Q132 / 160 · TS EAMCET 2013 · Physics
Two particles $A$ and $B$ of masses ' $m$ ' and ' $2 m$ ' are suspended from massless springs of force constants $K_1$ and $K_2$. During their oscillation, if their maximum velocities are equal, then the ratio of amplitudes of $A$ and $B$ is
- A$\sqrt{\frac{K_1}{K_2}}$
- Correct answerB$\sqrt{\frac{K_2}{2 K_1}}$
- C$\sqrt{\frac{K_2}{K_1}}$
- D$\sqrt{\frac{2 K_1}{K_2}}$
Solution
We knows, maximum velocity $V_{\max }=A \omega=A \sqrt{\frac{K}{m}}$
Given,
$$
\begin{gathered}
K_1, m_1=m, K_2, m_2=2 m \\
\left(V_{\max }\right)_A=\left(V_{\max }\right)_B
\end{gathered}
$$
$$
\begin{aligned}
\quad A_A \sqrt{\frac{K_1}{M}} & =A_B \sqrt{\frac{K_2}{2 m}} \\
\Rightarrow \quad \frac{A_A}{A_B} & =\sqrt{\frac{K_2}{2 K_1}}
\end{aligned}
$$
Hence, the correct answer is option: 'B'.