Correct Answer is Option (D)

$\because$ $V_{max}=A\omega$

Given, ${\omega }_1{A}_1={\omega }_2{A}_2$

We know that $\omega =\sqrt{\frac{K}{m}}$

$\therefore$ $\sqrt{\frac{k_1}{m}}{A}_1=\sqrt{\frac{k_2}{m}}{A}_2$

$\Rightarrow$ $\frac{A_1}{A_2}=\sqrt{\frac{k_2}{k_1}}$

Correct Answer is Option (C)

$\frac{1}{K_{eq}}=\frac{1}{K_1}+\frac{1}{K_1}$

$\Rightarrow K_{eq}=\frac{K_1\times K_1}{K_1+K_2}=\frac{K_1^2}{2K_1}=\frac{K_2}{2}$

$\therefore$ $T^{\prime }=2\pi \sqrt{\frac{m}{K_{eq}}}$

$=2\pi \sqrt{\frac{m}{\frac{K_1}{2}}}$

$=2\pi \sqrt{\frac{2m}{K_1}}$

$=\sqrt{2}T$ [ where $T=2\pi \sqrt{\frac{m}{K_1}}$]

Correct Answer is Option (B)

Due to parallel combination K_eff = 2k + 2k = 4k

$\because$ $T=2\pi \sqrt{\frac{m}{k_{eff}}}$

$=2\pi \sqrt{\frac{m}{4k}}$

$T=\pi \sqrt{\frac{m}{k}}$

Correct Answer is Option (C)

Let T be the time period of oscillation, then

$T=2\pi \sqrt{\frac{M}{k_{eq}}}$

$\therefore$ $T=2\pi \sqrt{\frac{M}{2k}}$ [$\because$ $k_{eq}=k+k$]


and frequency $(f)=\frac{1}{T}=\frac{1}{2\pi }\sqrt{\frac{2k}{M}}$

Correct Answer is Option (D)

Given, initial amplitude = A

Velocity at mean position, v = A$\omega$

Applying conservation of momentum at mean position, we get

M₁v₁ = M₂v₂

MA$\omega$ = (M + m)v'

$\Rightarrow v^{\prime }=\frac{MA\omega }{M+m}=\frac{MA\sqrt{\frac{k}{M}}}{M+m}$

$\therefore$ $v^{\prime }=A^{\prime }{\omega }^{\prime }=A^{\prime }\sqrt{\frac{k}{M+m}}$

$\Rightarrow A^{\prime }=\frac{MA\sqrt{\frac{k}{M}}}{M+m}\times \sqrt{\frac{M+m}{k}}$

$A^{\prime }=\sqrt{\frac{M}{M+m}}A$