Shm , calculation of amplitude of spring mass system

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SUMMARY

The discussion focuses on calculating the amplitude of a spring-mass system, specifically addressing a scenario where a spring stretches 2 cm under a 10 N force and 10 cm when a 2 kg mass is attached. The force constant of the spring is determined to be 500 N/m using the formula k = F/x. The amplitude of vibration is derived through conservation of energy principles, leading to an initial incorrect calculation of 1.8 m, which is corrected to 0.06 m. Participants emphasize the importance of considering the spring's equilibrium position in energy calculations.

PREREQUISITES
  • Understanding of Hooke's Law and spring constant (k)
  • Familiarity with conservation of energy principles
  • Knowledge of gravitational potential energy and elastic potential energy equations
  • Basic mechanics of oscillatory motion
NEXT STEPS
  • Study the derivation and application of Hooke's Law in spring systems
  • Learn about conservation of energy in mechanical systems
  • Explore the concept of equilibrium in spring-mass systems
  • Investigate the relationship between amplitude and energy in oscillatory motion
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Students studying physics, particularly those focusing on mechanics and oscillations, as well as educators seeking to clarify concepts related to spring-mass systems and energy conservation.

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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