Math Problem Statement

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Explain example simply

Solution

The example you provided discusses a spring-mass system and solves for the motion of the mass after it is released. Here's a simplified explanation:

  1. Physical Setup:

    • A spring with stiffness k=700N/mk = 700 \, \text{N/m} is hanging vertically.
    • A mass of M=0.05kgM = 0.05 \, \text{kg} is attached to the spring.
    • When the mass is pulled down and released, it oscillates up and down.
  2. Newton’s Second Law:

    • Using F=Md2xdt2F = M \frac{d^2x}{dt^2}, the forces acting on the mass are:
      • The spring force: kx-kx, opposing displacement.
    • The net force gives the equation of motion: Md2xdt2=kx.M \frac{d^2x}{dt^2} = -kx.
  3. Second-Order Differential Equation:

    • Rearranging, we get: d2xdt2+kMx=0.\frac{d^2x}{dt^2} + \frac{k}{M} x = 0.
    • Substituting k=700N/mk = 700 \, \text{N/m} and M=0.05kgM = 0.05 \, \text{kg}: d2xdt2+100x=0.\frac{d^2x}{dt^2} + 100x = 0.
    • This is a second-order linear differential equation describing simple harmonic motion.
  4. Solving the Equation:

    • The characteristic equation is: m2+100=0    m=±10i.m^2 + 100 = 0 \implies m = \pm 10i.
    • Complex roots indicate oscillatory motion. The general solution is: x(t)=c1cos(10t)+c2sin(10t),x(t) = c_1 \cos(10t) + c_2 \sin(10t), where c1c_1 and c2c_2 are constants determined by initial conditions.
  5. Initial Conditions:

    • At t=0t = 0, the displacement x(0)=0.05mx(0) = 0.05 \, \text{m}, and velocity x˙(0)=0\dot{x}(0) = 0:
      • Substituting these, the constants are found to be: c1=0.05,c2=0.c_1 = 0.05, \quad c_2 = 0.
  6. Particular Solution:

    • The specific solution becomes: x(t)=0.05cos(10t).x(t) = 0.05 \cos(10t).
  7. Resulting Motion:

    • The mass undergoes simple harmonic motion:
      • Amplitude: 0.05m0.05 \, \text{m}.
      • Frequency: 1.59cycles/sec1.59 \, \text{cycles/sec}.
      • Period: 0.628sec0.628 \, \text{sec}.

In short:

The spring-mass system oscillates with a cosine function, showing periodic, smooth up-and-down motion due to the spring's restoring force.


Do you want more details on any specific step? Here are some related questions:

  1. What is the significance of the characteristic equation?
  2. How do the initial conditions influence the solution?
  3. Why does the motion follow a cosine function?
  4. What determines the frequency and amplitude of oscillation?
  5. Can damping or air resistance be included in this type of problem?

Tip: Always check units carefully when substituting values into formulas to avoid calculation errors!

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Math Problem Analysis

Mathematical Concepts

Second-Order Differential Equations
Simple Harmonic Motion
Newton’s Laws of Motion

Formulas

Newton's second law: F = Ma
Equation of motion for a spring: M(d²x/dt²) = -kx
General solution: x(t) = c1 cos(ωt) + c2 sin(ωt)

Theorems

Solution of second-order linear differential equations
Characteristics of harmonic motion

Suitable Grade Level

Undergraduate Level (Physics/Mathematics)