Shm , calculation of amplitude of spring mass system

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In a spring-mass system, the force constant of the spring is calculated to be 500 N/m based on a 2 cm stretch from a 10 N force. The amplitude of vibration is initially calculated to be 1.8 m, but the correct answer is 0.06 m. The discussion emphasizes the importance of considering the spring's potential energy at equilibrium and using conservation of energy principles for accurate calculations. Participants suggest checking the spring potential energy at equilibrium to resolve discrepancies in the amplitude calculation. The conversation highlights the need for clarity in energy equations and the significance of equilibrium conditions in spring systems.
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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