Question

A horizontal spring executes S.H.M. with amplitude 'A₁', when mass 'm₁' is attached to it, When it passes through mean position another mass 'm₂' is placed on it. Both masses move together with amplitude 'A₂'. Therefore A₂ : A₁ is ______ 

Options

• `[(m_1 + m_2)/m_2]^{1/2}`

• `[(m_1 + m_2)/m_1]^{1/2}`

• `[m_2/(m_1 + m_2)]^{1/2}`

• `[m_1/(m_1 + m_2)]^{1/2}`

Answer 1

A horizontal spring executes S.H.M. with amplitude 'A₁', when mass 'm₁' is attached to it, When it passes through mean position another mass 'm₂' is placed on it. Both masses move together with amplitude 'A₂'. Therefore A₂ : A₁ is `underline([m_1/(m_1 + m_2)]^{1/2})`.

Explanation:

At mean position f_net = 0

∴ Applying conservation of momentum

m₁v₁ = (m₁ + m₂)v₂

m₁ω₁A₁ = (m₁ + m₂)ω₂A₂

But `omega_1 = sqrt(k/m_1)`

`omega_2 = sqrt(k/(m_1 + m_2))A_2`

∴ `m_1sqrt(k/m_1) A_1 = (m_1 + m_2)sqrt(k/(m_1 + m_2))A_2` 

`A_1/A_2 = sqrt((m_1 + m_2)/m_1)`

`A_2/A_1 = sqrt(m_1/(m_1 + m_2))`