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A horizontal spring executes S.H.M. with amplitude 'A 1 ', when mass ' $\mathrm{m}_{1}{ }^{"}$ is attached to it. When it passes through mean position another mass ' $\mathrm{m}_{2}$ ' is placed on it. Both masses move together with amplitude ' $A_{2}$ '. Therefore $A_{2}: A_{1}$ is
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The correct answer is:
$\left[\frac{\mathrm{m}_{1}}{\mathrm{~m}_{1}+\mathrm{m}_{2}}\right]^{1 / 2}$
$\mathrm{U}_{1}=\frac{1}{2} \mathrm{kx}_{1}^{2} \quad \therefore \mathrm{x}_{1}=\sqrt{\frac{2 \mathrm{U}_{1}}{\mathrm{k}}}$
$\mathrm{U}_{2}=\frac{1}{2} \mathrm{kx}_{2}^{2} \quad \therefore \mathrm{x}_{2}=\sqrt{\frac{2 \mathrm{U}_{2}}{\mathrm{k}}}$
$\therefore \mathrm{U}=\frac{1}{2} \mathrm{k}\left(\mathrm{x}_{1}+\mathrm{x}_{2}\right)^{2}$
$\therefore \mathrm{x}_{1}+\mathrm{x}_{2}=\sqrt{\frac{2 \mathrm{U}}{\mathrm{k}}}$
$\quad \sqrt{\frac{2 \mathrm{U}_{1}}{\mathrm{k}}}+\sqrt{\frac{2 \mathrm{U}_{2}}{\mathrm{k}}}=\sqrt{\frac{2 \mathrm{U}}{\mathrm{k}}}$
$\therefore \quad \sqrt{\mathrm{U}}=\sqrt{\mathrm{U}_{1}}+\sqrt{\mathrm{U}_{2}}$
$\mathrm{U}_{2}=\frac{1}{2} \mathrm{kx}_{2}^{2} \quad \therefore \mathrm{x}_{2}=\sqrt{\frac{2 \mathrm{U}_{2}}{\mathrm{k}}}$
$\therefore \mathrm{U}=\frac{1}{2} \mathrm{k}\left(\mathrm{x}_{1}+\mathrm{x}_{2}\right)^{2}$
$\therefore \mathrm{x}_{1}+\mathrm{x}_{2}=\sqrt{\frac{2 \mathrm{U}}{\mathrm{k}}}$
$\quad \sqrt{\frac{2 \mathrm{U}_{1}}{\mathrm{k}}}+\sqrt{\frac{2 \mathrm{U}_{2}}{\mathrm{k}}}=\sqrt{\frac{2 \mathrm{U}}{\mathrm{k}}}$
$\therefore \quad \sqrt{\mathrm{U}}=\sqrt{\mathrm{U}_{1}}+\sqrt{\mathrm{U}_{2}}$
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