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Chapter 3: Problem 6

A mass of \(5 \mathrm{kg}\) stretches a spring \(10 \mathrm{cm} .\) The mass is acted on by an external force of \(10 \mathrm{sin}(t / 2) \mathrm{N}\) (newtons) and moves in a medium that imparts a viscous force of \(2 \mathrm{N}\) when the speed of the mass is \(4 \mathrm{cm} / \mathrm{sec} .\) If the mass is set in motion from its equilibrium position with an initial velocity of \(3 \mathrm{cm} / \mathrm{sec}\), formulate the initial value problem describing the motion of the mass.

Short Answer

Expert verified
Additionally, the mass is set in motion from its equilibrium position with an initial velocity of 3 cm/s.

Step by step solution

01

Determine the spring constant (k)

To find the spring constant (k), we can use the relationship between the spring force and the mass's displacement: \(F = -kx\). In this problem, a mass of \(5 kg\) stretches the spring \(10 cm\) (\(10 cm = 0.1 m\)), which implies the force due to the spring is equal to the gravitational force on the 5 kg mass: \(F = mg\). Solving for k, we have $$k = \frac{mg}{x} = \frac{(5 kg)(9.81 m/s^2)}{0.1 m} \approx 490.50 N/m.$$
02

Determine the damping coefficient (c)

We know that the viscous force is \(2N\) when the speed of the mass is \(4 cm/s\) (\(4 cm/s = 0.04 m/s\)). The viscous force can also be expressed as \(-c\frac{dx}{dt}\), where c is the damping coefficient, and \(\frac{dx}{dt}\) is the velocity. Solving for c, we have: $$c = \frac{F_{viscous}}{\frac{dx}{dt}} = \frac{2 N}{0.04 m/s} = 50 Ns/m.$$
03

Formulate the equation of motion

Using Newton's second law of motion (F=ma), we can write the equation of motion for the mass considering the spring, external, and viscous forces acting on it: $$m\frac{d^2x}{dt^2} = -kx - c\frac{dx}{dt} + F_{external}.$$ Substitute the values of mass (m), spring constant (k), and damping coefficient (c) into the equation: $$5\frac{d^2x}{dt^2} = -490.50x - 50\frac{dx}{dt} + 10\sin(\frac{t}{2}).$$
04

Convert given initial conditions to initial value problem

We are given that the mass is set in motion from its equilibrium position with an initial velocity of \(3 cm/s\) (\(3 cm/s = 0.03 m/s\)). At the equilibrium position, the displacement x is 0, and the initial velocity is \(\frac{dx}{dt}=0.03 m/s\). So, we have the following initial conditions: $$x(0) = 0, \quad \frac{dx}{dt}(0) = 0.03 m/s.$$ Now, we can write the initial value problem for the motion of the mass: $$\begin{cases} 5\frac{d^2x}{dt^2} = -490.50x - 50\frac{dx}{dt} + 10\sin(\frac{t}{2}) \\ x(0) = 0 \\ \frac{dx}{dt}(0) = 0.03 m/s \end{cases}.$$

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Key Concepts

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

Initial Value Problem
An Initial Value Problem (IVP) in the context of differential equations is a problem where we are given a differential equation and the condition of the solution is specified at a particular time. For our spring-mass-damper system, we are interested in finding the position and velocity of the mass over time. To do this, we first establish the specific conditions of the system at the starting point.

In our problem, the mass is set in motion from its equilibrium position with an initial velocity of 3 cm/s. At time zero, these are specified as the initial conditions: displacement, \(x(0) = 0\), and velocity, \(\frac{dx}{dt}(0) = 0.03\) m/s.

The differential equation is then solved with these initial conditions to describe how the system evolves over time. This IVP setup allows us to track the motion from a known starting point, which is essential for solving many real-world dynamic systems.
Second Order Linear Differential Equation
A second order linear differential equation is a type of differential equation that involves the second derivative of a function. These are critical in modeling systems with inertial components, such as our spring-mass-damper system.

The general form used in our problem is: \[ m\frac{d^2x}{dt^2} + c\frac{dx}{dt} + kx = F_{external}. \]This equation includes the mass \(m\), damping coefficient \(c\), spring constant \(k\), function of displacement \(x\), and an external force. The terms \(m\frac{d^2x}{dt^2}\), \(c\frac{dx}{dt}\), and \(kx\) represent inertia, damping, and spring forces, respectively.

Solving this equation helps to understand the response of the system to external influences, including periodic forces.
Spring-Mass-Damper System
A spring-mass-damper system is a classic model used to study mechanical vibrations and dynamics. It consists of three key components: a mass that can move, a spring that exerts a force when compressed or stretched, and a damper that dissipates energy.

In our problem, the mass (5 kg) is connected to a spring that has a stiffness or spring constant (k), which was calculated to be \(490.50 \text{ N/m}\). This spring provides a restoring force that counteracts the displacement of the mass.
  • The mass offers inertia, opposing acceleration and deceleration.
  • The spring stores and releases energy, offsetting displacement.
  • The damper introduces resistance, dampening oscillations due to a viscous damping coefficient (c) of \(50 \text{ Ns/m}\).
Working together, these components define how the system reacts when subjected to forces, such as gravity or external influences like the sinusoidal force in this exercise.
Damping Coefficient
The damping coefficient is a critical parameter in the spring-mass-damper system, indicating how much resistive force is applied by the damper. It measures the viscous damping, which is why in our problem, it was determined based on the viscous force experienced by the moving mass.

The damping force is mathematically expressed as \(-c\frac{dx}{dt}\), where \(c\) is the damping coefficient. In our setup, with a viscous force of 2 N at a speed of 0.04 m/s, the damping coefficient was found to be \(50 \text{ Ns/m}\).

  • High damping leads to quick reduction in motion.
  • Low damping allows for prolonged oscillations.
Choosing the right damping coefficient helps control system behavior, especially in designing systems to minimize vibrations or oscillations for efficient operation.

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Most popular questions from this chapter

By choosing the lower limitofation in Eq. ( 28 ) inthe textas the initial point \(t_{0}\), show that \(Y(t)\) becomes $$ Y(t)=\int_{t_{0}}^{t} \frac{y_{1}(s) y_{2}(t)-y_{1}(s) y_{2}(s)}{y_{1}(s) y_{2}^{2}(s)-y_{1}^{\prime}(s) y_{2}(s)} g(s) d s $$ Show that \(Y(t)\) is asolution of the initial value problem $$ L[y], \quad y\left(t_{0}\right)=0, \quad y^{\prime}\left(t_{0}\right)=0 $$ Thus \(Y\) can be identific d writh \(v\) in Problem \(21 .\)

A series circuit has a capacitor of \(0.25 \times 10^{-6}\) farad, a resistor of \(5 \times 10^{3}\) ohms, and an inductor of 1 henry. The initial charge on the capacitor is zero. If a 12 -volt battery is connected to the circuit and the circuit is closed at \(t=0,\) determine the charge on the capacitor at \(t=0.001 \mathrm{sec},\) at \(t=0.01 \mathrm{sec},\) and at any time \(t .\) Also determine the limiting charge as \(t \rightarrow \infty\)

Use the method of Problem 33 to find a second independent solution of the given equation. \(x^{2} y^{\prime \prime}+x y^{\prime}+\left(x^{2}-0.25\right) y=0, \quad x>0 ; \quad y_{1}(x)=x^{-1 / 2} \sin x\)

determine \(\omega_{0}, R,\) and \(\delta\) so as to write the given expression in the form \(u=R \cos \left(\omega_{0} t-\delta\right)\) $$ u=-\cos t+\sqrt{3} \sin t $$

By combining the results of Problems 24 through \(26,\) show that the solution of the initial value problem $$ L[y]=\left(a D^{2}+b D+c\right) y=g(t), \quad y\left(t_{0}\right)=0, \quad y^{\prime}\left(t_{0}\right)=0 $$ where \(a, b,\) and \(c\) are constants, has the form $$ y=\phi(t)=\int_{t_{0}}^{t} K(t-s) g(s) d s $$ The function \(K\) depends only on the solutions \(y_{1}\) and \(y_{2}\) of the corresponding homogeneous equation and is independent of the nonhomogeneous term. Once \(K\) is determined, all nonhomogeneous problems involving the same differential operator \(L\) are reduced to the evaluation of an integral. Note also that although \(K\) depends on both \(t\) and \(s,\) only the combination \(t-s\) appears, so \(K\) is actually a function of a single variable. Thinking of \(g(t)\) as the input to the problem and \(\phi(t)\) as the output, it follows from Eq. (i) that the output depends on the input over the entire interval from the initial point \(t_{0}\) to the current value \(t .\) The integral in Eq. (i) is called the convolution of \(K\) and \(g,\) and \(K\) is referred to as the kernel.
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