A horizontal spring executes S.H.M. with amplitude 'A 1 ', when mass ' $\mathrm{m}_{1}{ }^{"}$ is attached…

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
  1. $\left[\frac{\mathrm{m}_{2}}{\mathrm{~m}_{1}+\mathrm{m}_{2}}\right]^{1 / 2}$
  2. $\left[\frac{\mathrm{m}_{1+} \mathrm{m}_{2}}{\mathrm{~m}_{1}}\right]^{1 / 2}$
  3. $\left[\frac{\mathrm{m}_{1}}{\mathrm{~m}_{1}+\mathrm{m}_{2}}\right]^{1 / 2}$
  4. $\left[\frac{\mathrm{m}_{1+} \mathrm{m}_{2}}{\mathrm{~m}_{2}}\right]^{1 / 2}$

Solution

At mean position \(f_{\text {net }}=0\)
\(\therefore\) Applying conservation of momentum
\(\begin{aligned}
& \mathrm{m}_1 \mathrm{v}_1=\left(\mathrm{m}_1+\mathrm{m}_2\right) \mathrm{v}_2 \\
& \mathrm{~m}_1 \omega_1 \mathrm{~A}_1=\left(\mathrm{m}_1+\mathrm{m}_2\right) \omega_2 \mathrm{~A}_2 \\
& \text {But } \omega_1=\sqrt{\frac{k}{m_1}} \\
& \omega_2=\sqrt{\frac{k}{m_1+m_2}} A_2 \\
& \therefore m_1 \sqrt{\frac{k}{m_1} A_1=\left(m_1+m_2\right) \sqrt{\frac{k}{m_1+m_2}} A_2} \\
& \frac{A_1}{A_2}=\sqrt{\frac{m_1+m_2}{m_1}} \\
& \frac{A_2}{A_1}=\sqrt{\frac{m_1}{m_1+m_2}}
\end{aligned}\)

Asked in: MHT CET 2020 (15 Oct Shift 1)

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