91Ó°ÊÓ

Understanding angular frequency is key to comprehending simple harmonic motion. It essentially refers to how quickly an object completes its oscillations in terms of its circular path analogy. Mathematically, angular frequency \( \omega \) is defined as \( \omega = \frac{2\pi}{T} \), where \( T \) is the period.
For our exercise, the period \( T \) is given as 0.45 seconds. Substituting this into the formula gives: \( \omega = \frac{2\pi}{0.45} \approx 13.96 \) rad/s.
This tells us that in the harmonic motion described, the object completes approximately 13.96 radians of its cycle in one second. This frequency also determines other properties of the motion, such as speed and acceleration.

Simple harmonic motion can be described by a specific type of equation that gives the position of the oscillating object over time. In our scenario, the object is released from a compressed position, so the position as a function of time \( y(t) \) is expressed in a cosine form. This is because the cosine wave starts at its maximum value, reflecting our starting point.
The equation takes the form: \[ y(t) = A \cos(\omega t) \] where \( A \) is the amplitude, or the maximum displacement from the equilibrium position. Here, \( A = 0.16 \) m and \( \omega \) is the angular frequency we calculated earlier.
This results in our equation of motion: \[ y(t) = 0.16 \cos(13.96 t) \] This formula can be used to determine the position \( y \) of the object at any given time \( t \). It visually depicts the oscillatory motion as it varies over time.

In simple harmonic motion, the maximum speed of the object occurs as it passes through the equilibrium position. This is because all of the object's potential energy is converted into kinetic energy at that point.
The formula for maximum speed \( v_{max} \) is given by: \[ v_{max} = A \omega \] where \( A \) is the amplitude and \( \omega \) is the angular frequency.
Substituting the known values gives: \[ v_{max} = 0.16 \times 13.96 \approx 2.23 \text{ m/s} \]
Thus, the object's maximum speed is approximately 2.23 meters per second as it swings through the equilibrium point.

In simple harmonic motion, the object experiences maximum acceleration when it is at the maximum displacement from the equilibrium position because this is where the restoring force is greatest.
The maximum acceleration \( a_{max} \) is given by the equation: \[ a_{max} = A \omega^2 \]
For our problem, substituting in the values provides: \[ a_{max} = 0.16 \times (13.96)^2 \approx 31.15 \text{ m/s}^2 \]
This indicates that the greatest acceleration of 31.15 meters per second squared occurs when the object is furthest from the center, contributing to the rapid change in velocity as it swings back towards equilibrium.