MHB Non-dimensional differential equation 1

ra_forever8
Messages
106
Reaction score
0
A body of constant mass is thrown vertically upwards from the ground. It can be shown that the appropriate non-dimensional differential equation for the height $x(t;u)$, reached at time $t\geq0$ is given by
\begin{equation} \frac{d^2x}{dt^2} = -1-\mu (\frac{dx}{dt})
\end{equation}
with corresponding initial conditions $x(0)=0, \frac{dx}{dt}(0) =1$, and where $0<\mu<<1.$
Deduce that the (non-dimensional) height at the highest point (where $\frac{dx}{dt} =0$) is given by
\begin{equation} h(\mu)= \frac{1}{\mu}- \frac{1}{\mu^2} log_e(1+\mu) \end{equation}

=>
It really hard for me to start

I was thinking do integration twice by doing the separation of variable:
\begin{equation} \frac{d^2x}{dt^2} = -1-\mu (\frac{dx}{dt})
\end{equation}
I got the general solution of \begin{equation}x(t)= \frac{log(t\mu +1) -t \mu}{\mu^2}\end{equation}

after that I do not know how to get the answer.
Please help me.
 
Mathematics news on Phys.org
Re: Non-dimensional differential equation

grandy said:
A body of constant mass is thrown vertically upwards from the ground. It can be shown that the appropriate non-dimensional differential equation for the height $x(t;u)$, reached at time $t\geq0$ is given by
\begin{equation} \frac{d^2x}{dt^2} = -1-\mu (\frac{dx}{dt})
\end{equation}
with corresponding initial conditions $x(0)=0, \frac{dx}{dt}(0) =1$, and where $0<\mu<<1.$
Deduce that the (non-dimensional) height at the highest point (where $\frac{dx}{dt} =0$) is given by
\begin{equation} h(\mu)= \frac{1}{\mu}- \frac{1}{\mu^2} log_e(1+\mu) \end{equation}

=>
It really hard for me to start

I was thinking do integration twice by doing the separation of variable:
\begin{equation} \frac{d^2x}{dt^2} = -1-\mu (\frac{dx}{dt})
\end{equation}
I got the general solution of \begin{equation}x(t)= \frac{log(t\mu +1) -t \mu}{\mu^2}\end{equation}

after that I do not know how to get the answer.
Please help me.

Your method is the right approach.

Let me show you the first step to find $v$.

Let $v=\frac {dx}{dt}$ and let $\dot v = \frac {d^2x}{dt^2}$.
Then:
\begin{array}{}
\dot v &=& -1 - \mu v \\
\frac{\dot v}{1 + \mu v} &=& -1 \\
\frac 1 \mu \ln(1+\mu v) &=& -t + C \\
v &=& C' e^{-\mu t} - \frac 1 \mu \\
\end{array}

Applying the boundary condition $v(0)=1$ yields:
$$v = \frac 1 \mu((\mu + 1)e^{-\mu t} - 1)$$

From here you can find the time $t$ at which $v=0$.
And you can also integrate again to find $x$.
 
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
I'm interested to know whether the equation $$1 = 2 - \frac{1}{2 - \frac{1}{2 - \cdots}}$$ is true or not. It can be shown easily that if the continued fraction converges, it cannot converge to anything else than 1. It seems that if the continued fraction converges, the convergence is very slow. The apparent slowness of the convergence makes it difficult to estimate the presence of true convergence numerically. At the moment I don't know whether this converges or not.
Back
Top