JEE Mains · Physics · STD 11 - 13. oscillations
A mass \(0.9\,kg\), attached to a horizontal spring, executes \(SHM\) with an amplitude \(A _{1}\). When this mass passes through its mean position, then a smaller mass of \(124\,g\) 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\(......\)
- A \(18\)
- B \(8\)
- C \(16\)
- D \(32\)
Answer & Solution
Correct Answer
(C) \(16\)
Step-by-step Solution
Detailed explanation
\(\frac{1}{2} kA ^{2}=\frac{ p ^{2}}{2\,m }\) \(\Rightarrow\left(\frac{ A _{1}}{ A _{2}}\right)^{2}=\frac{ m _{2}}{ m _{1}}=\frac{1024}{900}\) \(\Rightarrow \frac{ A _{1}}{ A _{2}}=\frac{32}{30}=\frac{16}{15}=\frac{16}{16-1}\) \(\therefore \alpha=16\)
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