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Q. 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 _____

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Solution:

$ \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$