NEET · Physics · STD 11 - 13. oscillations
A particle executes linear simple harmonic motion with an ampilitude of \(3\,cm\) . When the particle is at \(2\,cm\) from the mean position , the magnitude of its velocity is equal to that of its acceleration. Then its time period in seconds is
- A \(\frac{{4\pi }}{{\sqrt 5 }}\)
- B \(\frac{{2\pi }}{{\sqrt 5 }}\)
- C \(\;\frac{{\sqrt 5 }}{\pi }\)
- D \(\frac{{\sqrt 5 }}{{2\pi }}\)
Answer & Solution
Correct Answer
(A) \(\frac{{4\pi }}{{\sqrt 5 }}\)
Step-by-step Solution
Detailed explanation
\(\text { Given, } A=3 \mathrm{cm}, x=2 \mathrm{cm}\) The velocity of a particle in simple harmonic motion is given as
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