91Ó°ÊÓ

In Simple Harmonic Motion (S.H.M.), the velocity of a particle changes over time. The **S.H.M. velocity equation** helps us understand this change. It is given by:
\[ v = \omega\sqrt{A^2 - x^2} \]
Here:

• **v** represents the velocity.
• **\( \omega \)** is the angular frequency.
• **A** signifies the amplitude, i.e. the maximum displacement.
• **x** is the displacement from the center position.

What this equation tells us is that the velocity depends on both the amplitude and the displacement. When the particle is at maximum displacement (amplitude), the velocity is zero since the particle is momentarily stationary. When the particle is at the center, the velocity reaches its maximum value.
By substituting known values of **v**, **A**, and **x**, we can find the angular frequency **\( \omega \)**, as shown in the step-by-step solution.

The **angular frequency** \( \omega \) is a crucial factor in S.H.M. It describes how quickly the particle oscillates back and forth. The angular frequency is found using the equation:
\[ \omega = \frac{4}{2\sqrt{2}} \]
Simplifying the above expression, we get:
\[ \omega = \sqrt{2} \text{ rad/s} \]
Angular frequency connects the linear speed (or velocity) with the amplitude and displacement. It is directly proportional to the square root of the stiffness of the system and inversely proportional to the square root of the mass if we are considering a spring-mass system. In S.H.M, angular frequency is also related to the normal frequency (f) by the equation:
\[ \omega = 2\pi f \]
In our problem, knowing the angular frequency \( \omega = \sqrt{2} \text{ rad/s} \) is essential for determining the period or the time it takes for one complete oscillation.

The **period of oscillation** \( T \) is the time it takes for the particle to complete one full cycle of motion. It is linked with the angular frequency using the formula:
\[ T = \frac{2\pi}{\omega} \]
Substituting the value of the angular frequency \( \omega = \sqrt{2} \text{ rad/s} \), we get:
\[ T = \frac{2\pi}{\sqrt{2}} \approx \sqrt{2\pi} \]
Simplifying, it's evident that the period of oscillation is:
\[ T = \sqrt{2\pi} \]
The period tells us how long it takes for the particle to go from one extreme end, back to that end (passing through the center). This concept is fundamental when thinking about oscillatory systems, as it allows prediction of the motion's timing precisely. Hence, in our exercise, the period evaluated as \( \sqrt{2}\pi \) seconds matches option (a) from the given choices.