Question

These exercises deal with undamped vibrations of a spring-mass system, $$ m y^{\prime \prime}+k y=0, \quad y(0)=y_{0}, \quad y^{\prime}(0)=y_{0}^{\prime} . $$ Use a value of \(9.8 \mathrm{~m} / \mathrm{s}^{2}\) or \(32 \mathrm{ft} / \mathrm{sec}^{2}\) for the acceleration due to gravity. A 4-kg mass was attached to a spring and set in motion. A record of the displacements was made and found to be described by \(y(t)=25 \cos (2 t-\pi / 6)\), with displacement measured in centimeters and time in seconds. Determine the initial displacement \(y_{0}\), initial velocity \(y_{0}^{\prime}\), spring constant \(k\), and period \(T\) of the vibrations.

Short answer

The initial displacement is \(y_0 = 25\cos\left(\frac{\pi}{6}\right)\) cm, the initial velocity is \(y_{0}^{\prime} = 50\sin\left(\frac{\pi}{6}\right)\) cm/s, the spring constant is \(k = 16\) N/m, and the period of the vibrations is \(T = \pi\) seconds.

Step-by-step solution

Calculate initial displacement \(y_0\)

To find the initial displacement, we need to evaluate the given function \(y(t)\) at \(t=0\). $$ y(0) = 25\cos\left(2(0)-\frac{\pi}{6}\right) = 25\cos\left(-\frac{\pi}{6}\right) $$ Since \(\cos(-x)=\cos(x)\), we have $$ y(0) = 25\cos\left(\frac{\pi}{6}\right) = y_0 $$

Calculate initial velocity \(y_{0}^{\prime}\)

Next, let's find the derivative of the function \(y(t)\). $$ y^{\prime}(t)= \frac{d}{dt}(25\cos(2t-\frac{\pi}{6}))=-50\sin(2t-\frac{\pi}{6}) $$ Now, we will evaluate \(y^{\prime}(t)\) at \(t=0\) to find the initial velocity. $$ y^{\prime}(0)=-50\sin\left(2(0)-\frac{\pi}{6}\right)=-50\sin\left(-\frac{\pi}{6}\right) = y_{0}^{\prime} $$ Since \(\sin(-x)=-\sin(x)\), we have $$ y_{0}^{\prime}=50\sin\left(\frac{\pi}{6}\right) $$

Determine spring constant \(k\)

Using the given differential equation: \(my^{\prime\prime}(t)+ky(t)=0\), we can find the spring constant \(k\). We already have the given mass \(m=4\) kg. First, we need to find the second derivative of y(t), \(y^{\prime\prime}(t)\). $$ y^{\prime\prime}(t)=\frac{d^2}{dt^2}(25\cos(2t-\frac{\pi}{6}))=-100\cos(2t-\frac{\pi}{6}) $$ Now, we will plug in the values of \(y(t)\) and \(y^{\prime\prime}(t)\) into the given equation and solve for \(k\). $$ 4(-100\cos(2t-\frac{\pi}{6}))+k(25\cos(2t-\frac{\pi}{6}))=0 $$ Divide both sides by \(\cos(2t-\frac{\pi}{6})\) $$ -400 + 25k = 0 $$ Now, solve for \(k\) $$ k = \frac{400}{25} = 16 \,\text{N/m} $$

Calculate the period \(T\)

The displacement function has the form \(y(t)=25\cos(2t-\frac{\pi}{6})\), which we can rewrite in terms of frequency \(\omega\): $$ y(t)=25\cos(\omega t-\frac{\pi}{6}) $$ Comparing the two equations, we have \(\omega = 2\). We know the relation between angular frequency \(\omega\) and period \(T\) is given by \(\omega = \frac{2\pi}{T}\). We can solve for the period \(T\) as follows: $$ T = \frac{2\pi}{\omega} = \frac{2\pi}{2} = \pi \, \text{seconds} $$ In conclusion, the initial displacement is \(y_0 = 25\cos\left(\frac{\pi}{6}\right)\) cm, the initial velocity is \(y_{0}^{\prime} = 50\sin\left(\frac{\pi}{6}\right)\) cm/s, the spring constant is \(k = 16\) N/m, and the period of the vibrations is \(T = \pi\) seconds.

Key concepts

spring-mass system

A spring-mass system is a classic physics model that helps us understand various types of motion, especially vibrations. In this model, a mass is attached to a spring. When the spring is displaced from its rest position and then released, it exerts a force on the mass. This interaction creates a motion that can be described using differential equations.
There are several things to remember about a spring-mass system:

• The mass, denoted as \(m\), determines how much inertia the system has, affecting how quickly it accelerates.
• The spring itself has a spring constant \(k\). This tells us how stiff the spring is. A higher \(k\) means a stiffer spring, resulting in faster oscillations.
• The equilibrium position is where the spring is at rest with no net forces acting on it.

Understanding these components is crucial for solving problems related to vibrations and oscillations.

undamped vibrations

In a spring-mass system, if there is no energy loss over time, the vibrations are described as being undamped. This means that the system will continue to oscillate indefinitely with constant amplitude.
For our exercise, the motion of the mass is described by the equation \(my^{\prime\prime} + ky = 0\), which predicts these undamped vibrations. This is a second-order linear homogeneous differential equation. Since there's no damping (energy loss due to friction or air resistance), the motion can be sustained forever theoretically, although, in reality, some small amount of damping is almost always present.
In practical terms, this concept is useful in understanding natural phenomena where damping is negligible, and it sets the stage for further study in more complex, real-world systems involving damping.

initial conditions

Initial conditions are crucial in determining the specific behavior of a physical system after it is disturbed. They are necessary to solve differential equations related to motion, like those in a spring-mass system.
The initial conditions given in this problem are:

• Initial displacement \(y_0\), which is the position of the mass at time \(t = 0\).
• Initial velocity \(y_0'\), indicating how fast the mass is moving at \(t = 0\).

These values allow us to precisely predict the position and velocity of the system at any future time by plugging them into the solution of the differential equations. In our example, we calculated these using the functions describing the displacement and its derivative, leading to the specific form of the motion equation.

angular frequency

Angular frequency, denoted as \(\omega\), is a key concept in understanding oscillatory motion in a spring-mass system. It describes how quickly the system oscillates. Angular frequency is related directly to the period of oscillation and is expressed in terms of radians per second.
The formula \(\omega = \frac{2\pi}{T}\) links the angular frequency to the period \(T\). It shows how many cycles occur in a unit of time. In our problem, the angular frequency \(\omega\) was found to be \(2\), leading to a period \(T\) of \(\pi\) seconds.
Understanding \(\omega\) not only helps in predicting the time for one full oscillation but also in understanding the kinetic and potential energies involved at different times. Recognizing \(\omega\)'s role can clarify how quickly and frequently the mass moves back and forth through the equilibrium position.