Question 155: Oscillations

Question

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 :

Accepted Answer

The net force becomes zero at the mean point. Therefore, linear momentum must be conserved.
$$	herefore$$ $$M{v_1} = \left( {M + m} ight){v_2}$$
$$M{A_1}\sqrt {{k \over M}}  = \left( {M + m} ight){A_2}\sqrt {{k \over {m + M}}} $$
$$	herefore$$ $$\left( {V = A\sqrt {{k \over M}} } ight)$$
$${A_1}\sqrt M  = {A_2}\sqrt {M + m} $$
$$	herefore$$ $${{{A_1}} \over {{A_2}}} = \sqrt {{{m + M} \over M}} $$

Answer

$${{M + m} \over M}$$

Answer

$${\left( {{M \over {M + m}}} \right)^{{1 \over 2}}}$$

Answer

$${\left( {{{M + m} \over M}} \right)^{{1 \over 2}}}$$

Answer

$${M \over {M + m}}$$

Oscillations

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 :

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