A mass $0.9 \mathrm{~kg}$, attached to a horizontal spring, executes SHM with an amplitude $\mathrm{A}_{1}$. When this mass passes through its mean position, then a smaller mass of $124 \mathrm{~g}$ is placed over it and both masses move together with amplitude $A_{2}$. If the ratio $\frac{A_{1}}{A_{2}}$ is $\frac{\alpha}{\alpha-1}$, then the value of $\alpha$ will be ___________.
Answer (integer)
16
Solution
<p>$$(0.9){A_1}\sqrt {{K \over {0.9}}} = (0.9 + 0.124){A_2}\sqrt {{K \over {0.9 + 0.124}}} $$</p>
<p>${{{A_1}} \over {{A_2}}} = \sqrt {{{0.9 + 0.124} \over {0.9}}}$</p>
<p>$= \sqrt {{{1.024} \over {0.9}}}$</p>
<p>$= {\alpha \over {\alpha - 1}}$</p>
<p>$\alpha = 16$</p>
About this question
Subject: Physics · Chapter: Oscillations · Topic: Simple Harmonic Motion
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