JEE MAIN - Physics (2011 - No. 13)
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 $$\left( {{{{A_1}} \over {{A_2}}}} \right)$$ is :
$${{M + m} \over M}$$
$${\left( {{M \over {M + m}}} \right)^{{1 \over 2}}}$$
$${\left( {{{M + m} \over M}} \right)^{{1 \over 2}}}$$
$${M \over {M + m}}$$
Explanation
The net force becomes zero at the mean point. Therefore, linear momentum must be conserved.
$$\therefore$$ $$M{v_1} = \left( {M + m} \right){v_2}$$
$$M{A_1}\sqrt {{k \over M}} = \left( {M + m} \right){A_2}\sqrt {{k \over {m + M}}} $$
$$\therefore$$ $$\left( {V = A\sqrt {{k \over M}} } \right)$$
$${A_1}\sqrt M = {A_2}\sqrt {M + m} $$
$$\therefore$$ $${{{A_1}} \over {{A_2}}} = \sqrt {{{m + M} \over M}} $$
$$\therefore$$ $$M{v_1} = \left( {M + m} \right){v_2}$$
$$M{A_1}\sqrt {{k \over M}} = \left( {M + m} \right){A_2}\sqrt {{k \over {m + M}}} $$
$$\therefore$$ $$\left( {V = A\sqrt {{k \over M}} } \right)$$
$${A_1}\sqrt M = {A_2}\sqrt {M + m} $$
$$\therefore$$ $${{{A_1}} \over {{A_2}}} = \sqrt {{{m + M} \over M}} $$
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