Physik für Geodäsie 511.018 / Physik M 513.805
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$ \large \frac{dx}{dt}=$
$ \large a_x=\frac{F_x}{m}=\frac{dv_x}{dt}=$
$x(t_0)=$
$\Delta t=$
$v_x(t_0)=$
$N_{steps}$
$t_0=$
Plot:
= , =
An object moving in one-dimension can be described in terms of its position $x$ and and its velocity $v_x$. If the force on the object is known, then the motion can be described by two first order differential equations,
$\large \frac{dx}{dt}=v_x$ and $\large \frac{dv_x}{dt}=a_x=F_x(x,v_x,t)/m.$
Here $F$ is the force, $m$ is the mass, and $t$ is the time. These equations can be solved numerically. If the initial conditions for $x(t_0)$ and $v_x(t_0)$ are known, a good estimate for $x$ and $v_x$ a short time $\Delta t$ later is,
$\large x(t_0+\Delta t) \approx \frac{dv_x}{dt}|_{t_0}\Delta t$ and $\large v_x(t_0+\Delta t) \approx F(x(t_0),v_x(t_0),t_0)\Delta t/m.$
Once an estimate for the position and the velocity of the object at time $t_0 + \Delta t$ is calculated, they can be used to estimate the position and the velocity at time $t_0 + 2\Delta t$. The solution is produced step-by-step as the time is increased by a small increment $\Delta t$.
The form below can be used to numerically integrate these equations for a total number of $N_{steps}$ steps using a step size of $\Delta t$. The will display the a help text that specifies the variables and functions that can be put in the blue text box.
The numerical integration will become unstable if the time step is too long. The time step should be a factor of 10 to 100 times smaller than any characteristic time of the motion that is being calculated. If the orbit of the earth around the sun is being calculated, then a time step of 3 days (about 4E6 seconds) is reasonable. If the motion of an electron moving past a ion is being calculated, a time step of 1 ps is reasonable. If the routine becomes unstable, nothing will be plotted. In this case, try a shorter time step.
There are many routines available to perform numerical integration. The procedure outlined above is called the Euler method. However this page really uses the more accurate fourth order Runge-Kutta method with a fixed step size. Some numerical integration routines calculate the optimum time step automatically.
Eine Kugel wird vertikal nach oben mit einer Anfangsgeschwindigkeit $v_0=10$ m/s geworden. Wird Reibung vernachlässigt, wirkt nur die Gravitationskraft als einzige Kraft auf die Kugel. Die Beschleungigung, die die Kugel erfährt, ist die Erdbeschleunigung an der Erdoberfläche, -9.81 m/s². Die Beschleungigung hängt nicht von der Masse der Kugel ab. Die Bewegung folgt einer Linie, die wir als $x$-Achse vereinbaren. Die Gleichungen werden unten in die APP zur Lösung numerischer Differentialgleichungen 2. Ordnung geladen.
A ball is thrown vertically upward with an initial velocity of $v_0=10$ m/s. There is a velocity dependent drag force directed in the opposite direction to the velocity. The total force on the ball is gravity plus the drag force $F=-mg-bv_x$, where $F$ is the force, $m$ is the mass of the ball, $g=9.81$ m/s² is the acceleration of gravity at the earth's surface, $b$ is the drag force constant, and $v_x$ is the velocity. The acceleration of the ball is $a_x=-g-bv_x/m$. The motion is in a line which we can take to be the $x$-axis. The equations are loaded into the numerical second order differential equation solver below.
$m=$ 1 [kg]
$b=$ 0.4 [kg/s]
For long times, the ball falls with a constant terminal velocity $v_{\text{terminal}}=-mg/b=$ -24.5 m/s.
A mass $m$ is attached to a linear spring with a spring constant $k$. The spring is stretched 2 cm from its equilibrium position and the mass is released from rest. A drag force acts on the mass that is in the opposite direction as the velocity $F_{\text{drag}}=-bv_x$ where $b$ is the drag force constant. The acceleration of the mass is $a_x=-kx/m-bv_x/m$. The motion is in a line which we can take to be the $x$-axis. The equations are loaded into the numerical second order differential equation solver below.
$b=$ 0.2 [kg/s]
$k=$ 0.9 [N/m]
The period of the oscillations is $T=2\pi/\sqrt{\frac{k}{m}-\frac{b^2}{4m}}=$ 6.66 s.
When the total force is linearly proportional to the displacement from the equilibrium position, the motion will be harmonic.
$\vec{F}= -kx \,\hat{x}$.
Here $k$ is the effective spring constant.
For a linear spring, the force is proportional to the extension of the spring, $F=-kx$. For a hard spring, the force grows faster than linearly as the spring is extended. An example of a hard spring force is $F=-kx|x|$. For a soft spring, the force grows slower than linearly as the spring is extended. An example of a soft spring force is $F=-k\sin(x)$. If friction is neglected, the mass oscillates around the equilibrium position of the spring. For nonlinear springs, the oscillation frequency depends on the amplitude of the oscillations. For hard springs, the frequency increases with amplitude and for soft springs, the frequency decreases with amplitude.
Hard spring $F=-kx|x|$ Linear spring $F=-kx$ Soft spring $F=-k\sin(x)$
$m=$ kg $k=$ N/m
A mass $m$ is attached to a linear spring with a spring constant $k$. A drag force acts on the mass that is in the opposite direction as the velocity $F_{\text{drag}}=-bv_x$ where $b$ is the drag force constant. The system is driven by a periodic drive force $F_0\cos(\Omega t)$. The differential equation that describes the motion is,
$\large m\frac{d^2x}{dt^2}+b\frac{dx}{dt}+kx=F_0\cos(\Omega t).$
$F_0=$ 0.1 [N]
$\Omega=$ 1 [rad/s]
The resonance frequency is $\omega=\sqrt{k/m-b^2/4m^2}=$ 0.943 rad/s. The maximum response will be when the drive frequency equals the resonance frequency. The amplitude of the steady-state oscillations are $F_0/\sqrt{(k-m\Omega^2)^2+\Omega^2b^2}=$ 0.447 m.
For more details see: Resonance of a damped driven harmonic oscillator.
The differential equation that describes the motion of a driven pendulum is, [Fitzpatrick 2006],
$\large \frac{d^2\theta}{dt^2}+\frac{1}{q}\frac{d\theta}{dt}+\sin(\theta)=A\cos(\Omega t),$
where $q$ describes the damping, $A$ measures the torque that is used to drive the pendulum at frequency $\Omega$ and $\theta$ is the angle measured from vertical. At $\theta=0$ the pendulum hangs down and at $\theta=\pi$ the pendulum stands up. This equation is known to exhibit chaotic solutions.
More details can be found at Chaotic solutions to the driven pendulum.
A relaxation oscillator is a system that produces a periodic signal. Heart beats and the squeaking of fingernails on a blackboard are examples of relaxation oscillations. One relaxation oscillator that can be built with electronic components is called the Van der Pol oscillator. This system is described by the differential equation,
$\frac{d^2x}{dt^2}-\mu (1-x^2)\frac{dx}{dt}+x=0.$
For small values of $x$, the damping term $ \mu (1-x^2)\frac{dx}{dt}$ is negative and the signal grows until $x>1$ when the damping becomes positive and the signal decays again. The frequency of the oscillations can be controlled by changing $\mu$. In an electronic circuit this can be achieved by using a variable resistor.
$\mu=$ 1
A parametric oscillator is a system that induces oscillations by periodically modulating some parameter of the system. A typical example is a child on a swing. For small amplitude oscillations of a swing, the restoring force is $-\frac{mg}{l}x$ where $m$ is the mass, $l$ is the length of the swing, and $g=9.81$ m/s² is the acceleration of gravity at the earth's surface. If the length of the swing is modulated periodically, this system is described by the differential equation,
$m\frac{d^2x}{dt^2}+b \frac{dx}{dt}+\frac{mg}{l(1-A\cos(\omega t))}x=0.$
Large amplitude oscillations are induced when the modulation is about twice the resonance frequency. For most parameters, no parametric amplification is observed.
$l=$ 0.5 [N/m]
$A=$ 0.4 [N]
$\omega=$ 8 [rad/s]
The resonance frequency is $\omega_0=\sqrt{g/l-b^2/4m^2}=$ 4.43 rad/s.