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A 0.430 kg mass is oscillating
on a spring of spring constant
85.0 N/m. Find the frequency
of the oscillation.
(Unit = Hz)​

Respuesta :

Space

Answer:

[tex]\displaystyle T = 0.446894 \ Hz[/tex]

General Formulas and Concepts:

Simple Harmonic Motion

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

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

Explanation:

Step 1: Define

[Given] m = 0.430 kg

[Given[ k = 85.0 N/m

[Solve] T

Step 2: Find Oscillation

  1. Substitute in variables [Period of a Spring Formula]:                                     [tex]\displaystyle T = 2\pi \sqrt{\frac{0.430 \ kg}{85.0 \ N/m}}[/tex]
  2. Evaluate:                                                                                                           [tex]\displaystyle T = 0.446894 \ Hz[/tex]

Topic: AP Physics 1 Algebra-Based

Unit: SMH

Answer:

2.24 Hz

Explanation:

To find frequency, use the frequency equation of SHM, which is: 1/(2π)√k/m.

So if you plug in the numbers it would turn out to be 2.24 Hz.

1/(2π)√85.0/0.430 = 2.237 = 2.24 Hz (The answer would be 2.24 Hz when you round it to 3 significant figures)

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