Ein schwingungsfähiges mechanisches System. bestehend aus einer Blattfeder mit der Federkonstanten \(c\) und einer Schwingmasse \(m\), befindet sich fest verankert auf einem reibungsfrei beweglichen Fahrgestell (Bild V-70). Unterliegt das Fahrwerk einer konstanten Beschleunigung \(a\) in der eingezeichneten Richtung. so genüg das Weg-Zeit-Gesetz \(x=x(t)\) des schwingungsfähigen Systems nach dem Newtonschen Grundgesetz der Mechanik der folgenden linearen Differentialgleichung 2. Ordnung mit konstanten Koeffizienten: \(m \ddot{x}=-c x+m a\) oder \(\ddot{x}+\omega_{0}^{2} x=a \quad\left(\omega_{0}^{2}=c / m\right)\) Lôsen Sie diese Gleichung für die Anfangswerte \(x(0)=0, v(0)=\dot{x}(0)=0\) und skizzieren Sie den zeitlichen Verlauf der Schwingung.

Step by step solution

01

Rewrite the Differential Equation

Given the differential equation: \[\ddot{x} + \omega_0^2 x = a\] where \(\omega_0^2 = \frac{c}{m}\), rewrite it with the substitution \(\omega_0^2 = \frac{c}{m}\).

02

Solve the Homogeneous Equation

Solve the homogeneous part of the differential equation: \[\ddot{x} + \omega_0^2 x = 0\] The characteristic equation is: \[r^2 + \omega_0^2 = 0 \Rightarrow r = \pm i\omega_0\] Thus, the general solution to the homogeneous part is: \[x_h(t) = A\cos(\omega_0 t) + B\sin(\omega_0 t)\]

03

Solve the Particular Solution

Since the non-homogeneous term is a constant 'a', assume a particular solution of the form: \[x_p = C\] Substitute \(x_p\) into the non-homogeneous differential equation: \[\ddot{x}_p + \omega_0^2 x_p = a\] \[0 + \omega_0^2 C = a \Rightarrow C = \frac{a}{\omega_0^2}\] Thus, the particular solution is: \[x_p = \frac{a}{\omega_0^2}\]

04

Form the General Solution

Combine the homogeneous and particular solutions to form the general solution: \[x(t) = x_h(t) + x_p(t)\] \[x(t) = A\cos(\omega_0 t) + B\sin(\omega_0 t) + \frac{a}{\omega_0^2}\]

05

Apply Initial Conditions

Use the initial conditions \(x(0) = 0\), \(v(0) = \dot{x}(0) = 0\) to find A and B:\(x(0) = 0 \Rightarrow A + \frac{a}{\omega_0^2} = 0 \Rightarrow A = -\frac{a}{\omega_0^2}\)\(\dot{x}(t) = -A\omega_0\sin(\omega_0 t) + B\omega_0\cos(\omega_0 t)\)\(\dot{x}(0) = 0 \Rightarrow B\omega_0 = 0 \Rightarrow B = 0\)

06

Write the Final Solution

Plug the values of A and B back into the general solution: \[x(t) = -\frac{a}{\omega_0^2}\cos(\omega_0 t) + \frac{a}{\omega_0^2}\]

07

Simplify the Solution and Sketch

Simplify the expression:\[x(t) = \frac{a}{\omega_0^2}(1 - \cos(\omega_0 t))\]Sketch the function which will show a periodic oscillation starting from zero and approaching \(\frac{2a}{\omega_0^2}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Newton's laws of motion

Newton's laws of motion are the backbone of classical mechanics and they describe how physical objects behave under various forces. In this context, we rely on Newton's second law. This states that the force acting on a mass is equal to the mass times its acceleration, or mathematically, \[ F = ma \]. This law helps us establish the differential equation for the oscillating system: \[ m \ddot{x} = -cx + ma \]. By dividing through by the mass \(m\) and rearranging terms, we get a simpler form: \[ \ddot{x} + \omega_0^2 x = a \].

homogeneous differential equation

A homogeneous differential equation refers to an equation where the right-hand side is zero. In our case, we consider the equation \[ \ddot{x} + \omega_0^2 x = 0 \]. Homogeneous equations are essential because solving them gives us general solutions that can be combined with particular solutions of non-homogeneous equations. For our system, solving the homogeneous equation allows us to find that the characteristic equation \[ r^2 + \omega_0^2 = 0 \] has roots \[ r = \pm i \omega_0 \]. This leads to a general solution involving sine and cosine functions: \[ x_h(t) = A\cos(\omega_0 t) + B\sin(\omega_0 t) \].

particular solution

The particular solution of a differential equation addresses the non-homogeneous part. For \[ \ddot{x} + \omega_0^2 x = a \], we assume a solution of the form: \[ x_p = C \]. Substituting this into our equation gives \[ \omega_0^2 C = a \] which simplifies to \[ C = \frac{a}{\omega_0^2} \]. This particular solution adds a constant term to our general solution, reflecting the system’s response to the external acceleration 'a' given in the problem..

initial conditions

Initial conditions specify the state of a system at time \(t = 0\). They are crucial for finding unique values for constants in the solutions of differential equations. In this case, given \[ x(0) = 0 \] and \[ \dot{x}(0) = 0 \], we substitute these into our combined general and particular solution: \[ x(t) = A\cos(\omega_0 t) + B\sin(\omega_0 t) + \frac{a}{\omega_0^2} \] to determine \(A\) and \(B\). This results in \[ A = \frac{a}{\omega_0^2}\] and \[ B = 0 \], making our final solution: \[ x(t) = \frac{a}{\omega_0^2}(1 - \cos(\omega_0 t)) \].

oscillatory motion

Oscillatory motion describes the repetitive back and forth movement of the system. In our scenario, the system oscillates due to the presence of a harmonic (sinusoidal) term in the solution. The solution \[ x(t) = \frac{a}{\omega_0^2}(1 - \cos(\omega_0 t)) \] indicates this motion has an amplitude shifted by a constant term. The solution's form shows that the system starts at zero, oscillates sinusoidally, but approaches a steady state displacement \[ \frac{a}{\omega_0^2} \] due to the continuous acceleration 'a' driving the motion. This results in a periodic but stable movement pattern..