{Total Displacement}}{T} = \frac{0}{T} = 0

Hence, over one complete vibration, the average velocity of a particle executing SHM is zero. This matches with the given correct answer.

For better comprehension, let’s see why other options are incorrect:

• \frac{A \omega^2}{2} and \frac{A_\omega}{2}: These values do not represent average velocity over a complete cycle; they might relate to partial motion properties within a phase.
• A_\omega: This represents peak velocity, not average velocity.

In conclusion, the correct answer is zero, which logically follows from the periodic nature of SHM.