Question

A mass \(m=1.00 \mathrm{~kg}\) in a spring-mass system with \(k=1.00 \mathrm{~N} / \mathrm{m}\) is observed to be moving to the right, past its equilibrium position with a speed of \(1.00 \mathrm{~m} / \mathrm{s}\) at time \(t=0\) a) Ignoring all damping, determine the equation of motion. b) Suppose the initial conditions are such that at time \(t=0,\) the mass is at \(x=0.500 \mathrm{~m}\) and moving to the right with a speed of \(1.00 \mathrm{~m} / \mathrm{s}\). Determine the new equation of motion. Assume the same spring constant and mass.

Short answer

Answer: a) The equation of motion for the scenario with no damping is \(x(t) = 1.00\sin(t)\). b) The new equation of motion with given initial conditions is \(x(t) = 0.500 \cos(t) + 1.00\sin(t)\).

Step-by-step solution

Write down the general equation for a spring-mass system

The equation for a spring-mass system with no damping is given by the simple harmonic motion (SHM) equation: $$ \ddot{x} + \omega^2 x = 0 $$ where \(\omega^2\) is the angular frequency squared and is equal to \(\frac{k}{m}\), x is the displacement from equilibrium and \(\dot{x}\) and \(\ddot{x}\) are the velocity and acceleration, respectively.

Determine the angular frequency squared

Given the mass \(m = 1.00\,\text{kg}\) and spring constant \(k = 1.00\,\text{N/m}\), let's find the value of \(\omega^2\): $$ \omega^2 = \frac{k}{m} = \frac{1.00\,\text{N/m}}{1.00\,\text{kg}} = 1.00\, \text{1/s^2} $$

Solve the SHM equation for displacement as a function of time

The general solution to the SHM equation is given as: $$ x(t) = A \cos(\omega t) + B \sin(\omega t) $$ Now, let's find the constants A and B for each part of the problem.

Find the constants A and B for part a

For part a, given that the mass is moving to the right past its equilibrium position with a speed of \(1.00\, \text{m/s}\) at time \(t = 0\). Thus, the initial conditions are: \(x(0) = 0\) (since it's moving right past its equilibrium) and \(\dot{x}(0) = 1.00\, \text{m/s}\). Using the general solution for SHM, we get: $$ x(0) = A \cos(0) + B \sin(0) \Rightarrow A = 0 $$ Taking the derivative of the general solution for SHM, we get the velocity as a function of time: $$ v(t) = \dot{x}(t) = -A\omega\sin(\omega t) + B\omega\cos(\omega t) $$ Plugging in the initial conditions and given \(\omega = 1\), we get: $$ \dot{x}(0) = 1.00\, \text{m/s} = -A\sin(0) + B\cos(0) \Rightarrow B = 1.00 $$ Thus, the equation of motion for part a is: $$ x(t) = 1.00\sin(t) $$

Find the constants A and B for part b

For part b, the initial conditions are given as: \(x(0) = 0.500\, \text{m}\) and \(\dot{x}(0) = 1.00\,\text{m/s}\). Plugging these into the general solution for SHM, we obtain: $$ x(0) = 0.500\, \text{m} = A \cos(0) + B \sin(0) \Rightarrow A = 0.500 $$ Using the velocity as function of time and plugging the initial conditions for velocity, we get: $$ \dot{x}(0) = 1.00\,\text{m/s} = -A\sin(0) + B\cos(0) \Rightarrow B = 1.00 $$ Thus, the equation of motion for part b is: $$ x(t) = 0.500 \cos(t) + 1.00\sin(t) $$ To summarize the answers: a) The equation of motion ignoring damping is \(x(t) = 1.00\sin(t)\). b) The new equation of motion with given initial conditions is \(x(t) = 0.500 \cos(t) + 1.00\sin(t)\).

Key concepts

Simple Harmonic Motion

Simple Harmonic Motion (SHM) is a fundamental concept in physics describing a type of periodic oscillation that occurs when the restoring force is directly proportional and opposite to the displacement. Imagine plucking a guitar string; the vibration that ensues is a classic example of SHM. For a student seeking to grasp the essence of SHM, consider a mass bobbing up and down on a spring. The force trying to bring the mass back to its rest position increases the further away it gets.

In mathematical terms, SHM is governed by the equation: \[\begin{equation}\frac{d^2x}{dt^2} + \frac{k}{m} x = 0,\[\begin{equation}where \( x \) represents the displacement from equilibrium, \( k \) is the spring constant, \( m \) is the mass, and \( \frac{d^2x}{dt^2} \) signifies the acceleration. It's critical to digest every component of this equation, as it encapsulates the essence of the physical phenomena at play. The sine and cosine functions represent the solution to this equation, embodying the repetitive nature of SHM—always oscillating around the equilibrium.

Spring-Mass System

The spring-mass system, used in the given exercises, is a perfect embodiment of SHM. In the system, a mass \(m\) is attached to a spring with a spring constant \(k\). In isolation from friction and other external forces, this system will oscillate back and forth in a stunning demonstration of SHM. Why does this happen? It's due to the spring's desire to return to its natural length, combined with the mass's inertia.

For learners tackling the initial conditions of a problem, remember: they set the stage for the entire motion. When the mass is at rest at its equilibrium position, the initial conditions lead us to the constants in our general SHM solution, which are pivotal for crafting the equation of motion. The interplay between the initial displacement and the initial velocity guides us to a precise description of the mass's motion through time.

Angular Frequency

Angular frequency, denoted by the Greek letter \(\omega\), is a cornerstone of SHM. It measures how quickly the mass oscillates, with its unit being radians per second. Intuitively, angular frequency is akin to the tempo of a song—the rate at which the spring-mass system completes its dance back and forth. Here's the kicker: the angular frequency is deeply rooted in the properties of the spring and the mass, as shown by the relationship \(\omega^2 = \frac{k}{m}\).

It is crucial to appreciate that \(\omega\) is not an arbitrary value. In the textbook exercise, we compute it using the mass and the spring constant. This number then waltzes into our general SHM solution, influencing how the sine and cosine waves behave. The greater the \(\omega\), the faster the oscillation, period. Students should not just compute \(\omega\) but understand its profound impact on the motion's characteristics.