Question

A 16 -lb weight is placed upon the lower end of a coil spring suspended from the ceiling and comes to rest in its equilibrium position, thereby stretching the spring 8 in. At time \(t=0\) the weight is then struck so as to set it into motion with an initial velocity of \(2 \mathrm{ft} / \mathrm{sec}\), directed downward. The medium offers a resistance in pounds numerically equal to \(6 x^{\prime}\), where \(x^{\prime}\) is the instantaneous velocity in feet per second. Determine the resulting displacement of the weight as a function of time and graph this displacement.

Short answer

The displacement of the weight as a function of time is given by \(x(t) = e^{-6t}\left(\frac{1}{3}\sin(6t)\right)\). This function represents an exponentially damped oscillation with a damping factor of 6 and an angular frequency of 6 radians per second. The graph of the displacement function shows a damped sinusoidal behavior of the weight's motion over time.

Step-by-step solution

Determining the spring constant

The spring is stretched 8 inches (or \( \frac{2}{3} \) feet) by the 16-lb weight to reach its equilibrium position. We can use Hooke's Law to find the spring constant: \[F = kx\] Where: - \(F\) is the force exerted by the spring (equal to the weight) - \(k\) is the spring constant - \(x\) is the displacement from the equilibrium position In equilibrium, the force exerted by the spring balances the weight, so: \[16 = k\left(\frac{2}{3}\right)\] Solving for the spring constant, k: \[k = \frac{16}{\frac{2}{3}} = 24\]

Writing the equation of motion

Using Newton's 2nd Law, we can write the equation of motion for the weight as: \[m\frac{d^2x}{dt^2} = -kx - R\] Where: - \(m\) is the mass of the weight (equal to the weight divided by the acceleration due to gravity: \( \frac{16}{32} \) = 0.5 slugs) - \(x\) is the displacement from the equilibrium position - \(R\) is the resistance offered by the medium The resistance is given as \(6x'\), where \(x'\) is the velocity of the weight: \[R = 6\frac{dx}{dt}\] So the equation of motion becomes: \[0.5\frac{d^2x}{dt^2} = -24x - 6\frac{dx}{dt}\]

Solving the equation of motion

To solve the given equation of motion, we first rewrite it as a 2nd order linear differential equation: \[\frac{d^2x}{dt^2} + 12\frac{dx}{dt} + 48x = 0\] Next, we can use the method of characteristic equations where the polynomial is given by: \[r^2 + 12r + 48 = 0\] Solving this equation, we get complex roots \(r = -6 \pm 6i\). Hence, the general solution for the given equation is of the form: \[x(t) = e^{-6t}(A\cos(6t) + B\sin(6t))\] We apply the initial conditions to find the values of A and B: 1) At \(t=0\), the weight is at its equilibrium position, so \(x(0) = 0\): \[0 = e^{0}(A\cos(0) + B\sin(0))\] \[0 = A\] 2) At \(t=0\), the weight has an initial velocity of 2 ft/s directed downward, so \( \frac{dx}{dt}(0)=2\): \[2 = -6A\sin(0) + 6B\cos(0)\] \[2 = 6B\] \[B = \frac{1}{3}\] Thus, the displacement function is given by: \[x(t) = e^{-6t}\left(\frac{1}{3}\sin(6t)\right)\]

Graphing the resulting displacement

Finally, we can graph the resulting displacement function, \(x(t) = e^{-6t}\left(\frac{1}{3}\sin(6t)\right)\). The graph shows an exponentially damped oscillation with a damping factor of \(6\) and an angular frequency of \(6\) radians per second. Plot the function to visualize the damping oscillation of the weight over time. This can be done using graphing tools like Desmos, Geogebra, or any other similar tool.

Key concepts

Hooke's Law

Hooke's Law is a principle of physics that describes the behavior of springs. It states that the force exerted by a spring is directly proportional to the displacement of the spring from its equilibrium position. Mathematically, it is expressed as:

• \( F = kx \)

Here, \( F \) is the force applied by the spring, \( k \) is the spring constant (a measure of the stiffness of the spring), and \( x \) is the displacement from its equilibrium position. In the given exercise, the 16-lb weight stretches the spring by 8 inches to reach an equilibrium position. Hooke's Law is used here to determine the spring constant \( k \), a crucial value required to understand the system's behavior.
By substituting the weight's force (16 pounds) and the displacement (\(\frac{2}{3}\) feet) into the formula, we can solve for \( k \) to get \( k = 24 \). This means the spring is relatively stiff, a necessary parameter for constructing the equation of motion.

Equation of Motion

The equation of motion is a fundamental concept used to describe the dynamics of mechanical systems under the influence of forces. It is derived using Newton's Second Law of Motion, which states that the force acting on an object is equal to the mass of the object times its acceleration. For this particular exercise, the equation of motion for the weight is expressed as:

• \( m\frac{d^2x}{dt^2} = -kx - R \)

In this equation:

• \( m \) represents the mass of the object (0.5 slugs in this case)
• \( k \) is the spring constant (24, as derived from Hooke's Law)
• \( R \) is the resistance exerted by the medium (6 times the velocity \( \frac{dx}{dt} \))

The equation \( 0.5\frac{d^2x}{dt^2} = -24x - 6\frac{dx}{dt} \) is a second-order differential equation that embodies the spring's restoring force and the damping effect of resistance. Solving this equation provides insight into how the weight moves over time, factoring in effects like resistance and spring stiffness.

Damping Oscillation

Damping oscillation occurs when an oscillating system, like our spring-mass system, experiences a force (damping) that diminishes its motion over time. This damping effect is common in many physical systems and is crucial for understanding how real-world systems behave.
The resistance in the exercise acts as a damping force, proportional to the velocity of the moving weight by a factor of 6. Mathematically, this is included in the equation of motion:

• \( 6\frac{dx}{dt} \)

This damping force reduces the amplitude of oscillation over time, leading to what is called a damped harmonic motion. The solution for this type of motion is:

• \( x(t) = e^{-6t}\left(\frac{1}{3}\sin(6t)\right) \)

The exponential factor \( e^{-6t} \) represents the damping effect where amplitude decreases exponentially as time progresses. Meanwhile, \( \sin(6t) \) describes the oscillation itself incorporating an angular frequency of 6 radians/second. Together, these parameters show how the weight's motion gradually diminishes while continuing to oscillate.

Characteristic Equations

Characteristic equations play a pivotal role in solving linear differential equations, particularly those involving oscillatory systems like springs. For our exercise, the differential equation is:

• \( \frac{d^2x}{dt^2} + 12\frac{dx}{dt} + 48x = 0 \)

To solve this, we convert it into a characteristic equation:

• \( r^2 + 12r + 48 = 0 \)

Solving this characteristic equation yields complex roots \( r = -6 \pm 6i \). The presence of imaginary roots indicates that the motion involves oscillations, with an exponential damping factor due to the real part of the root (-6 in this case).
From the roots, we can construct the general solution:

• \( x(t) = e^{-6t}(A\cos(6t) + B\sin(6t)) \)

This demonstrates both damping and oscillatory behavior mathematically. The terms \( A \) and \( B \) are constants determined by initial conditions, reflecting the specific behavior of the system at \( t=0 \). By substituting initial values, we solve for these constants, leading to the final displacement equation of the system.