MHT CET · Physics · Oscillations
A particle executes a linear S.H.M. In two of its positions the velocities are \(V_1, V_2\) and accelerations are \(a_1\) and \(a_2\) respectively \(\left(0 < a_1 < a_2\right)\). The distance between the positions is
- A \(\frac{V_1^2-V_2^2}{a_1-a_2}\)
- B \(\frac{V_2^2-V_1^2}{a_1-a_2}\)
- C \(\frac{V_1^2-V_2^2}{a_1+a_2}\)
- D \(\frac{V_2^2-V_1^2}{\left(a_1^2+a_2^2\right)}\)
Answer & Solution
Correct Answer
(C) \(\frac{V_1^2-V_2^2}{a_1+a_2}\)
Step-by-step Solution
Detailed explanation
\(V^2 = \omega^2 A^2 - a^2/\omega^2\) \(V_1^2 = \omega^2 A^2 - a_1^2/\omega^2\)
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