Question

A coil spring is such that a 25 -lb weight would stretch it 6 in. The spring is suspended from the ceiling, a 16 -lb weight is attached to the end of it, and the weight then comes to rest in its equilibrium position. It is then pulled down 4 in. below its equilibrium position and released at \(t=0\) with an initial velocity of \(2 \mathrm{ft} / \mathrm{sec}\), directed upward. (a) Determine the resulting displacement of the weight as a function of the time. (b) Find the amplitude, period, and frequency of the resulting motion. (c) At what time does the weight first pass through its equilibrium position and what is its velocity at this instant?

Short answer

a) The displacement function is: \(x(t) = -0.333\cos(10t) + 0.2\sin(10t)\) b) Amplitude: 0.385 ft, Period: \(\frac{\pi}{5}\) s, Frequency: \(\frac{5}{\pi}\) Hz c) Time when weight first passes through equilibrium: 0.491 s, Velocity at this instant: 5.963 ft/s

Step-by-step solution

Determine the spring constant

Using Hooke's Law, we can determine the spring constant (k) using the given information. Hooke's Law states that the force exerted by a spring is proportional to its displacement from its equilibrium position: \(F = - kx\) where, F is the force exerted, k is the spring constant, and x is the displacement from the equilibrium position. According to the problem, a 25-lb weight stretches the spring by 6 in, which is equivalent to 0.5 ft. So, we can write the equation for this condition: 25 = k * 0.5 Solving for k, we get: k = 50 lb/ft

Determine the equation of motion and displacement function

Now, we know that the equation of motion for a simple harmonic oscillator is given by: \(m\ddot{x} + kx = 0\) where x is the displacement, \(m\) is the mass attached to the spring, and \(\ddot{x}\) is the acceleration. We are given that the mass is 16 lbs, so we can convert it to mass (m) in slugs using the conversion factor (1 slug = 32.174 lb). However, since we're using force (16 lbs), it will cancel out: \(16\ddot{x} + 50x = 0\) Now, we can find the displacement function by solving the differential equation above: \(x(t) = A\cos(\omega t) + B\sin(\omega t)\) where A and B are constants, \(\omega\) is the angular frequency, and t is the time. We are given the initial conditions: - The weight is pulled down 4 inches (0.333 ft) below its equilibrium position, - The initial velocity is 2 ft/s upward, Using these initial conditions, we can find A and B: For \(t=0\), \(x(0) = A\) and \(\dot{x}(0) = 2\). Using the initial condition \(x(0) = -0.333\) and \(\dot{x}(0) = 2\), we get: \(A=-0.333\) \(\omega B = 2\) To find B, we first need to find the angular frequency, \(\omega\).

Determine amplitude, period, and frequency

For a simple harmonic oscillator, the angular frequency (\(\omega\)) is given by: \(\omega = \sqrt{\frac{k}{m}}\) Using the known values of k (50 lb/ft) and m (16 lbs), we can find \(\omega\): \(\omega = \sqrt{\frac{50}{0.5}} = 10 \frac{\text{rad}}{\text{s}}\) Now we can find the value for B with \(\omega\) = 10 rad/s: \(B = \frac{2}{10} = 0.2\) So, the displacement function becomes: \(x(t) = -0.333\cos(10t) + 0.2\sin(10t)\) Now we will find the amplitude (R), period (T), and frequency (f) of the motion. Amplitude R is given by: \(R = \sqrt{A^2 + B^2} = \sqrt{(-0.333)^2 + (0.2)^2} = 0.385 \text{ ft}\) The period T is given by: \(T = \frac{2\pi}{\omega} = \frac{2\pi}{10} = \frac{\pi}{5} \text{s}\) The frequency f is given by: \(f = \frac{1}{T} = \frac{5}{\pi} \text{ Hz}\)

Find the time when weight first passes through the equilibrium position

To find the time when the weight first passes through its equilibrium position, we need to find the moment when the displacement function, \(x(t)\), is equal to zero: \(x(t) = -0.333\cos(10t) + 0.2\sin(10t) = 0\) Solve for t: \(t = 0.491 \text{s}\) Now, we will find the velocity at this instant using the derivative of the displacement function: \(\dot{x}(t) = 10[-0.333\sin(10t) + 0.2\cos(10t)]\) Evaluate \(\dot{x}(t)\) at \(t=0.491\) s: \(\dot{x}(0.491) = 5.963 \text{ ft/s}\) Results: a) The displacement function is given by: \(x(t) = -0.333\cos(10t) + 0.2\sin(10t)\) b) The amplitude is 0.385 ft, the period is \(\frac{\pi}{5}\) s, and the frequency is \(\frac{5}{\pi}\) Hz. c) The weight first passes through its equilibrium position at 0.491 s, and its velocity at this instant is 5.963 ft/s.

Key concepts

Understanding Hooke's Law

Hooke's Law is a fundamental principle that applies to many problems in physics, particularly those that involve springs. It states that the force exerted by a spring is directly proportional to the displacement from its equilibrium position and is expressed by the formula:
\[ F = -kx \]
where \( F \) is the force in pounds, \( k \) is the spring constant in pounds per foot, and \( x \) is the displacement in feet. The negative sign indicates that the force exerted by the spring is in the opposite direction of the displacement.

In the case of the textbook exercise, a 25-pound weight stretches the spring by 6 inches, which translates to 0.5 feet. Using Hooke's Law, we can calculate the spring constant \( k \) by rearranging the formula:
\[ k = \frac{F}{x} \]
Substituting the known values, we obtain the spring constant \( k = 50 \text{ lb/ft} \), a crucial figure needed to solve for the harmonic motion of the spring-mass system. Understanding Hooke's Law and how to apply it correctly can significantly improve one's ability to tackle problems related to spring dynamics.

Exploring Simple Harmonic Motion

Simple harmonic motion is a type of periodic motion where the restoring force is directly proportional to the displacement and acts in the direction opposite to that of displacement. It's the motion experienced by an object when Hooke's Law is in effect, such as a mass attached to a spring. The motion is sinusoidal in time and demonstrates a single resonant frequency. Mathematically, the motion can be described by the equation:
\[ m\ddot{x} + kx = 0 \]
Here, \( m \) symbolizes the mass of the object, \( k \) represents the spring constant, and \( x \) is the displacement from the equilibrium. \( \ddot{x} \) denotes the acceleration, which is the second derivative of displacement with respect to time.

By solving this equation with given initial conditions, in this problem being the displacement and initial velocity, we arrive at the displacement function in terms of sine and cosine components. When the weight in our exercise is pulled down and released, it exhibits simple harmonic motion described by the function \( x(t) \) with parameters that can be deduced from initial conditions and the spring constant.

Understanding Angular Frequency

Angular frequency is a measure that describes how rapidly an object oscillates in simple harmonic motion. It is related to the period and frequency of the motion and can be calculated using the spring constant and mass of the object. The formula for angular frequency, denoted by \( \omega \), is:
\[ \omega = \sqrt{\frac{k}{m}} \] where \( k \) is the spring constant and \( m \) is the mass converted to slugs if needed.

For the exercise, the angular frequency is found by taking the square root of the spring constant divided by the mass. This gives us \( \omega = 10 \text{ rad/s} \). Angular frequency is an essential component in predicting the behavior of the spring-mass system over time and is instrumental in defining both the displacement equation and the derivatives used to find velocity. Knowing the angular frequency allows us to determine the system's period \( T \) and frequency \( f \) which describe the oscillation's timescale. The amplitude of the motion, another fundamental aspect, is the maximum extent of the vibration and can also be computed with angular frequency being known.