Oscillations and mechanical energy

Click For Summary

Homework Help Overview

The discussion revolves around a mass-spring system oscillating with a specified amplitude, spring constant, and mass. Participants are exploring the calculation of the mechanical energy of the system, which involves understanding the relationship between potential and kinetic energy.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster seeks guidance on how to initiate the problem. Some participants reference the equation for energy stored in a spring and question whether the amplitude is equivalent to the displacement used in the equation. Others clarify that at maximum amplitude, the velocity is zero, leading to a focus on potential energy.

Discussion Status

The discussion is active, with participants providing insights into the relevant equations and clarifying concepts related to potential energy at maximum amplitude. There is no explicit consensus, but helpful guidance has been offered regarding the interpretation of the amplitude in the context of the problem.

Contextual Notes

Participants are navigating the definitions and relationships between variables in the context of mechanical energy, with some uncertainty regarding the use of amplitude and velocity in their calculations.

Knfoster
Messages
45
Reaction score
0

Homework Statement



A mass-spring system oscillates with an amplitude of 3.30 cm. If the spring constant is 231 N/m and the mass is 537 g, determine the mechanical energy of the system.


Homework Equations



Mechanical energy is potential energy plus kinetic energy

The Attempt at a Solution



How do I go about starting this problem?
 
Physics news on Phys.org
Are you familiar with the equation for the energy stored in a compressed (or extended) spring, in terms of k and delta-x?
 
1/2*k*delta x but I'm not given delta x am I? or is the amplitude delta x? and if it is then what do I use for velocity in the 1/2 mv^2 ?
 
it's 1/2*k*delta x squared. The max amplitude is your delta x, and since you're considering the situation at max amplitude, the velocity is zero, and so you have just potential energy.

Good luck!
Arjun
 
Thank you
 

Similar threads

Replies
17
Views
2K
Replies
3
Views
1K
Replies
2
Views
1K
Replies
5
Views
1K
Replies
7
Views
1K
Replies
29
Views
3K
  • · Replies 1 ·
Replies
1
Views
8K
Replies
13
Views
2K
Replies
1
Views
1K
  • · Replies 18 ·
Replies
18
Views
3K