Chapter 14 - Oscillations - Exercises and Problems - Page 406: 72

See a detailed answer below.

First, we need to write the mentioned equations. $\bullet$ Equation 14.54: $$\dfrac{d^2x}{dt^2}+\dfrac{b}{m}\dfrac{dx}{dt}+\dfrac{k}{m}x=0\tag 1$$ $\bullet$ Equation 14.55: $$x_{(t)} = Ae^{-bt/2m} \cos(\omega t + \phi_0)\tag 2$$ $\bullet$ Equation 14.56: $$\omega=\sqrt{\dfrac{k}{m}- \dfrac{b^2}{4m^2}}\tag 3$$ The author asked us to prove that (2) is a solution of (1) if, and only if, the angular frequency is given by (3). Taking the first derivative of $x_{(t)}$ in (2): $$\dfrac{dx}{dt} =\dfrac{d}{dt}\left[ Ae^{-bt/2m} \cos(\omega t + \phi_0)\right]$$ $$\dfrac{dx}{dt} =\dfrac{ -bA}{2m}e^{-bt/2m} \cos(\omega t + \phi_0) -A\omega e^{-bt/2m} \sin(\omega t + \phi_0) $$ $$\dfrac{dx}{dt} =e^{-bt/2m}\left(\dfrac{ -bA}{2m} \cos(\omega t + \phi_0) -A\omega \sin(\omega t + \phi_0) \right)\tag 4$$ Taking the second derivative; $$\dfrac{d^2x}{dt^2} =\dfrac{d}{dt}\left[e^{-bt/2m}\left(\dfrac{ -bA}{2m} \cos(\omega t + \phi_0) -A\omega \sin(\omega t + \phi_0) \right)\right]$$ $$\dfrac{d^2x}{dt^2} =\dfrac{-b}{2m} e^{-bt/2m}\left(\dfrac{ -bA}{2m} \cos(\omega t + \phi_0) -A\omega \sin(\omega t + \phi_0) \right) +\\ e^{-bt/2m}\left(\dfrac{ -(-\omega)bA}{2m} \sin(\omega t + \phi_0) -A\omega^2 \cos(\omega t + \phi_0) \right) $$ $$\dfrac{d^2x}{dt^2} =\dfrac{b^2A}{4m^2} e^{-bt/2m} \cos(\omega t + \phi_0) +\dfrac{ bA\omega }{2m} e^{-bt/2m} \sin(\omega t + \phi_0) +\\ \dfrac{ bA\omega}{2m} e^{-bt/2m} \sin(\omega t + \phi_0) - A\omega^2 e^{-bt/2m} \cos(\omega t + \phi_0) $$ $$\dfrac{d^2x}{dt^2} =e^{-bt/2m}\left[\dfrac{b^2A}{4m^2} \cos(\omega t + \phi_0) - A\omega^2 \cos(\omega t + \phi_0) +\\ \dfrac{ bA\omega }{2m} \sin(\omega t + \phi_0) + \dfrac{ bA\omega}{2m} \sin(\omega t + \phi_0) \right]$$ $$\dfrac{d^2x}{dt^2} =e^{-bt/2m}\left[\cos(\omega t + \phi_0)\left(\dfrac{b^2A}{4m^2} - A\omega^2 \right)+ \\ \sin(\omega t + \phi_0) \left(\dfrac{ bA\omega }{2m} + \dfrac{ bA\omega}{2m} \right) \right]$$ $$\dfrac{d^2x}{dt^2} =e^{-bt/2m}\left[\cos(\omega t + \phi_0)\left(\dfrac{b^2A}{4m^2} - A\omega^2 \right)+ \\ \sin(\omega t + \phi_0) \left(\dfrac{ bA\omega }{ m} \right) \right]\tag 5$$ Plugging (4) and (5) into (1); $$ e^{-bt/2m}\left[\cos(\omega t + \phi_0)\left(\dfrac{b^2A}{4m^2} - A\omega^2 \right)+ \\ \sin(\omega t + \phi_0) \left(\dfrac{ bA\omega }{ m} \right) \right]+\\ \dfrac{b}{m}e^{-bt/2m}\left(\dfrac{ -bA}{2m} \cos(\omega t + \phi_0) -A\omega \sin(\omega t + \phi_0) \right)+\\ \dfrac{k}{m}x=0 $$ $$ e^{-bt/2m} \left[\cos(\omega t + \phi_0)\left(\dfrac{b^2A}{4m^2} - A\omega^2 \right)+ \sin(\omega t + \phi_0) \left(\dfrac{ bA\omega }{ m} \right) \\ +\dfrac{ -b^2A}{2m^2} \cos(\omega t + \phi_0) - \dfrac{bA\omega}{m} \sin(\omega t + \phi_0) \right] +\\ \dfrac{k}{m}x=0 $$ The term of $ \dfrac{bA\omega}{m} \sin(\omega t + \phi_0) $ cancel. $$ e^{-bt/2m}\cos(\omega t + \phi_0) \left[ \dfrac{b^2A}{4m^2} - A\omega^2 +\dfrac{ -b^2A}{2m^2} \right] + \dfrac{k}{m}x=0 $$ Plugging $x$ from (2); $$ e^{-bt/2m}\cos(\omega t + \phi_0) \left[ \dfrac{b^2A}{4m^2} - A\omega^2 +\dfrac{ -b^2A}{2m^2} \right] + \dfrac{k}{m} Ae^{-bt/2m} \cos(\omega t + \phi_0)=0 $$ $$A e^{-bt/2m}\cos(\omega t + \phi_0) \left[ \dfrac{b^2 }{4m^2} - \omega^2 +\dfrac{ -b^2 }{2m^2} + \dfrac{k }{m} \right] =0 $$ Dividing both sides by $A e^{-bt/2m}\cos(\omega t + \phi_0)$, so $$ \dfrac{b^2 }{4m^2} - \omega^2 +\dfrac{ -b^2 }{2m^2} + \dfrac{k }{m} =0 $$ Solving for $\omega^2$; $$ \omega^2 = \dfrac{b^2 -2b^2 }{4m^2} + \dfrac{k }{m} = \dfrac{-b^2 }{4m^2} + \dfrac{k }{m} $$ Therefore, $$ \omega =\sqrt{\dfrac{k }{m} - \dfrac{ b^2 }{4m^2} } $$ which is equation (3) as the author asked.