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The frequency of oscillation tells us how many cycles an oscillator completes in one second. This is a key aspect of simple harmonic motion as it helps us understand the rhythm of oscillations. In our exercise, we started with determining the angular frequency, represented as \( \omega \), using the formula:\[ \omega = \sqrt{\frac{k}{m}} \]where \( k \) is the spring constant. After finding the angular frequency, the frequency \( f \) can be calculated using:\[ f = \frac{\omega}{2\pi} \]This conversion is crucial because it takes us from the angular speed to the oscillations per second. In our particular example, after finding the mass, the calculated frequency came out to about \(5.58 \, \mathrm{Hz}\). It means the block-spring system oscillates about 5.58 times each second. Understanding this can give tangible insights into how fast or slow an oscillating system moves back and forth.

The mass of the block in harmonic motion plays a crucial role in determining the system's dynamics. In simple harmonic motion, the block's mass directly influences the overall inertia of the system.To find the mass, we rearranged the acceleration equation:\[ a = -\omega^2 x \]Using the positional and acceleration data given, we first solve for \( \omega \), the angular frequency, and then transform it to find the mass \( m \) by restating this in terms of \( m \):\[ \omega = \sqrt{\frac{425}{m}} \]By squaring both sides and isolating \( m \), we calculated the approximate mass to be around \(0.345 \, \mathrm{kg}\). This calculation is essential because mass affects how tightly or loosely an oscillator moves around its equilibrium point. A lower mass generally means quicker oscillations, while a higher mass implies slower oscillations.

The amplitude of motion is a measure of how far the block travels from its equilibrium point during oscillations. It represents the maximum extent of displacement and is a fundamental characteristic of oscillatory motion.To determine the amplitude \( A \), we used the velocity relation of simple harmonic motion:\[ v^2 = \omega^2 (A^2 - x^2) \]Given the velocity \( v = -13.6 \, \mathrm{m/s} \), along with our calculated \( \omega^2 \) and position \( x = 0.100 \, \mathrm{m} \), we proceed by solving for \( A^2 \) and thus for \( A \):\[ A^2 = \frac{184.96}{1230} + 0.01 \]By simplifying it, the amplitude results in approximately \(0.387 \, \mathrm{m}\). Amplitude is critical as it indicates the maximum potential energy stored in the system and shapes how vigorous or subtle the oscillations are. It ultimately defines the bounds within which the block oscillates around its equilibrium position.