MHT CET · Physics · Oscillations
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
- A \(\left[\frac{\mathrm{m}_{2}}{\mathrm{~m}_{1}+\mathrm{m}_{2}}\right]^{1 / 2}\)
- B \(\left[\frac{\mathrm{m}_{1+} \mathrm{m}_{2}}{\mathrm{~m}_{1}}\right]^{1 / 2}\)
- C \(\left[\frac{\mathrm{m}_{1}}{\mathrm{~m}_{1}+\mathrm{m}_{2}}\right]^{1 / 2}\)
- D \(\left[\frac{\mathrm{m}_{1+} \mathrm{m}_{2}}{\mathrm{~m}_{2}}\right]^{1 / 2}\)
Answer & Solution
Correct Answer
(C) \(\left[\frac{\mathrm{m}_{1}}{\mathrm{~m}_{1}+\mathrm{m}_{2}}\right]^{1 / 2}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{U}_{1}=\frac{1}{2} \mathrm{kx}_{1}^{2} \quad \therefore \mathrm{x}_{1}=\sqrt{\frac{2 \mathrm{U}_{1}}{\mathrm{k}}}\)
\(\mathrm{U}_{2}=\frac{1}{2} \mathrm{kx}_{2}^{2} \quad \therefore \mathrm{x}_{2}=\sqrt{\frac{2 \mathrm{U}_{2}}{\mathrm{k}}}\)
\(\therefore \mathrm{U}=\frac{1}{2} \mathrm{k}\left(\mathrm{x}_{1}+\mathrm{x}_{2}\right)^{2}\)
\(\therefore \mathrm{x}_{1}+\mathrm{x}_{2}=\sqrt{\frac{2 \mathrm{U}}{\mathrm{k}}}\)
\(\quad \sqrt{\frac{2 \mathrm{U}_{1}}{\mathrm{k}}}+\sqrt{\frac{2 \mathrm{U}_{2}}{\mathrm{k}}}=\sqrt{\frac{2 \mathrm{U}}{\mathrm{k}}}\)
\(\therefore \quad \sqrt{\mathrm{U}}=\sqrt{\mathrm{U}_{1}}+\sqrt{\mathrm{U}_{2}}\)
\(\mathrm{U}_{2}=\frac{1}{2} \mathrm{kx}_{2}^{2} \quad \therefore \mathrm{x}_{2}=\sqrt{\frac{2 \mathrm{U}_{2}}{\mathrm{k}}}\)
\(\therefore \mathrm{U}=\frac{1}{2} \mathrm{k}\left(\mathrm{x}_{1}+\mathrm{x}_{2}\right)^{2}\)
\(\therefore \mathrm{x}_{1}+\mathrm{x}_{2}=\sqrt{\frac{2 \mathrm{U}}{\mathrm{k}}}\)
\(\quad \sqrt{\frac{2 \mathrm{U}_{1}}{\mathrm{k}}}+\sqrt{\frac{2 \mathrm{U}_{2}}{\mathrm{k}}}=\sqrt{\frac{2 \mathrm{U}}{\mathrm{k}}}\)
\(\therefore \quad \sqrt{\mathrm{U}}=\sqrt{\mathrm{U}_{1}}+\sqrt{\mathrm{U}_{2}}\)
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