contestada

What is the time period of a simple harmonic oscillator with a mass of 0.3kg and a force constant of 5N/m?

Respuesta :

Space

Answer:

1.53906 seconds

General Formulas and Concepts:

Simple Harmonic Motion

Period of a Spring: [tex]\displaystyle T = 2 \pi \sqrt{\frac{m}{k}}[/tex]

  • T is period
  • m is mass (in kg)
  • k is spring constant (in N/m)

Explanation:

Step 1: Define

[Given] m = 0.3 kg

[Given] k = 5 N/m

[Solve] T

Step 2: Solve

  1. Substitute in variables [Period of a Spring]:                                                   [tex]\displaystyle T = 2 \pi \sqrt{\frac{0.3 \ kg}{5 \ N/m}}[/tex]
  2. Divide:                                                                                                              [tex]\displaystyle T = 2 \pi \sqrt{0.06}[/tex]
  3. Square root:                                                                                                      [tex]\displaystyle T = 2 \pi (0.244949)[/tex]
  4. Multiply:                                                                                                             [tex]\displaystyle T = 1.53906[/tex]

Topic: AP Physics 1 - Algebra Based

Unit: SHM (Simple Harmonic Motion)

Book: College Physics

Otras preguntas