91Ó°ÊÓ

In simple harmonic motion, the **amplitude** is a crucial component that describes the maximum extent of displacement from the equilibrium position. It essentially represents how far the object goes on either side of its resting point during oscillation.
In our exercise, the amplitude is extracted directly from the velocity function. The given velocity function \(v_{x}(t) = -(3.60 \mathrm{~cm/s}) \sin[(4.71 \mathrm{~rad/s}) t - (\pi/2)]\) has an amplitude of \( 3.60 \mathrm{~cm} \). This indicates that the maximum linear displacement from the equilibrium position is 3.60 centimeters. Remember, amplitude is always expressed in positive terms, showing the magnitude of maximum displacement irrespective of direction.

• It is represented with \( A \) in formulas.
• It influences the energy in the system—larger amplitudes mean more energy.

The **period** of simple harmonic motion refers to the time it takes for the oscillating object to complete one full cycle of motion. This is an intrinsic characteristic of the system and is inversely related to the angular frequency \( \omega \).
The formula to calculate the period \( T \) is:\[ T = \frac{2\pi}{\omega} \] For the function in our problem, \( \omega \) is \( 4.71 \mathrm{~rad/s} \).By substituting this \( \omega \) value into the period formula, we find: \[ T = \frac{2\pi}{4.71 \mathrm{~s}} \] This informs us how long it takes for the oscillation to complete. In practical terms, it lets you time when an object will return to the same position with the same motion. The period is usually measured in seconds.

• Understanding the period helps in predicting future positions in oscillatory motion.
• Every complete oscillation is identical within the same system.

The **maximum acceleration** in a motion is critical as it defines the peak rate of change of velocity the oscillating object can achieve. It reveals the points of force extremes within the cycle.
In simple harmonic motion, maximum acceleration \( a_{max} \) can be mathematically described as:\[ a_{max} = \omega^{2} \cdot A \] From the exercise, given \( \omega = 4.71 \mathrm{~rad/s} \) and \( A = 3.60 \mathrm{~cm} \), substitute these values to find:\[ a_{max} = (4.71 \mathrm{~rad/s})^2 \times 3.60 \mathrm{~cm} \]This computation will yield the extreme acceleration value in the system. Keep in mind that maximum acceleration points occur at the maximal displacement positions, i.e., the amplitudes.

• This metric shows how 'aggressive' or 'gentle' the motion is.
• Maximum acceleration has direct implications on the forces experienced by the system.

The **force constant** or spring constant \( k \) reflects the stiffness of the spring in a harmonic system. It tells us how much force is required to extend or compress the spring by a unit distance.
Determine the force constant using the mass \( m \) and the angular frequency \( \omega \) with:\[ k = m \cdot \omega^2 \] In this problem, with \( m = 0.5 \mathrm{~kg} \) and \( \omega = 4.71 \mathrm{~rad/s} \), we compute:\[ k = 0.5 \mathrm{~kg} \times (4.71 \mathrm{~rad/s})^2 \]The mathematical product provides the spring constant, which is typically measured in Newtons per meter (N/m).

• A higher force constant means a stiffer spring, responding less to the same force than a lower constant.
• The force constant plays a significant role in the period and frequency of oscillations.