Question

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

Answer 1

1.53906 seconds

General Formulas and Concepts:

Simple Harmonic Motion

Period of a Spring:
`\displaystyle T = 2 \pi \sqrt{(m)/(k)}`

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

Step-by-step 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]:
   `\displaystyle T = 2 \pi \sqrt{(0.3 \ kg)/(5 \ N/m)}`
2. Divide:
   `\displaystyle T = 2 \pi √(0.06)`
3. Square root:
   `\displaystyle T = 2 \pi (0.244949)`
4. Multiply:
   `\displaystyle T = 1.53906`

Topic: AP Physics 1 - Algebra Based

Unit: SHM (Simple Harmonic Motion)

Book: College Physics