A particle undergoes simple harmonic motion with amplitude \(25 \mathrm{cm}\) and maximum speed \(4.8 \mathrm{m} / \mathrm{s} .\) Find the (a) angular frequency, (b) period, and (c) maximum acceleration.

Understanding the angular frequency in simple harmonic motion (SHM) is crucial for comprehending the oscillatory dynamics of a particle. Angular frequency, denoted as \( \omega \), represents the rate at which an object moves through its cycle in radians per second. In SHM, it can be thought of as the 'speed' of the oscillation in terms of angular movement.

From the step-by-step solution, we see that angular frequency is related to the maximum velocity and amplitude via the formula \( \omega = \frac{v_{max}}{A} \). This formula succinctly ties together the amplitude of motion, the peak speed, and the angular aspect of SHM. In practice, after converting the amplitude from centimeters to meters, it was determined that the angular frequency for our particle was \( 19.2 \) radians per second. This quick angular 'turnover' rate helps us predict how rapidly the particle repeats its motion over a certain angular distance.

The period of simple harmonic motion, symbolized as \( T \), refers to the time it takes for the particle to complete one full cycle of its motion. It is the duration of one oscillation from start to finish and back again.

In the context of SHM, the period is intricately linked to the angular frequency. The formula to find the period is the reciprocal of the angular frequency, given as \( T = \frac{1}{\omega} \). A lower angular frequency indicates a longer period, meaning the particle takes more time to complete a cycle, whereas a higher angular frequency implies a shorter period with rapid cycles. In the studied example, with the calculated angular frequency of \( 19.2 \) rad/s, the period is remarkably short, at approximately \( 0.052 \) seconds, indicating a very swift oscillation.

Maximum acceleration is reached at the extremes of the SHM motion when the particle is at its maximum displacement from the equilibrium position. Regarding maximum acceleration, \( a_{max} \), the relationship with angular frequency and amplitude is given by the equation \( a_{max} = \omega^{2}A \). This equation shows that maximum acceleration increases with the square of the angular frequency – suggesting a highly sensitive dependence on how quickly the particle oscillates.

When we apply the provided values for angular frequency and amplitude from the exercise, the maximum acceleration calculated is \( 92.16 \) meters per second squared. This high value indicates that at the points of maximum displacement, there's a strong restoring force exerting upon the particle, as per Newton's second law, causing it to accelerate back towards the equilibrium position with substantial quickness.