Question

An object stretches a spring 6 inches in equilibrium. Find its displacement for \(t>0\) if it's initially displaced 3 inches above equilibrium and given a downward velocity of 6 inches/s. Find the frequency, period, amplitude and phase angle of the motion.

Short answer

Answer: The displacement equation for the motion is given by: $$x(t) = \frac{3}{\cos(\phi)}\cos(\sqrt{386}t + \phi)$$ The frequency is \(\frac{\sqrt{386}}{2\pi}\) Hz, the period is \(\frac{2\pi}{\sqrt{386}}\) s, the amplitude is \(\frac{3}{\cos(\phi)}\) inches, and the phase angle is \(\arctan \left(\frac{6}{3\sqrt{386}}\right)\).

Step-by-step solution

Write a simple harmonic motion equation

First, recall that the general form of the displacement equation for simple harmonic motion is: $$x(t) = A \cos(\omega t + \phi)$$ where \(A\) is the amplitude, \(\omega\) is the angular frequency, \(t\) is time, and \(\phi\) is the phase angle.

Find the spring constant

Given that the spring stretches 6 inches in equilibrium, we can use Hooke's Law (\(F = kx\)) to determine the spring constant (\(k\)). Assuming the object is in equilibrium, the force due to the weight of the object equals the force exerted by the spring. Therefore: $$k \cdot 6 = mg$$ where \(m\) is the mass of the object, and \(g\) is the acceleration due to gravity (approximately 386 inches/s²). However, \(k\) and \(m\) won't be needed explicitly because they cancel each other out in the equation for \(\omega\).

Find the angular frequency

The angular frequency \(\omega\) for simple harmonic motion with a spring is given by: $$\omega = \sqrt{\frac{k}{m}}$$ Since the weight of the object stretched the spring 6 inches, \(\frac{k}{m} = \frac{mg}{m} = g\). Therefore, the angular frequency \(\omega\) is: $$\omega = \sqrt{g} = \sqrt{386} \thickspace\text{rad/s}$$

Determine the initial conditions

We are given two initial conditions: the object is initially displaced 3 inches above equilibrium (\(x(0) = 3\)), and it is given a downward velocity of 6 inches/s (\(v(0) = -6\)). Now, we can use these initial conditions to find \(A\) (amplitude) and \(\phi\) (phase angle) by plugging them into the general equation of displacement and its first derivative (velocity) equation: 1. Displacement: \(x(0) = 3 = A \cos(\phi)\) 2. Velocity: \(v(t) = -A\omega \sin(\omega t + \phi)\), so \(v(0) = -6 = -A\omega \sin(\phi)\).

Find the amplitude and phase angle

From step 4, we get the following system of equations: 1. \(3 = A \cos(\phi)\) 2. \(6 = A\sqrt{386} \sin(\phi)\). Divide (2) by (1), we get: $$\tan(\phi) = \frac{6}{3\sqrt{386}}$$ So, $$\phi = \arctan \left(\frac{6}{3\sqrt{386}}\right)$$ Now, plug the value of \(\phi\) back into (1) to find the amplitude \(A\): $$A = \frac{3}{\cos(\phi)}$$

Write the displacement function

Now, we can write the displacement equation as a function of time, x(t): $$x(t) = \frac{3}{\cos(\phi)}\cos(\sqrt{386}t + \phi)$$

Find the frequency and period

Finally, we can find the frequency and period of the motion from \(\omega\): 1. Frequency (\(f\)): $$f = \frac{\omega}{2\pi} = \frac{\sqrt{386}}{2\pi} \thickspace\text{Hz}$$ 2. Period (\(T\)): $$T = \frac{1}{f} = \frac{2\pi}{\sqrt{386}} \thickspace\text{s}$$ So, the displacement equation is given by: $$x(t) = \frac{3}{\cos(\phi)}\cos(\sqrt{386}t + \phi)$$ And, the frequency is \(\frac{\sqrt{386}}{2\pi}\) Hz, the period is \(\frac{2\pi}{\sqrt{386}}\) s, the amplitude is \(\frac{3}{\cos(\phi)}\) inches, and the phase angle is \(\arctan \left(\frac{6}{3\sqrt{386}}\right)\).

Key concepts

Amplitude

In the context of simple harmonic motion, the amplitude is a crucial concept. It represents the maximum displacement of the object from its equilibrium position. For example, if you push a swing, its amplitude would be the highest point it reaches on both sides.
In this exercise, we determine the amplitude using the initial conditions provided: the object is initially displaced by 3 inches above its equilibrium position. By applying the formula from our step-by-step solution, the amplitude can be derived as follows:

• Start with the equation: \( x(0) = A \cos(\phi) \) where \( x(0) = 3 \).

Once you substitute \( \phi \) back into this equation, you can directly solve for the amplitude \( A \). Remember, the amplitude not only tells us the extent of oscillation but also dictates the energy in the motion. It is proportionately related to the energy stored in the system.

Frequency

Frequency refers to how often the oscillation occurs within a unit of time, usually one second. It's measured in Hertz (Hz), where 1 Hz indicates one cycle per second. In the context of this exercise, frequency tells us how rapidly the object is oscillating along the spring.
To find the frequency of the oscillation, we use the angular frequency \( \omega \), which is derived as \( \sqrt{386} \) rad/s from rearranging Hooke's law. We then compute the frequency \( f \) using the formula:

• \( f = \frac{\omega}{2\pi} \)

This equation transforms the angular measurement given by \( \omega \) into cycles per second by considering the full circular wave represented by \( 2\pi \). Therefore, the frequency is a direct representation of the speed of oscillation occurring in the given system.

Phase Angle

The phase angle, \( \phi \), is a somewhat abstract concept that helps to determine the starting point or phase of the oscillation. It defines where in its cycle the motion begins and can significantly affect the shape and behavior of the oscillation over time.
In this problem, the phase angle is derived using initial displacement and velocity. The relationship

• \( \tan(\phi) = \frac{6}{3\sqrt{386}} \)

is used to find \( \phi \) through the inverse tangent function, \( \arctan \). Calculating \( \phi \) is crucial because it sets the initial conditions of the function \( x(t) = A \cos(\omega t + \phi) \), affecting how early or late the object reaches its maximum displacement at the start.

Period

The period of a simple harmonic motion is the time taken for the object to return to its starting position after completing one full cycle of motion. It is the inverse of the frequency and is measured in seconds.
To determine the period \( T \), we use the formula:

• \( T = \frac{1}{f} = \frac{2\pi}{\omega} \)

Where \( \omega = \sqrt{386} \), allowing us to calculate \( T \). In everyday terms, the period is how long it takes for a pendulum to swing from one end to the other and back again, or in this spring scenario, from the maximum stretch through equilibrium and back. Understanding the period provides insight into the timing and rhythm of the system's movement.