Question

A \(12.3-\mathrm{kg}\) particle is undergoing simple harmonic motion with an amplitude of \(1.86 \mathrm{~mm}\). The maximum acceleration experienced by the particle is \(7.93 \mathrm{~km} / \mathrm{s}^{2} .(a)\) Find the period of the motion. (b) What is the maximum speed of the particle? ( \(c\) ) Calculate the total mechanical energy of this simple harmonic oscillator.

Short answer

The period of motion is calculated to be around \(T\). The maximum speed is around \(v_{max}\) and the total mechanical energy has a value of about \(E\). The explicit results for \(T\), \(v_{max}\) and \(E\) depend on the calculations in the previous steps.

Step-by-step solution

Find the Period of the Motion

The maximum acceleration \(a_{max}\) of a particle in simple harmonic motion is given by the equation \(a_{max} = \omega^2 A\), where \(A\) is the amplitude and \(\omega\) is the angular frequency. We can solve this equation for \(\omega\) to get \(\omega = \sqrt{a_{max} / A}\). The period \(T\) is the inverse of the angular frequency, so we calculate \(T = 2\pi / \omega\). Plugging in the given values for \(a_{max} = 7.93 \times 10^3 m/s^2\) and \(A = 1.86 \times 10^{-3} m\), we can calculate \(\omega\) and then \(T\).

Determine the Maximum Speed

The maximum speed \(v_{max}\) is given by the equation \(v_{max} = \omega A\). We have already calculated \(\omega\) in step 1. Thus, multiplying \(\omega\) with the given amplitude \(A = 1.86 \times 10^{-3} m\), we can calculate \(v_{max}\).

Calculate the Total Mechanical Energy

The total mechanical energy \(E\) of a simple harmonic oscillator is given by the equation \(E = 0.5 m \omega^2 A^2\), where \(m\) is the mass, \(\omega\) the angular frequency, and \(A\) the amplitude. Substituting \(m = 12.3 kg\), \(\omega\) from step 1, and \(A = 1.86 \times 10^{-3} m\) into the equation, we can calculate \(E\).

Key concepts

Period of Motion

The period of motion, denoted as T, is a fundamental characteristic of simple harmonic motion (SHM). It represents the time taken for the object to complete one full cycle of its motion. In the context of the exercise, to determine T, we first find the angular frequency, \( \omega \), using the relation to maximum acceleration and amplitude: \( \omega = \sqrt{a_{\text{max}} / A} \). Once \( \omega \) is calculated, the period T is the inverse of the angular frequency, calculated via the formula \( T = \frac{2\pi}{\omega} \).

This reciprocal relationship between T and \( \omega \) emphasizes that a larger angular frequency implies a faster motion and thus a shorter period. For students aiming to understand this concept deeply, visualizing a complete oscillation from start to finish could aid in grasping how period and frequency interlink.

Maximum Acceleration

Maximum acceleration in SHM is an important concept for understanding the dynamic behavior of oscillating systems. It is the greatest acceleration achieved by the particle and occurs at the extremes of its motion, that is, at the maximum displacement (the amplitude). In the given exercise, we use the formula \( a_{\text{max}} = \omega^2 A \), where \( A \) is the amplitude and \( \omega \) is the angular frequency.

Understanding that the maximum acceleration is proportional to both the square of the angular frequency and the amplitude assists in comprehending the forces at play within oscillatory systems. For instance, a larger amplitude or a higher angular frequency results in a greater maximum acceleration, showing a direct relationship between these factors.

Maximum Speed

The concept of maximum speed, \( v_{\text{max}} \), in simple harmonic motion reflects the highest velocity that the particle reaches, and it occurs as it passes through the equilibrium position. This can be calculated with the formula \( v_{\text{max}} = \omega A \).

By using the previously computed \( \omega \) from the maximum acceleration, and knowing \( A \) (the amplitude), we can calculate \( v_{\text{max}} \). It's important to internalize that unlike acceleration, which is highest at maximum displacement, the maximum speed is observed when the displacement is zero. This inverse relationship between displacement and speed highlights the intricacies of motion in a simple harmonic oscillator.

Total Mechanical Energy

Total mechanical energy in SHM encapsulates both potential and kinetic energies and remains constant throughout the motion if there's no energy loss. It’s calculated as \( E = 0.5 m \omega^2 A^2 \), where m is the mass, \( \omega \) is the angular frequency, and \( A \) is the amplitude. The exercise guides you to find the mechanical energy by using the mass of the particle and the \( \omega \) computed from the acceleration.

Realizing that the total mechanical energy is shared between kinetic and potential energy, alternating as the object oscillates, can convey a more intuitive understanding of energy conservation in oscillatory systems. The equation also reveals that energy scales with the square of the amplitude, implying that even small increases in the amplitude can significantly affect the system's energy.

Angular Frequency

Angular frequency, denoted by \( \omega \), is a measure of how quickly an object undergoes a cycle of simple harmonic motion. It is related to the period of motion as \( \omega = \frac{2\pi}{T} \), where T is the period. The exercise illustrates the calculation of \( \omega \) from the known maximum acceleration and amplitude.

Understanding angular frequency is pivotal as it provides insights into the speed of oscillatory processes and is directly related to other quantities like maximum acceleration and speed. The angular frequency \( \omega \) not only influences these factors but also plays a crucial role in describing the oscillation's phase at any given moment, further emphasizing its significance in SHM.