Two masses m1 and m2 are suspended together by a massless spring constant k when the masses are in equilibrium m1 is removed without disturbing the system
Answers
Answered by
2
The amplitude of oscillations is Δl = m1g / K
Explanation:
When mass m2 is alone present , extension i the string is l.
Then
m2g = Kl ....(1)
When mass m1 & m2 is both are present , extension i the string is l'.
(m1 + m2 )g = Kl' ......(2)
l' = l + Δl
Δl = amplitude of oscillation
From equation (1)and (2)
(m1 + m2 )g = m2g + KΔl
Δl = (m1 + m2 )g - m2g / K
Δl = m1g / K
The amplitude of oscillations is Δl = m1g / K
Also learn more
What is angular momentum???
https://brainly.in/question/6710222
Similar questions