Prove that the expression for $x\left(t\right)$in Equation$15.55$ is a solution to the equation of motion for a damped oscillator, Equation $15.54$, if and only if the angular frequency $\omega$is given by the expression in Equation $15.56$.

Damped Oscillations: During oscillations, the mechanical energy E in a real oscillating system diminishes because external forces, such as drag, limit the oscillations and convert mechanical energy to thermal energy. After that, the true oscillator and its motion are said to be in bed. If the damping force is given by $f=-bv$, where $v$is the oscillator's velocity and $b$is the damping constant, the oscillator's displacement is given by

$x\left(t\right)=A{e}^{-bt/2m}\cos (\omega t+\varphi )$

where $\omega$denotes the damped oscillator's angular frequency and k,

$\omega =\sqrt{\frac{k}{m}-\frac{b^2}{4m^2}}$

The angular frequency of an undamped oscillator $(b=0)$ is represented by$\sqrt{k/m}$. The solution to Equation$\left(1\right)$is the general equation of motion of a damped oscillator:

$\frac{d^{2}x}{d{t}^2}+\frac{b}{m}\frac{dx}{dt}+\frac{k}{m}x=0$

In Equation for$15.55$, the expression for$x\left(t\right)$ is:

$x\left(t\right)=A{e}^{-bt/2m}\cos (\omega t+\varphi )$

The equation for damped oscillator's motion is:

$\frac{d^{2}x}{d{t}^2}+\frac{b}{m}\frac{dx}{dt}+\frac{k}{m}x=0$

A damped oscillator's angular frequency is:

$\omega =\sqrt{\frac{k}{m}-\frac{b^2}{4m^2}}$

If the angular frequency is given by equation $\left(2\right)$, we must show that equation $\left(1\right)$is a solution to equation$\left(3\right)$

With regard to time, the first derivative of$x\left(t\right)$is:

$\frac{dx\left(t\right)}{dt}=\frac{d}{dt}\left[A{e}^{-bt/2m}\cos (\omega t+\varphi )\right]$

$=A\left[-\frac{b}{2m}{e}^{-bt/2m}\cos (\omega t+\varphi )-\omega e^{-b/2m}\sin (\omega t+\varphi )\right]$

$=A{e}^{-bt/2m}\left[-\frac{b}{2m}\cos (\omega t+\varphi )-\omega \sin (\omega t+\varphi )\right]$

where $x\left(t\right)$is given in Equation $\left(1\right)$

With regard to time, the second derivative of $x\left(t\right)$is:

$\frac{d^{2}x\left(t\right)}{d{t}^2}=\frac{d}{dt}\left[\frac{dx\left(t\right)}{dt}\right]$

Substitute for $dx\left(t\right)/dt$from Equation (4):

$\frac{d^{2}x\left(t\right)}{d{t}^2}=\frac{d}{dt}\left\{A\left[-\frac{b}{2m}{e}^{-bt/2m}\cos (\omega t+\varphi )-\omega e^{-bt/2m}\sin (\omega t+\varphi )\right]\right\}$

$=A\left\{\left[\frac{b^2}{4m^2}{e}^{-bt/2m}\cos (\omega t+\varphi )+\frac{\omega b}{2m}{e}^{-bt/2m}\sin (\omega t+\varphi )\right]\right.$

$\left.-\left[-\frac{b\omega }{2m}{e}^{-bt/2m}\sin (\omega t+\varphi )+{\omega }^2{e}^{-bt/2m}\cos (\omega t+\varphi )\right]\right\}$

$=A\left[\left(\frac{b^2}{4m^2}+{\omega }^2\right)e^{-bt/2m}\cos (\omega t+\varphi )+\frac{b\omega }{m}{e}^{-bt/2m}\sin (\omega t+\varphi )\right]$

$=A{e}^{-bt/2m}\left[\left(\frac{b^2}{4m^2}+{\omega }^2\right)\cos (\omega t+\varphi )+\frac{b\omega }{m}\sin (\omega t+\varphi )\right]$

Substitute for x, dx/dt, and $d^{2}x/d{t}^2$from Equations $\left(1\right)$through $\left(5\right)$for$x,dx/dt,d^{2}x/d{t}^2$in equation$\left(3\right)$

$A{e}^{-bt/2m}\left[\left(\frac{b^2}{4m^2}+{\omega }^2\right)\cos (\omega t+\varphi )+\frac{b\omega }{m}\sin (\omega t+\varphi )\right]$$+\frac{b}{m}A{e}^{-bt/2m}\left[-\frac{b}{2m}\cos (\omega t+\varphi )-\omega \sin (\omega t+\varphi )\right]$$+\frac{k}{m}\left[A{e}^{-bt/2m}\cos (\omega t+\varphi )\right]=0$

Divide all sides by $A{e}^{-bt/2m}\text{:}$

$\left(\frac{b^2}{4m^2}+{\omega }^2\right)\cos (\omega t+\varphi )+\frac{b\omega }{m}\sin (\omega t+\varphi )$$+\frac{b}{m}\left[-\frac{b}{2m}\cos (\omega t+\varphi )-\omega \sin (\omega t+\varphi )\right]$$+\frac{k}{m}\left[\cos \right(\omega t+\varphi \left)\right]=0$

$\left(\frac{b^2}{4m^2}+{\omega }^2\right)\cos (\omega t+\varphi )+\frac{b\omega }{m}\sin (\omega t+\varphi )$$-\frac{b^2}{2m^2}\cos (\omega t+\varphi )-\frac{b\omega }{m}\sin (\omega t+\varphi )$$+\frac{k}{m}\left[\cos \right(\omega t+\varphi \left)\right]=0$

$\left(\frac{b^2}{4m^2}-\frac{b^2}{2m^2}+{\omega }^2+\frac{k}{m}\right)\cos (\omega t+\varphi )=0$

${\omega }^2-\frac{b^2}{4m^2}+\frac{k}{m}=0$

solving for $\omega$, we obtain:

$\omega =\sqrt{\frac{b^2}{4m^2}-\frac{k}{m}}$

Hence, we have shown that Equation (1) is a solution to Equation (3) if the angular frequency is given by Equation (2).

$\frac{d^{2}x}{d{t}^2}+\frac{b}{m}\frac{dx}{dt}+\frac{k}{m}x=0$

$A{e}^{-bt/2m}\left[\left(\frac{b^2}{4m^2}+{\omega }^2\right)\cos (\omega t+\varphi )+\frac{b\omega }{m}\sin (\omega t+\varphi )\right]$$+\frac{b}{m}A{e}^{-bt/2m}\left[-\frac{b}{2m}\cos (\omega t+\varphi )-\omega \sin (\omega t+\varphi )\right]$$+\frac{k}{m}\left[A{e}^{-bt/2m}\cos (\omega t+\varphi )\right]=0$

$\left(\frac{b^2}{4m^2}+{\omega }^2\right)\cos (\omega t+\varphi )+\frac{b\omega }{m}\sin (\omega t+\varphi )$$-\frac{b^2}{2m^2}\cos (\omega t+\varphi )-\frac{b\omega }{m}\sin (\omega t+\varphi )$$+\frac{k}{m}\left[\cos \right(\omega t+\varphi \left)\right]=0$

$\left(\frac{b^2}{4m^2}-\frac{b^2}{2m^2}+{\omega }^2+\frac{k}{m}\right)\cos (\omega t+\varphi )=0$

${\omega }^2-\frac{b^2}{4m^2}+\frac{k}{m}=0$

solving for $\omega$, we obtain

$\omega =\sqrt{\frac{b^2}{4m^2}-\frac{k}{m}}$