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Shivang kohlii
- 19
- 1
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