Question

2. Show that the expression \[ x^{(2)}(t)=x_{0}+\frac{1}{2}\left[x_{\omega} e^{-i \omega t}+x_{\omega}^{} e^{i \omega t}\right]+\frac{1}{2}\left[x_{2 \omega} e^{-2 i \omega t}+x_{2 \omega}^{} e^{2 i \omega t}\right] \] is a solution of \[ \frac{d^{2} x}{d t^{2}}+\omega_{0}^{2} x+a x^{2}=\frac{-e}{2 m}\left[E_{\omega} e^{-i \omega t}+E_{\omega}^{*} e^{i \omega t}\right] \] that is valid up to second order in the electric-field strength.

          2. Show that the expression
\[
x^{(2)}(t)=x_{0}+\frac{1}{2}\left[x_{\omega} e^{-i \omega t}+x_{\omega}^{*} e^{i \omega t}\right]+\frac{1}{2}\left[x_{2 \omega} e^{-2 i \omega t}+x_{2 \omega}^{*} e^{2 i \omega t}\right]
\]
is a solution of
\[
\frac{d^{2} x}{d t^{2}}+\omega_{0}^{2} x+a x^{2}=\frac{-e}{2 m}\left[E_{\omega} e^{-i \omega t}+E_{\omega}^{*} e^{i \omega t}\right]
\]
that is valid up to second order in the electric-field strength.

2. Show that the expression

x^(2)(t)=x0+(1)/(2)[xω e^-i ω t+xω^* e^i ω t]+(1)/(2)[x2 ω e^-2 i ω t+x2 ω^* e^2 i ω t]

is a solution of

(d^2 x)/(d t^2)+ω0^2 x+a x^2=(-e)/(2 m)[Eω e^-i ω t+Eω^* e^i ω t]

that is valid up to second order in the electric-field strength.

Added by Krystal W.

The Physics of Vibrations and Waves

H. J. Pain 6th Edition

Chapter 13

Instant Answer

Solved by Expert Mukesh Devi

Step 1

The expression is: \[ x^{(2)}(t) = x_{0} + \frac{1}{2}\left[x_{\omega} e^{-i \omega t} + x_{\omega}^{*} e^{i \omega t}\right] + \frac{1}{2}\left[x_{2 \omega} e^{-2 i \omega t} + x_{2 \omega}^{*} e^{2 i \omega t}\right] \] The differential equation is: \[ Show more…

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2. Show that the expression \[ x^{(2)}(t)=x_{0}+\frac{1}{2}\left[x_{\omega} e^{-i \omega t}+x_{\omega}^{*} e^{i \omega t}\right]+\frac{1}{2}\left[x_{2 \omega} e^{-2 i \omega t}+x_{2 \omega}^{*} e^{2 i \omega t}\right] \] is a solution of \[ \frac{d^{2} x}{d t^{2}}+\omega_{0}^{2} x+a x^{2}=\frac{-e}{2 m}\left[E_{\omega} e^{-i \omega t}+E_{\omega}^{*} e^{i \omega t}\right] \] that is valid up to second order in the electric-field strength.