Question

A 4-lb weight is attached to the lower end of a coil spring suspended from the ceiling. The weight comes to rest in its equilibrium position, thereby stretching the spring 6 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. (a) Determine the resulting displacement and velocity of the weight as functions of the time. (b) Find the amplitude, period, and frequency of the motion. (c) Determine the times at which the weight is \(1.5 \mathrm{in}\). below its equilibrium position and moving downward. (d) Determine the times at which it is \(1.5\) in. below its equilibrium position

Short answer

The displacement and velocity functions for the weight are given by \(x(t) = \frac{1}{32} \sin(64t)\) and \(x'(t) = 2 \cos(64t)\), respectively. The amplitude, period, and frequency of the motion are \(A = \frac{1}{32}\,ft\), \(T = \frac{\pi}{32}\,s\), and \(f = \frac{32}{\pi}\, \frac{1}{s}\), respectively. There are no times at which the weight is 1.5 inches below its equilibrium position and moving either upward or downward.

Step-by-step solution

Determine the spring constant

Using Hooke's law, we are given the equation \(F = kx\) where F is the force applied on the spring, k is the spring constant and x is the displacement of the spring. In this case, the force applied on the spring is equal to the weight of the object i.e., \(F = 4 lb\). The spring is stretched by 6 inches, which is equivalent to 0.5 feet. Thus, we have \(F = (4)(32) = 128 lb\) and \(x = 0.5 ft\). Now, we can solve for the spring constant, k: \[ k = \frac{F}{x} = \frac{128}{0.5} = 256\, lb/ft \]

Determine the displacement and velocity as functions of time

Since the equation of motion for a spring is given by the second order, linear, homogenous, constant coefficient ordinary differential equation (ODE) as follows: \(mx''(t) + kx(t) = 0\), we can plug in the values of mass (m=4) and spring constant (k=256) to get the ODE: \[ 4x''(t) + 256x(t) = 0 \] Now dividing the ODE by 4, we get: \[x''(t) + 64x(t) = 0\] The above ODE has a general solution of the form: \[ x(t) = A \cos(\omega t) + B \sin(\omega t) \] where A and B are constants, and ω is the angular frequency. Since the weight was at its equilibrium position when struck, we have \(x(0) = 0\). This gives: \[ A \cos(0) + B \sin(0) = 0 \] From the above equation, we get \(A = 0\). Now to determine B, we need to find the velocity x'(t). Taking the derivative of the general solution w.r.t time, we have: \[ x'(t) = - Aω \sin(\omega t) + Bω \cos(\omega t) \] We are given that the initial velocity is 2 ft/s downward. Thus, \[ x'(0) = 2 = - A(0) + B(64) \] This gives us B = 1/32. Therefore, the displacement and velocity functions are: \[ x(t) = \frac{1}{32} \sin(64t) \] \[ x'(t) = 2 \cos(64t) \]

Calculate amplitude, period, and frequency

The amplitude, period, and frequency of a harmonic motion can be found using the coefficient and angular frequency in the sin or cos part of the displacement function. From \(x(t) = \frac{1}{32} \sin(64t)\), we have the amplitude A: \[ A = \frac{1}{32}\,ft \] The angular frequency ω is 64, which is related to the period T and frequency f as follows: \[ ω = 2πf = \frac{2π}{T} \] Solving for T and f, we have: \[ T = \frac{2π}{\omega} = \frac{\pi}{32}\,s \] \[ f = \frac{1}{T} = \frac{32}{\pi}\, \frac{1}{s} \]

Determine the times when the weight is 1.5 inches below its equilibrium position and moving downward

To find the time instances when the weight is 1.5 inches below its equilibrium position, we have to solve the following equation for t: \[ x(t) = -\frac{1.5}{12} \] Plug in our displacement equation: \[ \frac{1}{32}\sin(64t) = -\frac{1}{8} \] Now, solve for t (remembering the trigonometric identities): \[ \sin(64t) = -4 \rightarrow 64t = arcsin(-4) \] Since the arcsin function has a domain of \(-1 \leq x \leq 1\), there is no solution for this part.

Determine the times when the weight is 1.5 inches below its equilibrium position and moving upward

Similar to step 4, we need to solve for t when the weight is 1.5 inches below its equilibrium position. However, this time we are considering the case when the weight is moving upward, which means its velocity is positive. Therefore, we must also have \(x'(t) >0\). From our velocity equation: \[ x'(t) = 2\cos(64t) > 0 \] Now, we know that \(0 < t < T/4\), since we are looking for the first quarter of the motion where \(x'(t) > 0\). So, we need to find the time instances within that interval when \(x(t) = -\frac{1}{8}\). Since we couldn't find any solution for the case when the weight was moving downward, there will be no solution for this part either. In conclusion, the equations found for displacement and velocity are: \[ x(t) = \frac{1}{32} \sin(64t) \] \[ x'(t) = 2 \cos(64t) \] The amplitude, period, and frequency of the motion are \(A = \frac{1}{32}\,ft\), \(T = \frac{\pi}{32}\,s\), and \(f = \frac{32}{\pi}\, \frac{1}{s}\), respectively. There are no times at which the weight is 1.5 inches below its equilibrium position and moving either upward or downward.

Key concepts

Harmonic Motion

Harmonic motion is a type of periodic movement, where an object oscillates back and forth in a regular pattern over time. It's like the smooth, repetitive motion of a swing or a pendulum. This motion can often be described using functions like sine or cosine, which are perfect for showing repetitive patterns. In the context of our exercise, the weight attached to the spring exhibits this kind of motion, moving up and down as time progresses.

When a system undergoes harmonic motion, it goes through cycles that are repeated in specific time intervals. Each cycle includes one complete movement in one direction and returning back to the starting point. Understanding harmonic motion helps in analyzing systems like springs, which are often subject to oscillating forces. This is particularly useful in engineering and physics to design mechanisms that rely on cyclic motion, such as watches, meters, and even seismic devices used in earthquake detection. For students exploring how objects move within systems, knowing about harmonic motion can enhance their comprehension of dynamic mechanisms.

Hooke's Law

Hooke's Law is essential for studying how springs behave. It states that the force needed to extend or compress a spring is directly proportional to the distance it is stretched or compressed. Mathematically, this is written as:
\( F = kx \),
where \( F \) is the force applied, \( k \) is the spring constant, and \( x \) is the displacement of the spring from its natural (unstressed) position.

In practical terms, Hooke's Law means that if you pull a spring, the force required to stretch it further increases with every inch it is pulled. Thus, a spring is "springier" (or offers more resistance to stretching) if it has a larger spring constant. This law only applies until the spring reaches its elastic limit, beyond which it might not return to its original shape, much like a rubber band that becomes deformed if stretched too far.

Understanding Hooke's Law provides insight into designing structures that use springs, such as vehicle suspensions, trampolines, and wristwatches, where controlled motion and return to equilibrium are vital.

Spring Constant

The concept of a spring constant is crucial in understanding how springs work. It's a measure of a spring's stiffness. A higher spring constant means the spring is stiffer and requires more force to stretch or compress by a specific amount. The spring constant, denoted by \( k \), is derived from Hooke's Law where \( F = kx \).

In the exercise, calculating the spring constant helps understand how much the spring resists the weight pulling down. Here, we found the spring constant to be 256 lb/ft, indicating a relatively stiff spring. This large value for \( k \) affects how the spring and the attached weight will behave as it oscillates – less stretching under the same force translates into faster vibrations.

The spring constant is vital in engineering and construction, ensuring components perform as expected under forces they need to withstand. It's a key factor in selecting the right spring type for specific applications, from toy designs to automotive suspensions and industrial machinery.

Angular Frequency

Angular frequency is an important concept in the analysis of harmonic motion. It represents the rate of oscillation in radians per unit time, often denoted by \( \omega \). It determines how quickly an object moves back and forth in harmonic motion.

Mathematically, angular frequency is linked to the regular frequency \( f \) and the period \( T \) of the motion by the equations \( \omega = 2\pi f = \frac{2\pi}{T} \). This relationship gives you a way to connect the time-based description of motion (like period) with the spatial description (like the path of the motion) via radians.

In the given problem, the angular frequency \( \omega \) was found to be 64. This means that the weight completes the equivalent of 64 radians in one second, showing how swiftly it moves during each cycle of oscillation. Angular frequency is fairly common when dealing with circular and oscillatory motions, simplifying the mathematics that describe these physical phenomena. It plays a crucial role in technologies like radio wave transmissions, electrical circuits involving alternating current, and mechanical systems like pendulums and sensors.