AP EAMCET · PHYSICS · Oscillations
A particle is executing simple harmonic motion with amplitude A. At a distance ' \(x\) ' from the mean position, when the particle is moving towards extreme position it receives a blow in the direction of motion which instantaneously doubles its velocity. The new amplitude of the particle is (Frequency is constant during the motion)
- A \(A\)
- B \(\sqrt{A^2-x^2}\)
- C \(\sqrt{2 \mathrm{~A}^2-3 \mathrm{x}^2}\)
- D \(\sqrt{4 A^2-3 x^2}\)
Answer & Solution
Correct Answer
(D) \(\sqrt{4 A^2-3 x^2}\)
Step-by-step Solution
Detailed explanation
\( v = \omega \sqrt{A^2 - x^2} \) \( v' = 2v \) \( 2 \omega \sqrt{A^2 - x^2} = \omega \sqrt{(A')^2 - x^2} \) \( 4(A^2 - x^2) = (A')^2 - x^2 \) \( 4A^2 - 4x^2 = (A')^2 - x^2 \) \( (A')^2 = 4A^2 - 3x^2 \) \( A' = \sqrt{4A^2 - 3x^2} \)
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