AP EAMCET · PHYSICS · Oscillations
As shown in the figure, two blocks of masses \(m_1\) and \(\mathrm{m}_2\) are connected to a spring of force constant k . The blocks are slightly displaced in opposite directions to \(\mathrm{x}_1\), \(x_2\) distances and released. If the system executes simple harmonic motion, then the frequency of oscillation of the system ( \(\omega\) ) is
- A \(\left(\frac{1}{m_1}+\frac{1}{m_2}\right) k^2\)
- B \(\sqrt{\left(\frac{1}{m_1}+\frac{1}{m_2}\right) k^2}\)
- C \(\sqrt{\left(\frac{1}{m_1}+\frac{1}{m_2}\right)}\)
- D \(\sqrt{\left(\frac{1}{m_1}+\frac{1}{m_2}\right) k}\)
Answer & Solution
Correct Answer
(D) \(\sqrt{\left(\frac{1}{m_1}+\frac{1}{m_2}\right) k}\)
Step-by-step Solution
Detailed explanation
\(\mu=\frac{m_1 m_2}{m_1+m_2}\) \(\therefore\) Frequency of oscillation is…
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