JEE Main · 2022 · 27 Jul 2022 · Shift 1 · Elasticity & SHM

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}}

\frac{α}{α - 1}

, then the value of

α

will be

____

Answer: 16

Solution

When the smaller mass is placed on the larger mass at the mean position, momentum is conserved but mechanical energy is not (inelastic collision). However, the total energy in SHM is

E = \frac{1}{2} k A^{2} = \frac{p ^{2}}{2 m}

where

p

is the momentum at mean position.

At mean position, momentum

p

is the same for both cases (before and after adding mass). Using energy conservation at mean position:

\frac{1}{2} k A_{1}^{2} = \frac{p ^{2}}{2 m _{1}}

and

\frac{1}{2} k A_{2}^{2} = \frac{p ^{2}}{2 m _{2}}

Taking the ratio:

(\frac{A _{1}}{A _{2}})^{2} = \frac{m _{2}}{m _{1}} = \frac{900 + 124}{900} = \frac{1024}{900}

Therefore:

\frac{A _{1}}{A _{2}} = \frac{1024}{900} = \frac{32}{30} = \frac{16}{15} = \frac{16}{16 - 1}

Comparing with

\frac{α}{α - 1}

, we get

α = 16

JEE Main · 2022 · 27 Jul 2022 · Shift 1 · Integer answer

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JEE Main · 2022 · 27 Jul 2022 · Shift 1 · Elasticity & SHM

Solution

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