In describing the heat capacities of solids in Section 18.4...

In describing the heat capacities of solids in Section 18.4 we stated that the potential energy $U=\frac{1}{2} k x^{2}$ of a harmonic oscillator averaged over one period of the motion is equal to the kinetic energy $K=\frac{1}{2} m v^{2}$ averaged over one period. Prove this result using Eqs. $(14.13)$ and $(14.15)$ for the position and velocity of a simple harmonic oscillator. For simplicity, assume that the initial position and velocity make the phase angle $\phi$ equal to zero. (Hint: Use the trigonometric identities $\cos ^{2}(\theta)=[1+\cos (2 \theta)] / 2$ and $\sin ^{2}(\theta)=[1-\cos (2 \theta)] / 2 .$ What is the average value of $\cos (2 \omega t)$ over one period?

In describing the heat capacities of solids in Section 18.4 we stated that the potential energy $U=\frac{1}{2} k x^{2}$ of a harmonic oscillator averaged over one period of the motion is equal to the kinetic
energy $K=\frac{1}{2} m v^{2}$ averaged over one period. Prove this result using Eqs. $(14.13)$ and $(14.15)$ for the position and velocity of a simple harmonic oscillator. For simplicity, assume that the initial position and velocity make the phase angle $\phi$ equal to zero. (Hint: Use the trigonometric identities $\cos ^{2}(\theta)=[1+\cos (2 \theta)] / 2$ and $\sin ^{2}(\theta)=[1-\cos (2 \theta)] / 2 .$ What is the average value of $\cos (2 \omega t)$ over one period?

Key Concepts

Simple Harmonic Oscillator

A simple harmonic oscillator is a system in which the restoring force is directly proportional to the displacement and acts in the opposite direction, leading to sinusoidal oscillations. This model is fundamental in understanding periodic motion in many physical systems, including springs, pendulums (for small displacements), and vibrational modes in solids.

Potential and Kinetic Energy in Oscillatory Motion

In a simple harmonic oscillator, the potential energy, given by (1/2)kx², and the kinetic energy, (1/2)mv², both vary sinusoidally with time. Analyzing these energies is crucial for understanding energy conservation in oscillatory systems and for demonstrating that, when averaged over a full cycle, the contributions from both forms of energy are equal.

Time Averaging in Periodic Systems

Time averaging involves calculating the average of a time-dependent quantity over one complete period of oscillation. This method is particularly useful in systems with periodic behavior since it smooths out the instantaneous fluctuations and reveals the overall energy distribution or other averaged properties of the system.

Trigonometric Identities and Periodic Functions

Trigonometric identities, such as the ones for the squares of sine and cosine functions, are essential tools for simplifying expressions involving oscillatory functions. They allow the decomposition of products or squares of sine and cosine functions into forms that can be easily averaged over a period, exemplified by the fact that the average of cos(2?t) over one period is zero.

Transcript

00:02 Hello, so for the simple harmonic oscillator, the displacement x equals a which is the amplitude cause omega t, omega is the angular frequency t is the time, if you differentiate the displacement you get velocity and that gives you minus a omega sine omega t the negative is coming from the differential of the course and we also have the omega bing equal to root k over m where k is a constant m is the mass so to start the average value of cost two omega t because zero therefore sine squared omega t the average value of this equals the average value of that it gives you half so the average value of 2 omega t over 1 period is 0 that's why we have this expression here so from the question the average potential energy is given by half k x squared k is a constant x is the displacement and we know that x equals a equals omega t so i'm going to put in my expression there half k x squared x is a plus omega t everything squared that gives you half k a squared cause squared omega t but from this expression right here you can isolate the k so you have to square both sides and you get omega squared equals k over m so k equals m omega squared so i'm going to do the substitution so finally our average potential energy will be half times k is m omega squared and then i have a squared cross squared omega t so that would be the expression for the average potential energy go to the kinetic energy that's half mv squared that's the average kinetic energy average so this would be half and the mass the mass times v so we know that v is negative a omega sine omega t i'm going to put that in bracket negative a omega sine omega t uh now squared this is just half m so the negative will go away because of the squared so you get a squared omega squared uh sine squared omega t all right so uh the average potential energy can write as that but from here from this expression right here you know that sine squared omega t equals square omega t and is equal to half so each is equal to half so for the potential energy this is just going to be times half for the kinetic energy average kinetic energy it's also going to be half m a squared omega squared sine squared omega t is also half so finally my u a v is equal to 1 over 4 m squared omega squared and my kinetic energy is also 1 over 4 squared so you can see right here that the u the average potential energy equals the average kinetic energy is so you can say that u equals k like that.

06:14 Let's be mindful that in general, at any given time, the potential energy is not equal to the kinetic energy...

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