MHB Non-dimensional differential equation 1

AI Thread Summary
A non-dimensional differential equation describes the height of a body thrown vertically upwards, given by d²x/dt² = -1 - μ(dx/dt) with initial conditions x(0)=0 and dx/dt(0)=1, where 0<μ<<1. The maximum height h(μ) is derived as h(μ) = 1/μ - (1/μ²) log_e(1+μ). The discussion involves solving this equation through integration, with a suggestion to use separation of variables. The first step involves substituting v = dx/dt and finding v as a function of time, leading to the expression v = (1/μ)((μ + 1)e^(-μt) - 1). The next steps include determining when v=0 to find the time at which the maximum height is reached.
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 $x$.
 
Suppose ,instead of the usual x,y coordinate system with an I basis vector along the x -axis and a corresponding j basis vector along the y-axis we instead have a different pair of basis vectors ,call them e and f along their respective axes. I have seen that this is an important subject in maths My question is what physical applications does such a model apply to? I am asking here because I have devoted quite a lot of time in the past to understanding convectors and the dual...
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...
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...
Back
Top