- #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.
$$A$$
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}
$$A$$
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}
Last edited:
