A mass M , attached to a horizontal spring, executes S.H.M. with amplitude A_1 . When the mass M passes through its mean position then a smaller mass - Physics MCQ FIITJEE AIEEE 2011 (Set Q) 2026 | ExamTest.live

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A mass M, attached to a horizontal spring, executes S.H.M. with amplitude A_{1} . When the mass M passes through its mean position then a smaller mass m is placed over it and both of them move together with amplitude A_{2} . The ratio of \frac{A _{1}}{A _{2}} is:

Solution & Step-by-step Explanation

At the mean position, potential energy is zero.Initial velocity v_{1} = A_{1} ω_{1} = A_{1} k / M .Momentum is conserved when mass m is placed: M v_{1} = (M + m) v_{2} M (A_{1} k / M) = (M + m) (A_{2} k / (M + m)) A_{1} M = A_{2} M + m \frac{A _{1}}{A _{2}} = \frac{M + m}{M} = (\frac{M + m}{M})^{1/2}

Try it yourself before checking the explanation above.

A mass M, attached to a horizontal spring, executes S.H.M. with amplitude A_{1} . When the mass M passes through its mean position then a smaller mass m is placed over it and both of them move together with amplitude A_{2} . The ratio of \frac{A _{1}}{A _{2}} is:

\frac{M + m}{M}

(\frac{M}{M + m})^{1/2}

(\frac{M + m}{M})^{1/2}

\frac{M}{M + m}

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