- #1

NODARman

- 57

- 13

- Homework Statement
- .

- Relevant Equations
- .

Hi, I had those exercises and want to know if they're correct. Also, feedback/tips would be great from you, professionals.

$$A$$

\begin{equation}

m \ddot{x}+\alpha \dot{x}=-\kappa x

\end{equation}

but with

\begin{equation}

\alpha^2=4 m \kappa

\end{equation}

and after inserting, show that the general solution will be like this:

\begin{equation}

x(t)=\mathrm{e}^{-\alpha t / 2 m}[\mathcal{A} t+\mathcal{B}]

\end{equation}

Express A and B constants with the initial coordinates and velocity and analyze.

&m\ddot{x} + \dot{x}\sqrt{4mk} +kx=0\\

&x(t)=e^{-\frac{\alpha t}{2m}}[\mathcal {A}t+\mathcal{B}]\\

&\\

&\dot{x}(t)=-\frac{\alpha}{2m} e^{-\frac{\alpha t}{2m}} [\mathcal {A}t+\mathcal{B}]+e^{-\frac{\alpha t}{2m}}\mathcal{A}\\

&\ddot{x}(t)=-\frac{4\mathcal{A}\alpha m-\mathcal{A}{\alpha^{2}}t-\mathcal{B}{\alpha^{2}}}{4m^2}e^{-\frac{\alpha t}{2m}}\\

&\\

&x(0)=\mathcal{B}\equiv x_0 \\

&\dot{x}(0)=-\frac{\alpha}{2m} \mathcal{B}+\mathcal{A} = -\frac{\alpha}{2m} x_0 +\mathcal{A} \\

&\\

&\mathcal{A} = \dot{x}_0 + \frac{\alpha}{2m} x_0\\

&\\

&x(t)=e^{-\frac{\alpha t}{2m}}[\dot{x}_0 + \frac{\alpha}{2m} x_0 t+x_0] = e^{-\frac{\alpha t}{2m}}\left[\dot{x}_0 + \left(\frac{\alpha}{2m} t+1\right)x_0\right]\\

\end{aligned}

$$A$$

**1.**Let's consider the oscillator with a friction parameter...\begin{equation}

m \ddot{x}+\alpha \dot{x}=-\kappa x

\end{equation}

but with

\begin{equation}

\alpha^2=4 m \kappa

\end{equation}

and after inserting, show that the general solution will be like this:

\begin{equation}

x(t)=\mathrm{e}^{-\alpha t / 2 m}[\mathcal{A} t+\mathcal{B}]

\end{equation}

Express A and B constants with the initial coordinates and velocity and analyze.

**My solution:**\begin{aligned}&m\ddot{x} + \dot{x}\sqrt{4mk} +kx=0\\

&x(t)=e^{-\frac{\alpha t}{2m}}[\mathcal {A}t+\mathcal{B}]\\

&\\

&\dot{x}(t)=-\frac{\alpha}{2m} e^{-\frac{\alpha t}{2m}} [\mathcal {A}t+\mathcal{B}]+e^{-\frac{\alpha t}{2m}}\mathcal{A}\\

&\ddot{x}(t)=-\frac{4\mathcal{A}\alpha m-\mathcal{A}{\alpha^{2}}t-\mathcal{B}{\alpha^{2}}}{4m^2}e^{-\frac{\alpha t}{2m}}\\

&\\

&x(0)=\mathcal{B}\equiv x_0 \\

&\dot{x}(0)=-\frac{\alpha}{2m} \mathcal{B}+\mathcal{A} = -\frac{\alpha}{2m} x_0 +\mathcal{A} \\

&\\

&\mathcal{A} = \dot{x}_0 + \frac{\alpha}{2m} x_0\\

&\\

&x(t)=e^{-\frac{\alpha t}{2m}}[\dot{x}_0 + \frac{\alpha}{2m} x_0 t+x_0] = e^{-\frac{\alpha t}{2m}}\left[\dot{x}_0 + \left(\frac{\alpha}{2m} t+1\right)x_0\right]\\

\end{aligned}

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