Question

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)​

Answer 1

2.24 Hz

Step-by-step 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)

Answer 2

`\displaystyle T = 0.446894 \ Hz`

General Formulas and Concepts:

Simple Harmonic Motion

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

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

Step-by-step 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]:
   `\displaystyle T = 2\pi \sqrt{(0.430 \ kg)/(85.0 \ N/m)}`
2. Evaluate:
   `\displaystyle T = 0.446894 \ Hz`

Topic: AP Physics 1 Algebra-Based

Unit: SMH