Shm , calculation of amplitude of spring mass system

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Homework Help Overview

The discussion revolves around a spring-mass system where a spring stretches under the influence of a force and a mass. The original poster seeks to find the force constant of the spring and the amplitude of vibration after releasing the mass from rest. The problem involves concepts from mechanics, specifically related to spring dynamics and energy conservation.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to apply energy conservation principles to find the amplitude of vibration, questioning their calculations and seeking clarification on potential mistakes. Other participants suggest alternative methods, such as finding the extension at equilibrium, and inquire about the equations and points of consideration for using conservation of energy (COE).

Discussion Status

The discussion is ongoing, with participants exploring different methods to approach the problem. Some guidance has been offered regarding the equilibrium extension, but there is no explicit consensus on the best method to use. The original poster is encouraged to clarify their understanding of the energy conservation approach.

Contextual Notes

Participants note that the spring is not relaxed at equilibrium, which may affect the calculations. The original poster expresses a desire to understand the COE method despite suggestions for a simpler approach.

Shivang kohlii
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Homework Statement


In A spring mass system , the spring stretches 2 cm from its 's frelength when a force of 10 N is applied . This spring is stretched 10 cm from it's free length , when a body of mass m = 2 kg is attached to it and released from rest at time t = 0 . Find the A) force constant of spring
B) amplitude of vibration

Homework Equations


kx = mg
X is the extension from mean position
Elastic potential energy = 1/2 kx^2

The Attempt at a Solution


A) k = 10 / 0.02 = 500N/m
B ) considering the mean position as the point of extension of spring due to mass m ,
Energy at mean position ( X= 0 )= energy at maximum displacement( X= Amplitude , A )

Kinetic energy at mean position
=1/2 m×ω^2× A
= (1/2) × 2 × (k/m) × A
= 250 × A Joules

Gravitational potential energy at mean position ( taking base as the extreme position ) = mgA =
2 × 10 × A = 20A

Spring potential energy at mean position = 0

ii) kinetic energy at extreme position = gravitational potential energy at extreme position = 0
Elastic potential energy at extreme position = 1/2 × 500 × (A)^2 = 250 ×A^2

Equating the energies :
A = 1.8 but answer is 0.06 m

I am wondering if I made some mistake in the initial energy ? If yes , please point the mistake out with correction.. any help will be much appreciated ! Thanks
 
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Shivang kohlii said:
Spring potential energy at mean position = 0
The spring is not relaxed at equilibrium.

There is a much easier way. Just find the extension at equilibrium.
 
haruspex said:
The spring is not relaxed at equilibrium.

There is a much easier way. Just find the extension at equilibrium.
Yes I know the method through extension at equilibrium ... But I want to check out the method of COE for future purposes...
Can u please just tell me the equation that u would use and the two points that u would consider for solving through COE ?
 
Shivang kohlii said:
Can u please just tell me the equation that u would use and the two points that u would consider for solving through COE ?
Try again for the spring PE at equilibrium.
 

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