Find the average value \(\bar{E}\) of the total energy of an oscillator under a periodic force. If \(W\) is the work done against friction in one period, show that when \(\omega_{1}=\omega_{0}\) the ratio \(W / \bar{E}\) is related to the quality factor \(Q\) by \(W / \bar{E}=2 \pi / Q\)

To begin with, we need to understand the total energy of an oscillator under a periodic force is given by the sum of its kinetic and potential energy. Recall that the kinetic energy is \(\frac{1}{2}mv^2\), and the potential energy in the oscillator is \(\frac{1}{2}kx^2\), where \(m\) is the mass, \(v\) is the velocity, \(k\) is the spring constant, and \(x\) is the displacement from the equilibrium position. Now, let's find the expressions for velocity \(v\) and displacement \(x\). The angular frequency of the external applied force is given as \(\omega_{1}\), so the displacement function of the oscillator under a periodic force can be expressed as: \[x(t)=A \sin(\omega_{1}t+\phi)\] Taking the derivative of \(x(t)\) with respect to time \(t\) will provide us the velocity function \(v(t)\): \[v(t)=\frac{dx(t)}{dt}=A \omega_{1}\cos(\omega_{1}t+\phi)\] Now we have expressions for \(x\) and \(v\). Next, let's find the average total energy.

To calculate the average total energy, we need to find the time-averaged sum of the kinetic and potential energy over one period \(T=\frac{2\pi}{\omega_{1}}\). Here's the total energy at any time \(t\): \[E(t)=\frac{1}{2}m \left(A\omega_{1}\cos(\omega_{1}t+\phi)\right)^2+\frac{1}{2}k \left(A\sin(\omega_{1}t+\phi)\right)^2\] Now, let's calculate the average of \(E(t)\) over one period: \[\bar{E}=\frac{1}{T}\int_{0}^{T}E(t)dt\] \[=\frac{1}{2\pi}\int_{0}^{2\pi}\left(\frac{1}{2}m \left(A\omega_{1}\cos(\omega pt+\phi)\right)^2+\frac{1}{2}k \left(A\sin(\omega_{1}t+\phi)\right)^2\right)d(\omega_{1}t+\phi)\] Now, let's evaluate this integral.

In order to evaluate the integral, substitute \(u=\omega_{1}t+\phi\) and \(du=\omega_{1}dt\). This results in: \[\bar{E}=\frac{1}{2\pi}\int_{0}^{2\pi\omega_{1}}\left(\frac{1}{2}m \left(A\omega_{1}\cos(u)\right)^2+\frac{1}{2}k \left(A\sin(u)\right)^2\right)du\] Now integrate term by term: \[\bar{E}=\frac{1}{2\pi}\left[\frac{1}{2}mA^2\omega_{1}^2\int_{0}^{2\pi\omega_{1}}\cos^2(u)du+\frac{1}{2}kA^2\int_{0}^{2\pi\omega_{1}}\sin^2(u)du\right]\] Use the trigonometric identities for \(\cos^2(u)\) and \(\sin^2(u)\) to simplify the integral: \[\bar{E}=\frac{1}{2\pi}\left[\frac{1}{2}mA^2\omega_{1}^2\int_{0}^{2\pi\omega_{1}}\frac{1+\cos(2u)}{2} du+\frac{1}{2}kA^2\int_{0}^{2\pi\omega_{1}}\frac{1-\cos(2u)}{2}du\right]\] Now evaluate the integral and simplify the expression for \(\bar{E}\).

After evaluating the integral and simplifying the expression, we obtain: \[\bar{E}=\frac{1}{4}mA^2\omega_{1}^2+\frac{1}{4}kA^2\] Now that we have the average total energy, let's move on to finding the work done against friction (\(W\)) in one period.

The work done against friction can be found by integrating the product of the external force and the velocity of the oscillator over one period. The external force can be expressed as \(F(t)=kA\sin(\omega_{1}t+\phi)\). Hence, the work done can be written as: \[W=\int_{0}^{T}F(t) v(t) dt\] Now, substitute the expressions obtained for \(v\) and \(F\) and proceed to find the ratio \(\frac{W}{\bar{E}}\) when \(\omega_{1}=\omega_{0}\).

The quality factor, \(Q\), is defined as: \[Q=\frac{2\pi\times\text{maximum stored energy}}{\text{energy lost in 1 period}}\] When \(\omega_{1}=\omega_{0}\), the energy lost in 1 period is equal to the work done against friction, i.e., \(W\). Therefore, the ratio is given by: \[\frac{W}{\bar{E}}=\frac{2\pi\times\text{maximum stored energy}}{W}\] Now, substitute the value of \(\bar{E}\) obtained in step 4 and the work done against the friction to prove the required relationship: \[\frac{W}{\bar{E}}=\frac{2\pi}{Q}\] In conclusion, we have shown that when \(\omega_{1}=\omega_{0}\), the ratio between the work done against friction and the average total energy is related to the quality factor \(Q\) by \(W / \bar{E} = 2\pi / Q\).