Question

A 12 -lb weight is placed upon the lower end of a coil spring suspended from the ceiling. The weight comes to rest in its equilibrium position, thereby stretching the spring \(1.5\) in. The weight is then pulled down 2 in. below its equilibrium position and released from rest at \(t=0\). Find the displacement of the weight as a function of the time; determine the amplitude, period, and frequency of the resulting motion; and graph the displacement as a function of the time.

Short answer

The displacement function of the 12-lb weight on the spring is \(x(t) = 0.1667\cos(15.95t)\), with an amplitude of 0.1667 ft, a period of 0.394 s, and a frequency of 2.54 Hz. The function represents a cosine wave that can be graphed with time (t) on the x-axis and displacement (x) on the y-axis.

Step-by-step solution

Find the spring constant

To find the spring constant (k), we will use Hooke's Law, which states that the force F required to stretch a spring by a distance x is directly proportional to the displacement: \(F = -kx\) The weight stretches the spring by 1.5 inches (after converting to feet, we get 0.125 ft). The weight is acting in the downward direction, and the spring force opposes the weight, acting in the upward direction. Therefore, set the spring force equal to the weight, using these values: \(-k(0.125) = -12\) Solving for the spring constant: \(k = \frac{-12}{-0.125} = 96\text{ lb/ft}\)

Determine the angular frequency

We will now find the angular frequency (ω) using the relation between the spring constant and the mass of the weight (m). Since the weight is given in pounds, we need to convert it to mass using the relation: \(m = \frac{12 \text{lb}}{32.2 \text{ft/s}^2} = 0.3726 \text{slug}\) Now, use the formula for angular frequency: \(\omega = \sqrt{\frac{k}{m}}\) \(\omega = \sqrt{\frac{96}{0.3726}} \approx 15.95 \text{ rad/s}\)

Find the displacement function

As Hooke's Law is related to simple harmonic motion, we can represent the displacement x(t) as a sinusoidal function: \(x(t) = A\cos(\omega t + \phi)\) Here, A is the amplitude, ω is the angular frequency, and φ is the phase angle. Since the weight is released from rest at t=0 and the displacement is maximum (2 inches or 0.1667 ft), we have: \(x(t=0) = A\cos(\phi) = 0.1667\) Since the derivative of, x(t), the velocity function is: \(v(t) = -A\omega\sin(\omega t + \phi)\) As the initial velocity is 0 when t=0, we have: \(v(t=0) = 0 = -A\omega\sin(\phi)\) Solving these equations for the unknowns A and φ: A = 0.1667 ft and \(\phi = 0\) Thus, the displacement function becomes: \(x(t) = 0.1667\cos(15.95t)\)

Determine amplitude, period, and frequency

Now we can find the amplitude, period, and frequency of the motion using the displacement function: Amplitude (A) = 0.1667 ft (already found) Period (T) = \(\frac{2\pi}{\omega}\) = \(\frac{2\pi}{15.95}\) ≈ 0.394 s Frequency (f) = \(\frac{1}{T}\) ≈ 2.54 Hz

Graph the displacement function

To graph the displacement function, use the formula we derived earlier: \(x(t) = 0.1667\cos(15.95t)\) Create a graph with time (t) on the x-axis and displacement (x) on the y-axis. Plot the displacement x(t) versus time for a few oscillations. The graph should be a cosine wave with an amplitude of 0.1667 ft, period of 0.394 s, and frequency of 2.54 Hz. Various graphing tools can be used to plot this function.