For an oscillator under a periodic force \(F(t)=F_{1} \cos \omega_{1} t\), calculate the power, the rate at which the force does work. (Neglect the transient.) Show that the average power is \(P=m \gamma \omega_{1}^{2} a_{1}^{2}\), and hence verify that it is equal to the average rate at which energy is dissipated against the resistive force. Show that the power \(P\) is a maximum, as a function of \(\omega_{1}\), at \(\omega_{1}=\omega_{0}\), and find the values of \(\omega_{1}\) for which it has half its maximum value.

In physics, a periodic force refers to a force that acts on a system repeatedly at regular intervals of time. This oscillatory input is fundamental in understanding systems such as pendulums, springs, and circuits where repetitive motion or cycles are present. Most notably, it's essential in calculating the response of an oscillator, a system that exhibits periodic motion.

When dealing with oscillators, such as in the given exercise, the force might look something like this: \( F(t)=F_{1} \times \text{cos} \(\omega_{1} t\) \), where \( F_{1} \) is the amplitude of the force and \( \omega_{1} \) is the angular frequency of the force. It’s crucial for students to understand that the periodic force not only brings about motion but also dictates the frequency at which the system will oscillate, ultimately influencing the power output.

For a practical understanding, let's consider a child being pushed on a swing. If pushed periodically (at a constant rate), the swing moves back and forth at a predictable rhythm. Similarly, applying a periodic force in the context of oscillators can result in what's called resonance when the frequency of the external force matches the system's natural frequency, this leads to maximum energy transfer — an essential aspect demonstrated in the exercise solution.

Another major topic that comes into play with oscillators is energy dissipation. To dampen the mathematical jargon, imagine energy dissipation as the process by which vibrational energy in a system (like our oscillator) is slowly lost over time. It’s similar to a toy top that spins vigorously at first but slows down gradually due to the friction with the surface. The mechanism through which energy is systematically lost is often described as a 'resistive' or 'damping' force, and this force counteracts the motion of the oscillator.

Within the given exercise, we're dealing with a resistive force proportional to the velocity of the oscillator, which follows \( F_{resistive}=-m\text{γ}v(t) \). This negative sign indicates that the resistive force works in the opposite direction of motion. What it means for students is this: for every cycle of the oscillation, a bit of energy is 'consumed' by factors such as friction or air resistance. And, when the average power due to the external force equals the average rate at which energy is dissipated, this balances out and ensures the sustained motion of the oscillator at a steady amplitude.

Finally, wrapping our minds around the concept of natural frequency is crucial in understanding oscillatory systems. Natural frequency is the inherent rate at which a system oscillates when it's not being disturbed by outside forces. Think of it as the unique 'pitch' a glass might have when tapped – that specific tone is akin to its natural frequency.

In the oscillator exercise, the natural frequency \( \omega_{0} \) plays a pivotal role in determining the power transfer when subjected to a periodic force. It’s at this frequency that resonance occurs — the state where the system absorbs and dissipates energy most efficiently. A remarkable aspect of the exercise solution is showing that when the driving frequency \( \omega_{1} \) matches the natural frequency \( \omega_{0} \), the average power reaches its peak. Revealing this is akin to striking the precise note that makes the glass resonate and potentially shatter.

The exercise also takes into consideration the values of \( \omega_{1} \) at which the average power is half its maximum, often referred to as the 'half-power points'. These points are invaluable in real-life applications such as tuning a radio to a specific station or improving the acoustics of a concert hall. The half-power points influence bandwidth and help in understanding the range of frequencies over which a system effectively responds.