TS EAMCET · Physics · Oscillations
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 \(\sqrt{\frac{K_1}{K_2}}\)
- B \(\sqrt{\frac{K_2}{2 K_1}}\)
- C \(\sqrt{\frac{K_2}{K_1}}\)
- D \(\sqrt{\frac{2 K_1}{K_2}}\)
Answer & Solution
Correct Answer
(B) \(\sqrt{\frac{K_2}{2 K_1}}\)
Step-by-step Solution
Detailed explanation
We knows, maximum velocity \(V_{\max }=A \omega=A \sqrt{\frac{K}{m}}\) Given, \[ \begin{gathered} K_1, m_1=m, K_2, m_2=2 m \\ \left(V_{\max }\right)_A=\left(V_{\max }\right)_B \end{gathered} \]…
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