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Magazine Chapter 17: Problem 4
A force of \(13 \mathrm{N}\) is needed to keep a spring with a 2 -kg mass stretched \(0.25 \mathrm{m}\) beyond its natural length. The damping constant of the spring is \(c=8\). (a) If the mass starts at the equilibrium position with a velocity of \(0.5 \mathrm{m} / \mathrm{s}\), find its position at time \(t .\) (b) Graph the position function of the mass.
Short Answer
Expert verified
The position function is \( x(t) = 0.107 e^{-2t}\sin(4.69t) \).
Step by step solution
01
Understanding Hooke's Law
We know that Hooke's Law states that the force required to stretch or compress a spring by a distance (x) from its natural length is proportional to that distance. Mathematically, it is represented by:\[ F = kx \]where \( F \) is the force applied (13 N), and \( x \) is the distance (0.25 m). We can use this to find the spring constant \( k \).
02
Calculate Spring Constant k
From Hooke's Law:\[ 13 = k \times 0.25 \]Solving for \( k \), we find:\[ k = \frac{13}{0.25} = 52 \, \mathrm{N/m} \]
03
Write the Differential Equation for Damped Motion
The equation of motion for a damped spring-mass system is given by:\[ m\frac{d^2x}{dt^2} + c\frac{dx}{dt} + kx = 0 \]where \( m = 2 \) kg, \( c = 8 \), and \( k = 52 \). Substituting these values, we get:\[ 2\frac{d^2x}{dt^2} + 8\frac{dx}{dt} + 52x = 0 \]
04
Solve the Characteristic Equation
The characteristic equation of the differential equation is:\[ 2r^2 + 8r + 52 = 0 \]Using the quadratic formula \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 2 \), \( b = 8 \), and \( c = 52 \), we find:\[ r = \frac{-8 \pm \sqrt{8^2 - 4 \times 2 \times 52}}{4} \]\[ r = \frac{-8 \pm \sqrt{64 - 416}}{4} = \frac{-8 \pm \sqrt{-352}}{4} \]\[ r = \frac{-8 \pm 18.76i}{4} = -2 \pm 4.69i \]
05
Determine the General Solution for Motion
Since the roots are complex, the solution is of the form:\[ x(t) = e^{-2t}(C_1\cos(4.69t) + C_2\sin(4.69t)) \]We need to find constants \( C_1 \) and \( C_2 \) using initial conditions.
06
Apply Initial Conditions
The mass starts from the equilibrium position, so initially, \( x(0) = 0 \) and the velocity \( \frac{dx}{dt} = 0.5 \) m/s at \( t=0 \).Using \( x(0)=0 \), we have:\[ 0 = C_1 \]Using the derivative for velocity and initial condition \( \frac{dx}{dt}|_{t=0} = 0.5 \), compute\[ x'(t) = e^{-2t}(-2C_1\cos(4.69t) - 4.69C_1\sin(4.69t) + C_2\cos(4.69t)) - 2x(t) \]Evaluate at \( t=0 \):\[ 0.5 = 4.69C_2 \]Thus, \( C_2 = \frac{0.5}{4.69} \approx 0.107 \).
07
Final Position Function
Substituting \( C_1 = 0 \) and \( C_2 \approx 0.107 \) into the general solution gives:\[ x(t) = 0.107 e^{-2t}\sin(4.69t) \]
08
Graph the Position Function
Plot \( x(t) = 0.107 e^{-2t}\sin(4.69t) \) to visualize the damped oscillatory behavior of the system. The graph will show an oscillating function with exponentially decreasing amplitude over time.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Hooke's Law
To understand how springs work in physics, we often rely on Hooke's Law. This principle states that the force needed to extend or compress a spring by some distance is directly proportional to that distance. The formula is simple:
- \( F = kx \)
- where \( F \) is the force applied to the spring, \( k \) is the spring constant, and \( x \) is the displacement from its natural length.
- \( k = \frac{F}{x} = \frac{13}{0.25} = 52 \mathrm{N/m} \)
Damping Constant
In real-world scenarios, objects in motion tend to lose energy over time due to resistance forces such as friction or air resistance. This concept translates into what we call damping in springs.
- The damping constant, often symbolized as \( c \), quantifies this resistive force.
- While Hooke's law gives us a purely elastic model, adding the damping constant helps in modeling a more realistic scenario where energy loss gradually reduces the amplitude of oscillation over time.
Characteristic Equation
When analyzing damped harmonic motion, especially for springs, solving the characteristic equation is essential. This equation arises from the differential equation that describes the system, which in our case includes both the spring and the damping constant.
- The equation for our system is \( 2\frac{d^2x}{dt^2} + 8\frac{dx}{dt} + 52x = 0 \).
- To find the characteristic equation, we rewrite this as a quadratic for \( r \) in the form \( ar^2 + br + c = 0 \).
- \( r = -2 \pm 4.69i \)
Equation of Motion
The equation of motion describes how the system behaves over time. It is derived through solving the differential equation that characterizes the movement of the spring-mass system.
- Our solution, given the complex roots, suggests that the system undergoes damped oscillations: \( x(t) = e^{-2t}(C_1\cos(4.69t) + C_2\sin(4.69t)) \).
- Here, the exponential part \( e^{-2t} \) reflects the decay of motion due to damping, while the sine and cosine terms indicate oscillations.
- By setting initial conditions \( x(0) = 0 \) and initial velocity \( \frac{dx}{dt}(0) = 0.5 \) m/s, we calculate the constants: \( C_1 = 0 \) and \( C_2 \approx 0.107 \).
