Question

A mass weighing 3 Ib stretches a spring 3 in. If the mass is pushed upward, contracting the spring a distance of 1 in, and then set in motion with a downward velocity of \(2 \mathrm{ft}\) sec, and if there is no damping, find the position \(u\) of the mass at any time \(t .\) Determine the frequency, period, amplitude, and phase of the motion.

Short answer

Question: Determine the displacement function, frequency, period, amplitude, and phase of a mass-spring system without damping given the following initial conditions: the 3 Ib mass stretches the spring by 3 inches, it is pushed upward (contracting the spring) by 1 inch, and is set in motion with a downward velocity of 2 ft/s. Answer: The displacement function u(t) is given by: $$u(t) = 0.247 \cos(6.576t - 0.183)$$ The frequency (f) is approximately 1.046 Hz, the period (T) is approximately 0.955 s, the amplitude (A) is 0.247 ft, and the phase constant (φ) is -0.183 rad.

Step-by-step solution

Set up the differential equation for the mass-spring system

In a mass-spring system without damping, the second-order linear differential equation is given by: $$m\frac{d^2u}{dt^2}+k u = 0$$ where \(m\) is the mass, \(k\) is the spring constant, and \(u(t)\) is the displacement of the mass as a function of time \(t\). First, we find the spring constant \(k\): Given that 3 Ib mass stretches the spring by 3 in, we can use Hooke's Law to find \(k\). Hooke's Law states: $$F = ku$$ where \(F\) is the force exerted by the spring, \(k\) is the spring constant, and \(u\) is the displacement of the spring. We can use this equation to find \(k\), noting that 1 Ib weight corresponds to a force of 32.174 lb·ft/s² (as we will be using ft units in our problem). Therefore: $$k = \frac{3 \times 32.174}{(3/12)} = 128.696 \,\mathrm{lb \cdot ft/s^2}$$ Now we can write the differential equation for our system: $$m\frac{d^2u}{dt^2}+128.696 u = 0$$

Apply the initial conditions

We are given that the mass is pushed upward (contracting the spring) by 1 in and is set in motion with a downward velocity of 2 ft/s. In terms of the displacement function \(u(t)\), these initial conditions can be written as: $$u(0) = -\frac{1}{12} \quad \mathrm{and} \quad \frac{du}{dt}(0) = 2$$

Solve the differential equation for \(u(t)\) given the initial conditions

We know that the solution to the given mass-spring system is of the form: $$u(t) = A \cos(\omega t + \phi)$$ where \(\omega\) is the angular frequency, \(A\) is the amplitude, and \(\phi\) is the phase constant. The angular frequency \(\omega\) is given by: $$\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{128.696}{3}} \approx 6.576 \,\mathrm{rad/s}$$ Now we can plug in the initial conditions to find the amplitude \(A\) and phase constant \(\phi\). $$-\frac{1}{12} = A \cos(\phi) \implies A = -\frac{1}{12\cos(\phi)}$$ Differentiating \(u(t)\), we get: $$\frac{du}{dt}(t) = -6.576 A \sin(6.576t + \phi)$$ Applying the initial condition \(\frac{du}{dt}(0) = 2\): $$2 = -6.576 A \sin(\phi) \implies A = -\frac{2}{6.576\sin(\phi)}$$ Equating the expressions for \(A\), we get: $$-\frac{2}{6.576\sin(\phi)} = -\frac{1}{12\cos(\phi)}$$ Solve for \(\phi\): $$\phi \approx -0.183 \,\mathrm{rad}$$ Find the amplitude \(A\) using any one of the above expressions for \(A\): $$A \approx 0.247 \,\mathrm{ft}$$ Now we have the displacement function \(u(t)\): $$u(t) = 0.247 \cos(6.576t - 0.183)$$

Determine the frequency, period, amplitude, and phase

From the above equation, we can determine the following properties of the motion: Frequency (\(f\)): $$f = \frac{\omega}{2\pi} \approx 1.046 \,\mathrm{Hz}$$ Period (\(T\)): $$T = \frac{1}{f} \approx 0.955 \,\mathrm{s}$$ Amplitude (\(A\)): \(0.247\) ft Phase constant (\(\phi\)): \(-0.183\) rad In this exercise, we have derived the displacement function \(u(t)\) of a mass-spring system without damping, given the initial conditions of stretching and contracting the spring and the initial downward velocity. We have also calculated the frequency, period, amplitude, and phase of the motion.

Key concepts

Mass-Spring System

A mass-spring system consists of a mass attached to a spring that can move up and down. It is a classic example of harmonic motion, where the system oscillates back and forth around an equilibrium point. In our problem, the mass pulls on the spring, causing it to stretch or compress. The natural behavior of this system is dictated by the physical properties of the mass and the spring.

When analyzing such a system without damping (ignore resistance or friction), the equation governing the motion is a second-order differential equation:

• \[m\frac{d^2u}{dt^2}+k u = 0\]

Here, \(m\) is the mass, \(u\) is the displacement, and \(k\) is the spring constant. This equation shows how the balance between inertia (mass) and elasticity (spring constant) determines the behavior of the system.

Initial Conditions

Initial conditions are crucial in solving differential equations as they allow us to find a specific solution to a general problem. In our example, the mass is initially pushed upward by 1 inch and then released with a downward velocity of 2 ft/s. These conditions set the stage for understanding how the mass will behave over time.

Mathematically, they are written as:

• \(u(0) = -\frac{1}{12}\)
• \(\frac{du}{dt}(0) = 2\)

These conditions provide the displacement and velocity of the mass at time \(t=0\). By plugging these into the equation, we can solve for the constants that define the specific motion of this system.

Hooke's Law

Hooke's Law is a fundamental principle in physics that relates to the elasticity of springs. It states that the force exerted by a spring is directly proportional to its displacement:

• \(F = k u\)

Where \(F\) is the force, \(k\) is the spring constant, and \(u\) is the displacement from the equilibrium position. This linear relationship is only valid within the elastic limit of the spring.

In the context of our problem, Hooke's Law was used to find the spring constant \(k\):

• \(k = \frac{3 \times 32.174}{(3/12)} = 128.696 \,\mathrm{lb \cdot ft/s^2}\)

The spring constant \(k\) defines how stiff the spring is: larger values mean a stiffer spring.

Angular Frequency

Angular frequency \(\omega\) describes how quickly the mass-spring system oscillates. It is a crucial factor in understanding the dynamics of the system, measured in radians per second. The angular frequency is obtained using the formula:

• \(\omega = \sqrt{\frac{k}{m}}\)

For our specific problem, this becomes:

• \(\omega \approx 6.576 \,\mathrm{rad/s}\)

Angular frequency helps us understand the timing of the oscillations. It is directly related to the system's natural frequency, and in contrast to angular speed, it doesn't depend on time or path length. Instead, it only depends on the physical properties of the system like the spring constant \(k\) and mass \(m\).