Problem 15 - Precalculus with Calculus Previews

Question

The equation of simple harmonic motion of a spring/mass system is \(y(t)=\frac{5}{2} \sin (2 t-\pi / 3)\). Determine the initial displacement \(y_{0}\) and initial velocity \(v_{0}\) of the mass.

Step-by-Step Solution

Answer

Initial displacement \(y_0 = -\frac{5\sqrt{3}}{4}\); initial velocity \(v_0 = \frac{5}{2}\).

Step 1: Understand Initial Conditions

To determine the initial displacement and initial velocity, we need to evaluate the functions at time \(t = 0\). The initial displacement \(y_{0}\) is given directly from the function evaluated at \(t = 0\) as \(y(0)\). The initial velocity \(v_{0}\) is the time derivative of \(y(t)\) evaluated at \(t = 0\).

Step 2: Calculate the Initial Displacement

Substitute \(t = 0\) into the equation \(y(t) = \frac{5}{2} \sin(2t - \pi/3)\):\[ y(0) = \frac{5}{2} \sin(2 \cdot 0 - \pi/3) = \frac{5}{2} \sin(-\pi/3) \]The sine of \(-\pi/3\) is the same as \(-\sin(\pi/3)\). Therefore, \(-\sin(\pi/3) = -\frac{\sqrt{3}}{2}\).Thus, \(y(0) = \frac{5}{2} \cdot (-\frac{\sqrt{3}}{2}) = -\frac{5\sqrt{3}}{4}\).

Step 3: Differentiate to Find Velocity Function

Differentiate the function \(y(t) = \frac{5}{2} \sin(2t - \pi/3)\) with respect to \(t\) to find the velocity function \(v(t)\):\[ v(t) = \frac{d}{dt}\left(\frac{5}{2} \sin(2t - \pi/3)\right) = \frac{5}{2} \cdot 2 \cos(2t - \pi/3) \]This simplifies to:\[ v(t) = 5 \cos(2t - \pi/3) \]

Step 4: Calculate the Initial Velocity

Substitute \(t = 0\) into the velocity function:\[ v(0) = 5 \cos(2 \cdot 0 - \pi/3) = 5 \cos(-\pi/3) \]Since the cosine of \(-\pi/3\) is the same as \(\cos(\pi/3)\), and \(\cos(\pi/3) = \frac{1}{2}\), the initial velocity is:\[ v(0) = 5 \cdot \frac{1}{2} = \frac{5}{2} \]

Key Concepts

Understanding Initial Displacement

In simple harmonic motion, the initial displacement is crucial to understanding the initial position of the oscillating mass. It indicates where the mass begins its movement within the cycle of the motion. For the equation of a spring/mass system, the initial displacement is represented as \(y(0)\), which you find when you substitute \(t = 0\) into the displacement equation.

In our given function, \(y(t) = \frac{5}{2} \sin(2t - \pi/3)\), to find the initial displacement, replace \(t\) with zero. This simplifies to \(y(0) = \frac{5}{2} \sin(-\pi/3)\). Knowing that \(-\sin(\pi/3) = -\frac{\sqrt{3}}{2}\), we can calculate the initial displacement as:

• \(y(0) = \frac{5}{2} \cdot \left(-\frac{\sqrt{3}}{2}\right) = -\frac{5\sqrt{3}}{4}\)

This initial displacement value tells us that at \(t = 0\), the mass is positioned negatively along the axis, indicating it starts below the equilibrium point.

Determining Initial Velocity

The initial velocity of the mass in simple harmonic motion gives us insight into the speed and direction with which the mass starts its oscillation. We find the initial velocity by differentiating the displacement function with respect to time to obtain the velocity function \(v(t)\), and then evaluating this function at \(t = 0\).

Starting with the given function \(y(t) = \frac{5}{2} \sin(2t - \pi/3)\), differentiate it:\

• \(v(t) = \frac{d}{dt}\left(\frac{5}{2} \sin(2t - \pi/3)\right) = 5 \cos(2t - \pi/3)\)

Now, substitute \(t = 0\) into this velocity equation, \(v(0) = 5 \cos(-\pi/3)\). Since \(\cos(-\pi/3) = \cos(\pi/3) = \frac{1}{2}\), we find that:

• \(v(0) = 5 \times \frac{1}{2} = \frac{5}{2}\)

Thus, the initial velocity is \(\frac{5}{2}\), meaning that the mass begins its motion moving upwards with a velocity of \(\frac{5}{2}\) units per time unit.

Exploring the Spring/Mass System

In the context of simple harmonic motion, the spring/mass system is a classic example often used to illustrate how forces work in oscillating systems. This system consists of a mass attached to a spring, where the force exerted by the spring is proportional to the displacement from its equilibrium position following Hooke's Law.

The dynamics of such a system can be described by differential equations, which, when solved, depict the mass oscillating back and forth around the equilibrium point. The equation \(y(t) = \frac{5}{2} \sin(2t - \pi/3)\) is an example of a solution to these differential equations in the form of a sine function, which denotes the displacement of the mass relative to time.

• The frequency of oscillation is determined by the coefficient of \(t\) in the argument of the sine function, indicating how quickly the mass oscillates.
• The amplitude, \(\frac{5}{2}\), shows the maximum extent of oscillation from the equilibrium point.
• Lastly, the phase shift \(-\pi/3\) tells us how the sine curve is shifted horizontally, affecting the starting position in the cycle of motion.

Understanding these components helps to grasp the overall behavior and properties of spring/mass systems in harmonic motion.