Physics, asked by hameedabdul2639, 9 months ago

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 Fatimakincsem
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

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