What is the resulting angular frequency of the oscillation?

Click For Summary
SUMMARY

The discussion centers on calculating the angular frequency of a mass-spring system, specifically a 0.65-kg mass attached to a spring with a spring constant of 15 N/m. The correct formula for the period, T, is T = 2π√(m/k), which yields T = 0.2294 seconds. The angular frequency, ω, is then calculated using ω = 2π/T, resulting in ω = 27.39 rad/s. The error in the initial calculation stemmed from not correctly applying the mass in the period formula.

PREREQUISITES
  • Understanding of Hooke's Law and spring constants
  • Familiarity with the concepts of mass and oscillation
  • Knowledge of angular frequency and its relationship to period
  • Basic algebra and manipulation of equations
NEXT STEPS
  • Review the derivation of the period formula for mass-spring systems
  • Study the relationship between angular frequency and linear frequency
  • Explore examples of damped and driven oscillations
  • Learn about the effects of varying mass and spring constants on oscillation frequency
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and oscillatory motion, as well as educators looking for clear examples of mass-spring systems in action.

jimmyboykun
Messages
39
Reaction score
0

Homework Statement



A 0.65-kg mass is hanging from a spring with spring constant 15 N/m. Then the mass is displaced from the equilibrium by 2 cm and let go.

Homework Equations



angular frequency:ω=2∏/T

The Attempt at a Solution



I found T: 2∏sqrtm/k, 2∏sqrt0.02/15N/m= 0.229429488s

ω=2∏/0.229429488s=27.39rad/s

I got the answer wrong but the units are suppose to be in rad/s. I feel like 0.65kg plays a role in the equation, but don't know where to place it. Where exactly did I go wrong.
 
Physics news on Phys.org
jimmyboykun said:

Homework Statement



A 0.65-kg mass is hanging from a spring with spring constant 15 N/m. Then the mass is displaced from the equilibrium by 2 cm and let go.

Homework Equations



angular frequency:ω=2∏/T

The Attempt at a Solution



I found T: 2∏sqrtm/k, 2∏sqrt0.02/15N/m= 0.229429488s

ω=2∏/0.229429488s=27.39rad/s

I got the answer wrong but the units are suppose to be in rad/s. I feel like 0.65kg plays a role in the equation, but don't know where to place it. Where exactly did I go wrong.

Take a look at the formula you used to find the period, T. What are the variables involved?

(also, make use of parentheses to delineate the argument of a function like sqrt(...) so we can tell what you're taking the square root of )
 

Similar threads

  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 11 ·
Replies
11
Views
2K
Replies
5
Views
4K
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
3
Views
957
Replies
9
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K