Ein Pendel unterliege der periodischen Beschleunigung \(a(t)=-5 \cdot \cos t .\) Bestimmen Sie die Geschwindigkeits- Zeit-Funktion \(v=v(t)\) und die Weg-Zeit- Funktion \(s=s(t)\) fur die \(A\) nfangswerte \(s(0)=5, v(0)=0\).

Short Answer

Expert verified
The velocity function is \( v(t) = -5 \sin t \). The position function is \( s(t) = 5 \cos t \).

Step by step solution

01

- Integrate the acceleration to find the velocity function

The acceleration given is a(t) = -5 \cos t. To find the velocity function v(t), integrate the acceleration with respect to time t: \( v(t) = \int a(t) \mathrm{d}t = \int -5 \cos t \mathrm{d}t = -5 \int \cos t \mathrm{d}t = -5 \sin t + C \). Here, C is the integration constant.
02

- Determine the constant of integration for the velocity function

Use the initial condition v(0) = 0 to find C. Substitute t = 0 into the velocity function: \(0 = -5 \sin(0) + C = 0 + C \). Therefore, C = 0. The velocity function is \( v(t) = -5 \sin t \).
03

- Integrate the velocity to find the position function

Now, find the position function s(t) by integrating the velocity function: \( s(t) = \int v(t) \mathrm{d}t = \int -5 \sin t \mathrm{d}t = 5 \cos t + D \). Here, D is the integration constant.
04

- Determine the constant of integration for the position function

Use the initial condition s(0) = 5 to find D. Substitute t = 0 into the position function: \(5 = 5 \cos(0) + D = 5 + D \). Therefore, D = 0. The position function is \( s(t) = 5 \cos t \).

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Key Concepts

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

Pendel
A pendulum (Pendel in German) is a weight suspended from a pivot so that it can swing freely. When displaced, it undergoes periodic motion. The motion of a pendulum is a classic example in physics to study harmonic motion.

For small angles, the motion can be approximated as simple harmonic motion, where the restoring force is proportional to the displacement. This makes it useful for demonstrating principles of acceleration and integration.
In this exercise, we analyze the pendulum's acceleration and find its velocity and position as functions of time.
Beschleunigung
The acceleration (Beschleunigung in German) of the pendulum is given by the equation: \( a(t) = -5 \, \cos{t} \).

This periodic function describes how the pendulum's acceleration changes over time. The negative sign indicates that the acceleration is opposite to the direction of cosine, which aligns with the restoring force of a pendulum.
By understanding this relationship, we move on to find the velocity and position functions through integration.
Integration
Integration is the process of finding the antiderivative or the area under a curve. We start with the given acceleration function and integrate it with respect to time to find the velocity function. Given:\[ v(t) = \int a(t) \mathrm{d}t = \int -5 \, \cos{t} \mathrm{d}t = -5 \, \sin{t} + C \]
Where \( C \) is the constant of integration.

Using the initial condition \( v(0) = 0 \), the constant \( C \) is determined to be 0.
Thus, the velocity function is: \( v(t) = -5 \, \sin{t} \).
Next, we integrate the velocity function to find the position function:
\[ s(t) = \int v(t) \mathrm{d}t = \int -5 \, \sin{t} \mathrm{d}t = 5 \, \cos{t} + D \]
Using the initial condition \( s(0) = 5 \), the constant \( D \) is determined to be 0, thus the position function is: \( s(t) = 5 \, \cos{t} \).
Integration has allowed us to derive these important functions governing the motion of the pendulum.
Anfangswerte
Anfangswerte, or initial values, are crucial for solving differential equations, as they help determine the unknown constants after integration.

In this exercise, the given initial values are:
  • \( s(0) = 5 \)
  • \( v(0) = 0 \)
These values are substituted back into the integrated functions to find the constants of integration \( C \) and \( D \).
For the velocity function, \( v(0) = 0 \) helps us determine that \( C = 0 \).
For the position function, \( s(0) = 5 \) helps us determine that \( D = 0 \).
This clear verification process ensures that our integrated solutions accurately describe the motion of the pendulum from the start.

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Most popular questions from this chapter

Ein Massenpunkt bewege sich in der \(x, y\)-Ebene so, daB seine kartesischen Koordinaten \(x\) und \(y\) den folgenden Differentialgleichungen genügen: $$ \vec{x}=\dot{y}, \quad \ddot{y}=-\dot{x} $$ Bestimmen Sie die Bahnkurve für die Anfangswerte $$ x(0)=y(0)=0, \quad \dot{x}(0)=0 . \quad \dot{y}(0)=1 $$ Hinweis: Das Differentialgleichungssystem l?Bt sich mit Hilfe der Substitutionen \(u=\dot{x}\) und \(v=\dot{y}\) auf ein System linearer Differentialgleichungen \(t\). Orinung zurückf?hren. L?sen Sie zunächst dieses System. Durch Ricksubstitution und anschlieBende Integration erhalten Sie dann die gesuchten zeitabhängigen Koordinaten \(x=x(t)\) und \(y=y(t)\) des Massenpunktes.

Zeigen Sie: Die Funktionen $$ y_{1}(x)=\mathrm{e}^{2 x} \quad \text { und } \quad y_{2}(x)=x \cdot \mathrm{e}^{2 x} $$ bilden eine Fundamentalbasis der homogenen Differentialgleichung 2 . Ordnung \(y^{\prime \prime}-4 y^{\prime}+4 y=0\)

Lösen Sie die folgenden Anfangswertprobleme: a) \(y^{\prime \prime}+4 y^{\prime}+5 y=0, \quad y(0)=\pi, \quad y^{\prime}(0)=0\) b) \(y^{\prime \prime}+20 y^{\prime}+64 y=0, \quad y(0)=0, \quad y^{\prime}(0)=2\) c) \(4 \ddot{x}-4 \dot{x}+x=0, \quad x(0)=5, \quad \dot{x}(0)=-1\)

L?sen Sie die folgenden Differentialgleichungen 1. Ordnung durch Trennung der Variablen: a) \(x^{2} y^{\prime}=y^{2}\) b) \(y^{\prime}\left(1+x^{2}\right)=x y\) c) \(\quad y^{\prime}=(1-y)^{2}\) d) \(y^{\prime} \cdot \sin y=-x\)

Ein Stromkreis mit einem zeitabhängigen ohmschen Widerstand werde durch die Differentialgleichung 1. Ordnung $$ \frac{d i}{d t}+(2 \cdot \sin t) \cdot i=\sin (2 t) \quad(t \geqslant 0) $$ beschrieben. Ermitteln Sie den zeitlichen Verlauf der Stromstärke \(i\) durch Variation der Konstanten für den Anfangswert \(l(0)=0\).

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