Problem 25
Question
\(\bullet\) Write the general equation of motion for a mass that is on a horizontal frictionless surface and is connected to a spring at equilibrium (a) if the mass is initially pulled in the \(+x\) axis from the spring (stretched) and released, and (b) if the mass is pushed in the \(-x\) axis toward the spring (compressed) and released.
Step-by-Step Solution
Verified Answer
(a) \(x(t) = A \cos(\omega t)\); (b) \(x(t) = -A \cos(\omega t)\).
1Step 1: Define the System
Consider a mass-spring system where a mass, \(m\), is attached to a horizontal spring with spring constant \(k\). This system is frictionless.
2Step 2: Principle of Simple Harmonic Motion
The motion of the mass on a spring is an example of simple harmonic motion (SHM). The equation of motion for SHM is given by \(F = -kx\), where \(x\) is the displacement from equilibrium.
3Step 3: Derive the Equation of Motion
Using Newton's second law, \(F = ma\), we can set \(ma = -kx\). Rearranging gives the differential equation \(\frac{d^2x}{dt^2} = -\frac{k}{m}x\). This is the standard form for SHM.
4Step 4: Construct the General Solution for SHM
The general solution of the differential equation for SHM is \(x(t) = A \cos(\omega t + \phi)\), where \(A\) is the amplitude, \(\omega = \sqrt{\frac{k}{m}}\) is the angular frequency, and \(\phi\) is the phase constant.
5Step 5: Solve for Initial Conditions (a) Stretched Spring
For part (a), the spring is stretched, meaning the mass is pulled in the \(+x\) direction. Initially, \(x(0) = A\) and \(v(0) = 0\). Substituting these into the general solution, we find \(\phi = 0\), so the equation becomes \(x(t) = A \cos(\omega t)\).
6Step 6: Solve for Initial Conditions (b) Compressed Spring
For part (b), the spring is compressed, meaning the mass is pushed towards the spring, \(-x\). Initially, \(x(0) = -A\) and \(v(0) = 0\). Substituting these into the general solution, we find \(\phi = \pi\), so the equation becomes \(x(t) = -A \cos(\omega t)\) or equivalently \(x(t) = A \cos(\omega t + \pi)\).
Key Concepts
Equation of MotionMass-Spring SystemDifferential Equation
Equation of Motion
In a typical mass-spring system, the equation of motion describes how the mass moves over time due to the spring force. This equation is fundamental to understanding simple harmonic motion (SHM). To begin, let's recall that the force exerted by a spring is given by Hooke’s Law:
- \( F = -kx \)
- \( ma = -kx \)
- \( \frac{d^2x}{dt^2} = -\frac{k}{m}x \)
Mass-Spring System
A mass-spring system consists of a mass attached to a spring, which can either stretch or compress. This system is a classic example of Simple Harmonic Motion (SHM). When the mass is on a horizontal frictionless surface, its motion is purely influenced by the spring's force. - If you pull the mass to stretch the spring (as in the case of the mass moving in the \(+x\) direction), the spring stores potential energy.- If you push the mass, compressing the spring (mass moving in the \(-x\) direction), the energy is again stored but starts as more compressed relative to the equilibrium point. When released from any initial position, the mass will oscillate about the equilibrium position due to the interplay of potential energy in the spring and the kinetic energy of the mass. Through these oscillations, the system follows the equation of motion derived from the spring force, leading to a predictable, cyclic pattern.
Differential Equation
The motion of a mass in a mass-spring system is described by a differential equation. This mathematical equation is crucial for explaining how displacement varies with time. The differential equation we derive from balancing forces in SHM is:
- \( \frac{d^2x}{dt^2} = -\frac{k}{m}x \)
- \( x(t) = A \cos(\omega t + \phi) \)
Other exercises in this chapter
Problem 23
\(\bullet\) If the frequency of a mass-spring system is \(1.50 \mathrm{~Hz}\) and the mass on the spring is \(5.00 \mathrm{~kg}\), what is the spring constant?
View solution Problem 24
\(\bullet\) A breeze sets a suspended lamp into oscillation. If the period is \(1.0 \mathrm{~s}\), what is the distance from the ceiling to the lamp at the lowe
View solution Problem 26
\(\bullet\) The equation of motion for an oscillator in vertical \(\mathrm{SHM}\) is given by \(y=(0.10 \mathrm{~m}) \sin [(100 \mathrm{rad} / \mathrm{s}) t] .\
View solution Problem 27
\(\bullet\) The displacement of an object is given by \(y=(5.0 \mathrm{~cm}) \cos [(20 \pi \mathrm{rad} / \mathrm{s}) t] .\) What are the object's (a) amplitude
View solution