91Ó°ÊÓ

The point charge is q.

The mass is m.

The spring constant is k.

The initial energy is$U_0=\frac{1}{2}m{v}_0^2$

The natural frequency,${\mathrm{Ó\neg }}_0=\sqrt{\frac{k}{m}}$

The equation of the motion,

$x+y+x+{\mathrm{Ó\neg }}_0^{2}x=0,$

The expression to calculate the total radiated energy is given as follows.

${\mathit{U}}_{\mathbf{0}}\mathbf{=}\mathbf{∫}<P>\mathbf{·}\mathit{d}\mathit{a}$ …… (1)

The expression for the force is given by

$F=mx$

The radiated force is given by,

role="math" localid="1658841100943" $F_{\mathrm{rad}}=m\mathrm{Ï„}x$

The spring force is given by,

$F_{spring}=-kx$

The force on the system is given by,

$F=F_{spring}+F_{rad}$

Substitute mx for F, -kx for F_spring and $m\mathrm{Ï„}x$ for F_rad into above equation.

$mx=-kx+m\mathrm{Ï„}x$

Divide the above equation by m,

$x=-\frac{k}{m}x+\mathrm{Ï„}x$

Substitute ${\mathrm{Ó\neg }}_0^2$ for k/m into above equation.

$\begin{array}{rcl}x& =& -{\mathrm{Ó\neg }}_0^{2}x+\mathrm{Ï„}x\\ x+{\mathrm{Ó\neg }}_0^{2}x+\mathrm{Ï„}x& =& 0\end{array}$ …… (2)

Since the damping is small, therefore it oscillates at the natural frequency.

Thus,

$x=-{\mathrm{Ó\neg }}_0^{2}x$

Substitute ${\mathrm{Ó\neg }}_0^{2}x$for -x into equation (2).

$$\begin{gathered} x+{\mathrm{Ó\neg }}_0^{2}x+\mathrm{Ï„}{\mathrm{Ó\neg }}_0^{2}x=0 \\ x+\mathrm{Ï„}{\mathrm{Ó\neg }}_0^{2}x+{\mathrm{Ó\neg }}_0^{2}x=0 \end{gathered}$$ …… (3)

The damping factor is given by,

$\mathrm{γ}={\mathrm{Ó\neg }}_0^2\mathrm{Ï„}$

Substitute ${\mathrm{Ó\neg }}_0^2\mathrm{Ï„}$for $\mathrm{γ}$into equation (3).

$x+\mathrm{γ}x+{\mathrm{Ó\neg }}_0^{2}x=0$

The solution of the equation to motion is given by,

$x\left(t\right)=A{e}^{-\mathrm{γ}t/2}\sin {\mathrm{Ó\neg }}_{0}t$ ……. (4)

Differentiate the above equation with respect to time.

role="math" localid="1658842249857" $v\left(t\right)=-\frac{\mathrm{γ}}{2}A{e}^{-\mathrm{γ}t/2}\sin \left({\mathrm{Ó\neg }}_{0}t\right)+{\mathrm{Ó\neg }}_{0}A{e}^{-\mathrm{γ}t/2}\cos {\mathrm{Ó\neg }}_{0}t$

At $t=0,v\left(0\right)=v_0$

$\begin{array}{rcl}v\left(t\right)& =& -\frac{\mathrm{γ}}{2}A{e}^{-\mathrm{γ}t/2}\sin {\mathrm{Ó\neg }}_0\left(0\right)+{\mathrm{Ó\neg }}_{0}A{e}^{-\mathrm{γ}t/2}\cos {\mathrm{Ó\neg }}_0\left(0\right)\\ v_0& =& {\mathrm{Ó\neg }}_{0}A\\ A& =& \frac{v_0}{{\mathrm{Ó\neg }}_0}\end{array}$

Substitute $\frac{v_0}{{\mathrm{Ó\neg }}_0}$ for A into equation (4).

role="math" localid="1658842771258" $x\left(t\right)=\frac{v_0}{{\mathrm{Ó\neg }}_0}{e}^{-\mathrm{γ}t/2}\sin {\mathrm{Ó\neg }}_{0}t$

The expression for the power is given by,

$P=\frac{{\mathrm{μ}}_0{q}^2{a}^2}{6\mathrm{π}c}$

Substitute $-{\mathrm{Ó\neg }}_0^{2}x$for a into above equation.

$$\begin{gathered} P=\frac{{\mathrm{μ}}_0{q}^2}{6\mathrm{Ï€}c}{\left(-{\mathrm{Ó\neg }}_0^{2}x\right)}^2 \\ P=\frac{{\mathrm{μ}}_0{q}^2}{6\mathrm{Ï€}c}-{\mathrm{Ó\neg }}_0^4{x}^2 \end{gathered}$$

Substitute $\frac{v_0}{{\mathrm{Ó\neg }}_0}{e}^{-\mathrm{γ}t/2}\sin {\mathrm{Ó\neg }}_{0}t$for x into above equation.

role="math" localid="1658843158125" $$\begin{gathered} P=-\frac{{\mathrm{μ}}_0{q}^2}{6\mathrm{Ï€}c}{\mathrm{Ó\neg }}_0^4{\left(\frac{v_0}{{\mathrm{Ó\neg }}_0}{e}^{-\mathrm{γ}t/2}\sin {\mathrm{Ó\neg }}_{0}t\right)}^2 \\ P=\frac{{\mathrm{μ}}_0{q}^2\left({\mathrm{Ó\neg }}_0^2{v}_0^2\right)}{6\mathrm{Ï€}c}{e}^{-\mathrm{γ}t}{\sin }^2{\mathrm{Ó\neg }}_{0}t \end{gathered}$$

Average power over the full cycle keeping the $e^{-\mathrm{γ}t}$constant.

role="math" localid="1658843227983" $$\begin{gathered} P=\frac{{\mathrm{μ}}_0{q}^2\left({\mathrm{Ó\neg }}_0^2{v}_0^2\right)}{6\mathrm{Ï€}c}{e}^{-\mathrm{γ}t}\frac{1}{2} \\ P=\frac{{\mathrm{μ}}_0{q}^2\left({\mathrm{Ó\neg }}_0^2{v}_0^2\right)}{12\mathrm{Ï€}c}{e}^{-\mathrm{γ}t} \end{gathered}$$

Calculate the total energy,

Substitute $\frac{{\mathrm{μ}}_0{q}^2\left({\mathrm{Ó\neg }}_0^2{v}_0^2\right)}{12\mathrm{Ï€}c}{e}^{-\mathrm{γ}t}$ for P into equation (1).

role="math" localid="1658843599437" $\begin{array}{rcl}{U}_0& =& {∫}_0^{∞}\frac{{\mathrm{μ}}_0{q}^2\left({\mathrm{Ó\neg }}_0^2{v}_0^2\right)}{12\mathrm{Ï€}c}{e}^{-\mathrm{γ}t}dt\\ U_0& =& \frac{{\mathrm{μ}}_0{q}^2\left({\mathrm{Ó\neg }}_0^2{v}_0^2\right)}{12\mathrm{Ï€}c}{∫}_0^{∞}{e}^{-\mathrm{γ}t}dt\\ U_0& =& \frac{{\mathrm{μ}}_0{q}^2\left({\mathrm{Ó\neg }}_0^2{v}_0^2\right)}{12\mathrm{Ï€}c}{\left(\frac{1}{\mathrm{γ}}{e}^{-\mathrm{γ}t}dt\right)}_0^{∞}\\ U_0& =& \frac{{\mathrm{μ}}_0{q}^2\left({\mathrm{Ó\neg }}_0^2{v}_0^2\right)}{12\mathrm{Ï€}c}\end{array}$

Substitute ${\mathrm{Ó\neg }}_0^2\mathrm{Ï„}$for $\mathrm{γ}$into above equation.

role="math" localid="1658843809745" $\begin{array}{rcl}{U}_0& =& \frac{{\mathrm{μ}}_0{q}^2\left({\mathrm{Ó\neg }}_0^2{v}_0^2\right)}{12\mathrm{Ï€}c{\mathrm{Ó\neg }}_0^2\mathrm{Ï„}}\\ U_0& =& \frac{{\mathrm{μ}}_0{q}^2{v}_0^2}{12\mathrm{Ï€}c}\frac{1}{\mathrm{Ï„}}\end{array}$ ……. (5)

The reaction torque is given by,

$\mathrm{τ}=\frac{{\mathrm{μ}}_0{q}^2}{6\mathrm{π}mc}$

Substitute$\frac{{\mathrm{μ}}_0{q}^2}{6\mathrm{π}mc}$ for $\mathrm{τ}$into equation (5).

$\begin{array}{rcl}{U}_0& =& \frac{{\mathrm{μ}}_0{q}^2{v}_0^2}{12\mathrm{π}c}\frac{1}{{\displaystyle \frac{{\mathrm{μ}}_0{q}^2}{6\mathrm{π}mc}}}\\ U_0& =& \frac{1}{2}m{v}_0^2\end{array}$

Hence it is proved that the total radiated energy is equal to U₀.

The reaction torque is given by,

$\mathrm{τ}=\frac{{\mathrm{μ}}_0{q}^2}{6\mathrm{π}mc}$ …… (6)

The single oscillator has twice charge and twice mass.

Substitute 2m for m and 2q fir q into equation (6).

$$\begin{gathered} {\mathrm{τ}}_1=\frac{{\mathrm{μ}}_0{\left(2q\right)}^2}{6\mathrm{π}\left(2m\right)c} \\ {\mathrm{τ}}_1=2\mathrm{τ} \end{gathered}$$

From the above the torque is double. Since$∫Pdt$ goes $q^2/\mathrm{γ}$, it is also double.

The power radiated is indeed four times, but the oscillator died ways faster, therefore the total energy radiated is just a twice as one oscillator.