A mass 0.9 kg, attached to a horizontal spring, executes SHM with an amplitude A₁. 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 A₂. If the ratio (A₁)/(A₂) is (α)/(α-1), then the value of α will be ___________. Numerical answer type. Work it out, then open the answer.

Answer: 16. At the mean position momentum is conserved: Mv₁ = (M+m)v₂ ⇒ (v₁)/(v₂) = M+m/M. Amplitude A = vmaxω with ω = √(k/mass), so (ω₂)/(ω₁) = √(M/M+m). (A₁)/(A₂) = (v₁)/(v₂)·(ω₂)/(ω₁) = √(M+m/M) = √((1.024)/(0.9)) = 16/15 (α)/(α-1) = 16/15 ⇒ α = 16

A mass 0.9 kg, attached to a horizontal spring, executes SHM with an… | JEE Main 2022 (27 Jul, Shift 1)

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Spring Systems

A mass $0.9\text{ kg}$ , 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 $124\text{ g}$ 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 ___________.

Numerical answer type. Work it out, then open the answer.

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Answer: 16

At the mean position momentum is conserved: $Mv_1 = (M+m)v_2 \Rightarrow \frac{v_1}{v_2} = \frac{M+m}{M}$ .
Amplitude $A = \frac{v_{max}}{\omega}$ with $\omega = \sqrt{k/\text{mass}}$ , so $\frac{\omega_2}{\omega_1} = \sqrt{\frac{M}{M+m}}$ .
$\frac{A_1}{A_2} = \frac{v_1}{v_2}\cdot\frac{\omega_2}{\omega_1} = \sqrt{\frac{M+m}{M}} = \sqrt{\frac{1.024}{0.9}} = \frac{16}{15}$
$\frac{\alpha}{\alpha-1} = \frac{16}{15} \Rightarrow \alpha = 16$

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