KCET · Physics · Oscillations
For a particle executing simple harmonic motion (SHM), at its mean position
- A velocity is zero and acceleration is maximum.
- B velocity is maximum and acceleration is zero,
- C both velocity and acceleration are maximum.
- D both velocity and acceleration are zero.
Answer & Solution
Correct Answer
(B) velocity is maximum and acceleration is zero,
Step-by-step Solution
Detailed explanation
We know that velocity \(v\) and acceleration of a particle in SHM.
\(v=\omega \sqrt{A^2-y^2}\) and \(a=-\omega^2 y\)
\(\therefore\) At mean position, \(y=0\)
\(\therefore v=\omega \sqrt{A^2-0}\)
\(\Rightarrow v_{\max }=\omega A\) and \(a=0\)
\(v=\omega \sqrt{A^2-y^2}\) and \(a=-\omega^2 y\)
\(\therefore\) At mean position, \(y=0\)
\(\therefore v=\omega \sqrt{A^2-0}\)
\(\Rightarrow v_{\max }=\omega A\) and \(a=0\)
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